id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-1800 | Change of Base of Logarithm | :$\log_b x = \dfrac {\log_a x} {\log_a b}$ | Let:
:$y = \log_b x \iff b^y = x$
:$z = \log_a x \iff a^z = x$
Then:
{{begin-eqn}}
{{eqn | l = z
| r = \map {\log_a} {b^y}
}}
{{eqn | r = y \log_a b
| c = Logarithms of Powers
}}
{{eqn | r = \log_b x \log_a b
}}
{{end-eqn}}
Hence the result.
{{qed}} | :$\log_b x = \dfrac {\log_a x} {\log_a b}$ | Let:
:$y = \log_b x \iff b^y = x$
:$z = \log_a x \iff a^z = x$
Then:
{{begin-eqn}}
{{eqn | l = z
| r = \map {\log_a} {b^y}
}}
{{eqn | r = y \log_a b
| c = [[Logarithms of Powers]]
}}
{{eqn | r = \log_b x \log_a b
}}
{{end-eqn}}
Hence the result.
{{qed}} | Change of Base of Logarithm/Proof 1 | https://proofwiki.org/wiki/Change_of_Base_of_Logarithm | https://proofwiki.org/wiki/Change_of_Base_of_Logarithm/Proof_1 | [
"Change of Base of Logarithm",
"Logarithms"
] | [] | [
"Logarithm of Power"
] |
proofwiki-1801 | Change of Base of Logarithm | :$\log_b x = \dfrac {\log_a x} {\log_a b}$ | We begin with an equation that relates two exponentials with two different bases, $a$ and $b$, both positive and neither equal to one:
:$b^y = a^z$
{{WLOG}}, select and solve for $b$:
:$b = a^{z/y}$
Take logarithm's base $a$ of both sides:
{{begin-eqn}}
{{eqn | l = \log_a b
| r = \map {\log_a} {a^{z/y} }
}}
{{eqn... | :$\log_b x = \dfrac {\log_a x} {\log_a b}$ | We begin with an equation that relates two exponentials with two different bases, $a$ and $b$, both positive and neither equal to one:
:$b^y = a^z$
{{WLOG}}, select and solve for $b$:
:$b = a^{z/y}$
Take logarithm's base $a$ of both sides:
{{begin-eqn}}
{{eqn | l = \log_a b
| r = \map {\log_a} {a^{z/y} }
}}
{... | Change of Base of Logarithm/Proof 2 | https://proofwiki.org/wiki/Change_of_Base_of_Logarithm | https://proofwiki.org/wiki/Change_of_Base_of_Logarithm/Proof_2 | [
"Change of Base of Logarithm",
"Logarithms"
] | [] | [
"Logarithm of Power"
] |
proofwiki-1802 | Derivative of Function to Power of Function | Let $\map u x, \map v x$ be real functions which are differentiable on $\R$.
Then:
:$\map {\dfrac \d {\d x} } {u^v} = v u^{v - 1} \map {\dfrac \d {\d x} } u + u^v \paren {\ln u} \map {\dfrac \d {\d x} } v$ | {{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {u^v}
| r = \map {\dfrac \d {\d x} } {\map \exp {v \ln u} }
| c = {{Defof|Power to Real Number}}
}}
{{eqn | r = \map \exp {v \ln u} \map {\dfrac \d {\d x} } {v \ln u}
| c = Chain Rule for Derivatives and Derivative of Exponential Function
}}
{{eqn | r... | Let $\map u x, \map v x$ be [[Definition:Real Function|real functions]] which are [[Definition:Differentiable on Interval|differentiable]] on $\R$.
Then:
:$\map {\dfrac \d {\d x} } {u^v} = v u^{v - 1} \map {\dfrac \d {\d x} } u + u^v \paren {\ln u} \map {\dfrac \d {\d x} } v$ | {{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {u^v}
| r = \map {\dfrac \d {\d x} } {\map \exp {v \ln u} }
| c = {{Defof|Power to Real Number}}
}}
{{eqn | r = \map \exp {v \ln u} \map {\dfrac \d {\d x} } {v \ln u}
| c = [[Chain Rule for Derivatives]] and [[Derivative of Exponential Function]]
}}
{... | Derivative of Function to Power of Function | https://proofwiki.org/wiki/Derivative_of_Function_to_Power_of_Function | https://proofwiki.org/wiki/Derivative_of_Function_to_Power_of_Function | [
"Derivative of Function to Power of Function",
"Differential Calculus",
"Exponential Function"
] | [
"Definition:Real Function",
"Definition:Differentiable Mapping/Real Function/Interval"
] | [
"Derivative of Composite Function",
"Derivative of Exponential Function",
"Product Rule for Derivatives",
"Derivative of Composite Function"
] |
proofwiki-1803 | Liouville's Theorem (Complex Analysis) | Let $f: \C \to \C$ be a bounded entire function.
Then $f$ is constant. | By assumption, there is $M \ge 0$ such that $\cmod {\map f z} \le M$ for all $z \in \C$.
For any $R \in \R: R > 0$, consider the function:
:$\map {f_R} z := \map f {R z}$
Using the Cauchy Integral Formula, we see that:
:$\ds \cmod {\map {f_R'} z} = \frac 1 {2 \pi} \cmod {\int_{\map {C_1} z} \frac {\map {f_R} w} {\par... | Let $f: \C \to \C$ be a [[Definition:Bounded Mapping|bounded]] [[Definition:Entire Function|entire]] [[Definition:Complex Function|function]].
Then $f$ is [[Definition:Constant Mapping|constant]]. | By assumption, there is $M \ge 0$ such that $\cmod {\map f z} \le M$ for all $z \in \C$.
For any $R \in \R: R > 0$, consider the function:
:$\map {f_R} z := \map f {R z}$
Using the [[Cauchy Integral Formula]], we see that:
:$\ds \cmod {\map {f_R'} z} = \frac 1 {2 \pi} \cmod {\int_{\map {C_1} z} \frac {\map {f_R} w}... | Liouville's Theorem (Complex Analysis)/Proof 1 | https://proofwiki.org/wiki/Liouville's_Theorem_(Complex_Analysis) | https://proofwiki.org/wiki/Liouville's_Theorem_(Complex_Analysis)/Proof_1 | [
"Liouville's Theorem (Complex Analysis)",
"Liouville's Theorem",
"Complex Analysis"
] | [
"Definition:Bounded Mapping",
"Definition:Entire Function",
"Definition:Complex Function",
"Definition:Constant Mapping"
] | [
"Cauchy's Integral Formula",
"Definition:Circle",
"Definition:Circle/Radius",
"Definition:Constant Mapping"
] |
proofwiki-1804 | Liouville's Theorem (Complex Analysis) | Let $f: \C \to \C$ be a bounded entire function.
Then $f$ is constant. | By assumption, there is $M \ge 0$ such that $\cmod {\map f z} \le M$ for all $z \in \C$.
Let $r > 0$.
consider:
:$D_r = \set {z \in \C: \cmod z \le r}$
Then, consider the parameterization $\gamma_r : \closedint 0 {2 \pi} \to \partial D_r$ given by:
:$\map {\gamma_r} t := r e^{2 \pi i t}$
For all $z \in \C$, we have:... | Let $f: \C \to \C$ be a [[Definition:Bounded Mapping|bounded]] [[Definition:Entire Function|entire]] [[Definition:Complex Function|function]].
Then $f$ is [[Definition:Constant Mapping|constant]]. | By assumption, there is $M \ge 0$ such that $\cmod {\map f z} \le M$ for all $z \in \C$.
Let $r > 0$.
consider:
:$D_r = \set {z \in \C: \cmod z \le r}$
Then, consider the [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterization]] $\gamma_r : \closedint 0 {2 \pi} \to \partial D_r$ giv... | Liouville's Theorem (Complex Analysis)/Proof 2 | https://proofwiki.org/wiki/Liouville's_Theorem_(Complex_Analysis) | https://proofwiki.org/wiki/Liouville's_Theorem_(Complex_Analysis)/Proof_2 | [
"Liouville's Theorem (Complex Analysis)",
"Liouville's Theorem",
"Complex Analysis"
] | [
"Definition:Bounded Mapping",
"Definition:Entire Function",
"Definition:Complex Function",
"Definition:Constant Mapping"
] | [
"Definition:Directed Smooth Curve/Parameterization/Complex Plane",
"Cauchy's Integral Formula",
"Triangle Inequality for Integrals/Complex",
"Zero Derivative implies Constant Complex Function",
"Definition:Constant Mapping"
] |
proofwiki-1805 | Riemann Removable Singularities Theorem | Let $U \subset \C$ be a domain.
Let $z_0 \in U$.
Let $f: U \setminus \set {z_0} \to \C$ be holomorphic.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f$ extends to a holomorphic function $f: U \to \C$}}
{{item|(2):|$f$ extends to a continuous function $f: U \to \C$}}
{{item|(3):|$f$ is bounded in a deleted neighborhood of $z... | {{WLOG}}, we may assume that:
:$U = \Bbb D := \set {z \in \C: \cmod z < 1}$
and that $z_0 = 0$.
Otherwise, restrict $f$ to a suitable disk centered at $z_0$, and precompose with a suitable affine map.
$(1) \implies (2)$ by Complex-Differentiable Function is Continuous.
$(2) \implies (3)$ by Continuous Function is Loca... | Let $U \subset \C$ be a domain.
Let $z_0 \in U$.
Let $f: U \setminus \set {z_0} \to \C$ be [[Definition:Holomorphic Function|holomorphic]].
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f$ [[Definition:Extension of Mapping|extends]] to a [[Definition:Holomorphic Function|holomorphic function]] $f: U \to \C$}}
{{item|(2):|... | {{WLOG}}, we may assume that:
:$U = \Bbb D := \set {z \in \C: \cmod z < 1}$
and that $z_0 = 0$.
Otherwise, restrict $f$ to a suitable disk centered at $z_0$, and [[Definition:Precomposition|precompose]] with a suitable [[Definition:Affine Mapping|affine map]].
$(1) \implies (2)$ by [[Complex-Differentiable Functio... | Riemann Removable Singularities Theorem | https://proofwiki.org/wiki/Riemann_Removable_Singularities_Theorem | https://proofwiki.org/wiki/Riemann_Removable_Singularities_Theorem | [
"Complex Analysis",
"Continuous Functions",
"Removable Singularities"
] | [
"Definition:Holomorphic Function",
"Definition:Extension of Mapping",
"Definition:Holomorphic Function",
"Definition:Extension of Mapping",
"Definition:Continuous Complex Function",
"Definition:Bounded Mapping",
"Definition:Deleted Neighborhood/Complex Analysis"
] | [
"Definition:Precomposition",
"Definition:Affine Transformation",
"Complex-Differentiable Function is Continuous",
"Continuous Implies Locally Bounded",
"Cauchy's Residue Theorem",
"Definition:Holomorphic Function",
"Definition:Holomorphic Function",
"Morera's Theorem",
"Cauchy's Residue Theorem",
... |
proofwiki-1806 | Gaussian Integral | :$\ds \int_{-\infty}^\infty e^{-x^2} \rd x = \sqrt \pi$ | {{begin-eqn}}
{{eqn | l = \int_{-\infty}^\infty e^{-x^2} \rd x
| r = 2 \int_0^\infty e^{-x^2} \rd x
| c = Definite Integral of Even Function
}}
{{eqn | r = \frac {2 \sqrt \pi} 2
| c = Integral to Infinity of Exponential of -t^2
}}
{{eqn | r = \sqrt \pi
}}
{{end-eqn}}
{{qed}}
Category:Gaussian Integral... | :$\ds \int_{-\infty}^\infty e^{-x^2} \rd x = \sqrt \pi$ | {{begin-eqn}}
{{eqn | l = \int_{-\infty}^\infty e^{-x^2} \rd x
| r = 2 \int_0^\infty e^{-x^2} \rd x
| c = [[Definite Integral of Even Function]]
}}
{{eqn | r = \frac {2 \sqrt \pi} 2
| c = [[Integral to Infinity of Exponential of -t^2]]
}}
{{eqn | r = \sqrt \pi
}}
{{end-eqn}}
{{qed}}
[[Category:Gaussi... | Gaussian Integral | https://proofwiki.org/wiki/Gaussian_Integral | https://proofwiki.org/wiki/Gaussian_Integral | [
"Gaussian Integral",
"Definite Integrals involving Exponential Function"
] | [] | [
"Definite Integral of Even Function",
"Integral to Infinity of Exponential of -t^2",
"Category:Gaussian Integral",
"Category:Definite Integrals involving Exponential Function"
] |
proofwiki-1807 | Indiscrete Topology is not Metrizable | Let $S$ be a set with more than one element.
The indiscrete topology on $S$ is not metrizable. | In order for $S$ to be a metrizable space, there needs to be a metric $d$ on $S$, so that $\struct {S, d}$ is a metric space.
As $S$ has more than one element, $\exists x, y \in S: x \ne y$.
Then:
:$\epsilon = \map d {x, y} > 0$
So the open $\epsilon$-ball $\map {B_{\epsilon / 2} } x$ is $d$-open.
Hence $\map {B_{\epsi... | Let $S$ be a [[Definition:Set|set]] with more than one [[Definition:Element|element]].
The [[Definition:Indiscrete Topology|indiscrete topology]] on $S$ is not [[Definition:Metrizable Topology|metrizable]]. | In order for $S$ to be a [[Definition:Metrizable Space|metrizable space]], there needs to be a [[Definition:Metric|metric]] $d$ on $S$, so that $\struct {S, d}$ is a [[Definition:Metric Space|metric space]].
As $S$ has more than one [[Definition:Element|element]], $\exists x, y \in S: x \ne y$.
Then:
:$\epsilon = \ma... | Indiscrete Topology is not Metrizable | https://proofwiki.org/wiki/Indiscrete_Topology_is_not_Metrizable | https://proofwiki.org/wiki/Indiscrete_Topology_is_not_Metrizable | [
"Indiscrete Topology",
"Examples of Metrizable Spaces"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Indiscrete Topology",
"Definition:Metrizable Space"
] | [
"Definition:Metrizable Space",
"Definition:Metric Space/Metric",
"Definition:Metric Space",
"Definition:Element",
"Definition:Open Ball",
"Definition:Open Set/Metric Space",
"Definition:Topology",
"Definition:Topology Induced by Metric",
"Definition:Topology Induced by Metric",
"Definition:Indiscr... |
proofwiki-1808 | Composite of Continuous Mappings is Continuous | Let $T_1, T_2, T_3$ be topological spaces.
Let $f: T_1 \to T_2$ and $g: T_2 \to T_3$ be continuous mappings.
Then the composite mapping $g \circ f: T_1 \to T_3$ is continuous. | Let $U \in T_3$ be open in $T_3$.
As $g$ is continuous, $g^{-1} \sqbrk U \in T_2$ is open in $T_2$.
As $f$ is continuous, $f^{-1} \sqbrk {g^{-1} \sqbrk U} \in T_1$ is open in $T_1$.
By Preimage of Subset under Composite Mapping, $f^{-1} \sqbrk {g^{-1} \sqbrk U} = \paren {g \circ f}^{-1} \sqbrk U$.
Hence the result.
{{q... | Let $T_1, T_2, T_3$ be [[Definition:Topological Space|topological spaces]].
Let $f: T_1 \to T_2$ and $g: T_2 \to T_3$ be [[Definition:Continuous Mapping (Topology)|continuous mappings]].
Then the [[Definition:Composition of Mappings|composite mapping]] $g \circ f: T_1 \to T_3$ is [[Definition:Continuous Mapping (Top... | Let $U \in T_3$ be [[Definition:Open Set (Topology)|open in $T_3$]].
As $g$ is [[Definition:Continuous Mapping (Topology)|continuous]], $g^{-1} \sqbrk U \in T_2$ is [[Definition:Open Set (Topology)|open in $T_2$]].
As $f$ is [[Definition:Continuous Mapping (Topology)|continuous]], $f^{-1} \sqbrk {g^{-1} \sqbrk U} \in... | Composite of Continuous Mappings is Continuous | https://proofwiki.org/wiki/Composite_of_Continuous_Mappings_is_Continuous | https://proofwiki.org/wiki/Composite_of_Continuous_Mappings_is_Continuous | [
"Composite of Continuous Mappings is Continuous",
"Continuous Mappings (Topology)",
"Composite Mappings"
] | [
"Definition:Topological Space",
"Definition:Continuous Mapping (Topology)",
"Definition:Composition of Mappings",
"Definition:Continuous Mapping (Topology)"
] | [
"Definition:Open Set/Topology",
"Definition:Continuous Mapping (Topology)",
"Definition:Open Set/Topology",
"Definition:Continuous Mapping (Topology)",
"Definition:Open Set/Topology",
"Preimage of Subset under Composite Mapping"
] |
proofwiki-1809 | Continuous Mapping is Continuous on Induced Topological Spaces | Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\tau_{d_1}$ and $\tau_{d_2}$ be the topologies induced by the metrics $d_1$ and $d_2$.
Let $T_1 = \struct {A_1, \tau_{d_1} }$ and $T_2 = \struct {A_2, \tau_{d_2} }$ be the resulting topological spaces.
Let $f: A_1 \to A_2$ be a mapping... | Follows directly from:
:the open set definition of continuity on a metric space
:the definition of continuity on a topological space.
{{qed}} | Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
Let $\tau_{d_1}$ and $\tau_{d_2}$ be the [[Definition:Topology Induced by Metric|topologies induced by the metrics]] $d_1$ and $d_2$.
Let $T_1 = \struct {A_1, \tau_{d_1} }$ and $T_2 = \struct {A_2, \tau_{d_2} }... | Follows directly from:
:the [[Definition:Metric Space Continuity by Open Set|open set definition]] of continuity on a [[Definition:Metric Space|metric space]]
:the definition of [[Definition:Continuous Mapping (Topology)|continuity]] on a [[Definition:Topological Space|topological space]].
{{qed}} | Continuous Mapping is Continuous on Induced Topological Spaces | https://proofwiki.org/wiki/Continuous_Mapping_is_Continuous_on_Induced_Topological_Spaces | https://proofwiki.org/wiki/Continuous_Mapping_is_Continuous_on_Induced_Topological_Spaces | [
"Topology",
"Metric Spaces",
"Continuous Mappings"
] | [
"Definition:Metric Space",
"Definition:Topology Induced by Metric",
"Definition:Topological Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Metric Space)",
"Definition:Continuous Mapping (Topology)"
] | [
"Definition:Continuous Mapping (Metric Space)/Space/Definition 2",
"Definition:Metric Space",
"Definition:Continuous Mapping (Topology)",
"Definition:Topological Space"
] |
proofwiki-1810 | Synthetic Basis and Analytic Basis are Compatible | Let $\struct {S, \tau}$ be a topological space.
Then $\BB$ is an analytic basis for $\tau$ {{iff}} $\tau$ is the topology on $S$ generated by the synthetic basis $\BB$. | === Necessary Condition ===
Suppose that $\BB$ is an analytic basis for $\tau$.
We proceed to check the axioms for $\BB$ to be a synthetic basis on $S$.
{{Open-set-axiom|3}} states that $S \in \tau$.
Therefore, by the definition of an analytic basis:
:$\ds \exists \SS \subseteq \BB: S = \bigcup \SS$
By Equivalent Condi... | Let $\struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Then $\BB$ is an [[Definition:Analytic Basis|analytic basis]] for $\tau$ {{iff}} $\tau$ is the [[Definition:Topology Generated by Synthetic Basis|topology on $S$ generated by the synthetic basis $\BB$]]. | === Necessary Condition ===
Suppose that $\BB$ is an [[Definition:Analytic Basis|analytic basis]] for $\tau$.
We proceed to check the axioms for $\BB$ to be a [[Definition:Synthetic Basis|synthetic basis]] on $S$.
{{Open-set-axiom|3}} states that $S \in \tau$.
Therefore, by the definition of an [[Definition:Analyt... | Synthetic Basis and Analytic Basis are Compatible | https://proofwiki.org/wiki/Synthetic_Basis_and_Analytic_Basis_are_Compatible | https://proofwiki.org/wiki/Synthetic_Basis_and_Analytic_Basis_are_Compatible | [
"Synthetic Bases",
"Analytic Bases"
] | [
"Definition:Topological Space",
"Definition:Basis (Topology)/Analytic Basis",
"Definition:Topology Generated by Synthetic Basis"
] | [
"Definition:Basis (Topology)/Analytic Basis",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Basis (Topology)/Analytic Basis",
"Equivalent Conditions for Cover by Collection of Subsets",
"Definition:Cover of Set",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Basis (Topology)/An... |
proofwiki-1811 | Picard's Existence Theorem | Let $\map f {x, y} : \R^2 \to \R$ be continuous in a region $D \subseteq \R^2$.
Let $\exists M \in \R: \forall x, y \in D: \size {\map f {x, y} } < M$.
Let $\map f {x, y}$ satisfy in $D$ the Lipschitz condition in $y$:
:$\size{\map f {x, y_1} - \map f {x, y_2} } \le A \size {y_1 - y_2}$
where $A$ is independent of $x, ... | Let us define the following series of functions:
{{begin-eqn}}
{{eqn | l = \map {y_0} x
| r = b
| c =
}}
{{eqn | l = \map {y_1} x
| r = b + \int_a^x \map f {t, \map {y_0} t} \rd t
| c =
}}
{{eqn | l = \map {y_2} x
| r = b + \int_a^x \map f {t, \map {y_1} t} \rd t
| c =
}}
{{eqn | ... | Let $\map f {x, y} : \R^2 \to \R$ be [[Definition:Continuous Real Function|continuous]] in a [[Definition:Region of Plane|region]] $D \subseteq \R^2$.
Let $\exists M \in \R: \forall x, y \in D: \size {\map f {x, y} } < M$.
Let $\map f {x, y}$ satisfy in $D$ the [[Definition:Lipschitz Condition (Real Function)|Lipschi... | Let us define the following series of [[Definition:Real Function|functions]]:
{{begin-eqn}}
{{eqn | l = \map {y_0} x
| r = b
| c =
}}
{{eqn | l = \map {y_1} x
| r = b + \int_a^x \map f {t, \map {y_0} t} \rd t
| c =
}}
{{eqn | l = \map {y_2} x
| r = b + \int_a^x \map f {t, \map {y_1} t} ... | Picard's Existence Theorem/Proof 1 | https://proofwiki.org/wiki/Picard's_Existence_Theorem | https://proofwiki.org/wiki/Picard's_Existence_Theorem/Proof_1 | [
"Picard's Existence Theorem",
"First Order ODEs"
] | [
"Definition:Continuous Real Function",
"Definition:Region/Plane",
"Definition:Lipschitz Continuity/Real Function",
"Definition:Quadrilateral/Rectangle",
"Definition:First Order Ordinary Differential Equation",
"Definition:Unique"
] | [
"Definition:Real Function",
"Definition:Quadrilateral/Rectangle",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Definition:Induction Hypothesis",
"Definition:Lipschitz Continuity/Real Function",
"Definition:Induction Hypothesis",
"Definition:Basis for the Induction",
... |
proofwiki-1812 | Picard's Existence Theorem | Let $\map f {x, y} : \R^2 \to \R$ be continuous in a region $D \subseteq \R^2$.
Let $\exists M \in \R: \forall x, y \in D: \size {\map f {x, y} } < M$.
Let $\map f {x, y}$ satisfy in $D$ the Lipschitz condition in $y$:
:$\size{\map f {x, y_1} - \map f {x, y_2} } \le A \size {y_1 - y_2}$
where $A$ is independent of $x, ... | Let us define the following series of functions:
{{begin-eqn}}
{{eqn | l = \map {y_0} x
| r = b
| c =
}}
{{eqn | l = \map {y_1} x
| r = b + \int_a^x \map f {t, \map {y_0} t} \rd t
| c =
}}
{{eqn | l = \map {y_2} x
| r = b + \int_a^x \map f {t, \map {y_1} t} \rd t
| c =
}}
{{eqn | ... | Let $\map f {x, y} : \R^2 \to \R$ be [[Definition:Continuous Real Function|continuous]] in a [[Definition:Region of Plane|region]] $D \subseteq \R^2$.
Let $\exists M \in \R: \forall x, y \in D: \size {\map f {x, y} } < M$.
Let $\map f {x, y}$ satisfy in $D$ the [[Definition:Lipschitz Condition (Real Function)|Lipschi... | Let us define the following [[Definition:Partial Sum|series]] of [[Definition:Real Function|functions]]:
{{begin-eqn}}
{{eqn | l = \map {y_0} x
| r = b
| c =
}}
{{eqn | l = \map {y_1} x
| r = b + \int_a^x \map f {t, \map {y_0} t} \rd t
| c =
}}
{{eqn | l = \map {y_2} x
| r = b + \int_a^... | Picard's Existence Theorem/Proof 2 | https://proofwiki.org/wiki/Picard's_Existence_Theorem | https://proofwiki.org/wiki/Picard's_Existence_Theorem/Proof_2 | [
"Picard's Existence Theorem",
"First Order ODEs"
] | [
"Definition:Continuous Real Function",
"Definition:Region/Plane",
"Definition:Lipschitz Continuity/Real Function",
"Definition:Quadrilateral/Rectangle",
"Definition:First Order Ordinary Differential Equation",
"Definition:Unique"
] | [
"Definition:Series/Sequence of Partial Sums",
"Definition:Real Function",
"Definition:Sequence",
"Definition:Line/Curve",
"Principle of Mathematical Induction",
"Definition:Sequence",
"Definition:Telescoping Series",
"Definition:Variable/Real",
"Definition:Parameter",
"Definition:Inequality",
"T... |
proofwiki-1813 | Continuity Test using Basis | Let $\struct {X_1, \tau_1}$ and $\struct {X_2, \tau_2}$ be topological spaces.
Let $f: X_1 \to X_2$ be a mapping.
Let $\BB$ be an analytic basis for $\tau_2$.
Suppose that:
:$\forall B \in \BB: f^{-1} \sqbrk B \in \tau_1$
where $f^{-1} \sqbrk B$ denotes the preimage of $B$ under $f$.
Then $f$ is continuous. | Let $U \in \tau_2$.
By the definition of an analytic basis, it follows that:
:$\ds \exists \AA \subseteq \BB: U = \bigcup \AA$
Hence:
{{begin-eqn}}
{{eqn | l = f^{-1} \sqbrk U
| r = f^{-1} \sqbrk {\bigcup \AA}
}}
{{eqn | r = \bigcup_{B \mathop \in \AA} f^{-1} \sqbrk B
| c = Preimage of Union under Mapping: ... | Let $\struct {X_1, \tau_1}$ and $\struct {X_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
Let $f: X_1 \to X_2$ be a [[Definition:Mapping|mapping]].
Let $\BB$ be an [[Definition:Analytic Basis|analytic basis]] for $\tau_2$.
Suppose that:
:$\forall B \in \BB: f^{-1} \sqbrk B \in \tau_1$
where $f^... | Let $U \in \tau_2$.
By the definition of an [[Definition:Analytic Basis|analytic basis]], it follows that:
:$\ds \exists \AA \subseteq \BB: U = \bigcup \AA$
Hence:
{{begin-eqn}}
{{eqn | l = f^{-1} \sqbrk U
| r = f^{-1} \sqbrk {\bigcup \AA}
}}
{{eqn | r = \bigcup_{B \mathop \in \AA} f^{-1} \sqbrk B
| c = [... | Continuity Test using Basis | https://proofwiki.org/wiki/Continuity_Test_using_Basis | https://proofwiki.org/wiki/Continuity_Test_using_Basis | [
"Topological Bases",
"Continuous Mappings"
] | [
"Definition:Topological Space",
"Definition:Mapping",
"Definition:Basis (Topology)/Analytic Basis",
"Definition:Preimage/Mapping/Subset",
"Definition:Continuous Mapping (Topology)/Everywhere"
] | [
"Definition:Basis (Topology)/Analytic Basis",
"Preimage of Union under Mapping/General Result",
"Definition:Continuous Mapping (Topology)/Everywhere"
] |
proofwiki-1814 | Sub-Basis for Real Number Line | Let the real number line $\R$ be considered as a topology under the usual (Euclidean) topology.
Then:
:$\BB := \set {\openint \gets a, \openint b \to: a, b \in \R}$ is a sub-basis for $\R$. | Let $\openint c d$ be an open real interval.
Then by definition:
:$\openint c d = \openint \gets d \cap \openint c \to$
and so $\openint c d$ is the intersection of two elements of $\BB$.
From Open Sets in Real Number Line, any open set of $\R$ is the union of countably many open real intervals.
The result follows by d... | Let the [[Definition:Real Number Line|real number line]] $\R$ be considered as a [[Definition:Topology|topology]] under the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]].
Then:
:$\BB := \set {\openint \gets a, \openint b \to: a, b \in \R}$ is a [[Definition:Sub-Basis|sub-basis]] for... | Let $\openint c d$ be an [[Definition:Open Real Interval|open real interval]].
Then by definition:
:$\openint c d = \openint \gets d \cap \openint c \to$
and so $\openint c d$ is the [[Definition:Set Intersection|intersection]] of two [[Definition:Element|elements]] of $\BB$.
From [[Open Sets in Real Number Line]], ... | Sub-Basis for Real Number Line | https://proofwiki.org/wiki/Sub-Basis_for_Real_Number_Line | https://proofwiki.org/wiki/Sub-Basis_for_Real_Number_Line | [
"Topological Bases",
"Real Numbers"
] | [
"Definition:Real Number/Real Number Line",
"Definition:Topology",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Sub-Basis"
] | [
"Definition:Real Interval/Open",
"Definition:Set Intersection",
"Definition:Element",
"Open Sets in Real Number Line",
"Definition:Open Set/Real Analysis",
"Definition:Set Union",
"Definition:Countable Set",
"Definition:Real Interval/Open",
"Definition:Sub-Basis"
] |
proofwiki-1815 | Synthetic Basis formed from Synthetic Sub-Basis | Let $X$ be a set.
Let $\SS$ be a synthetic sub-basis on $X$.
Define:
:$\ds \BB = \set {\bigcap \FF: \FF \subseteq \SS, \text{$\FF$ is finite} }$
Then $\BB$ is a synthetic basis on $X$.
Thus, by this construction, an arbitrary collection of subsets of $X$ can form a synthetic basis and thus generate a topology on $X$. | We consider $X$ as the universal set.
Thus, in accordance with Intersection of Empty Set, we take the convention that:
:$\ds \bigcap \O = X \in \BB$
By Set is Subset of Union: General Result, it follows that:
:$\ds X \subseteq \bigcup \BB$
That is, axiom $(\text B 1)$ for a synthetic basis is satisfied.
We have that $\... | Let $X$ be a [[Definition:Set|set]].
Let $\SS$ be a [[Definition:Synthetic Sub-Basis|synthetic sub-basis]] on $X$.
Define:
:$\ds \BB = \set {\bigcap \FF: \FF \subseteq \SS, \text{$\FF$ is finite} }$
Then $\BB$ is a [[Definition:Synthetic Basis|synthetic basis]] on $X$.
Thus, by this construction, an [[Definition:A... | We consider $X$ as the [[Definition:Universal Set|universal set]].
Thus, in accordance with [[Intersection of Empty Set]], we take the convention that:
:$\ds \bigcap \O = X \in \BB$
By [[Set is Subset of Union/General Result|Set is Subset of Union: General Result]], it follows that:
:$\ds X \subseteq \bigcup \BB$
Th... | Synthetic Basis formed from Synthetic Sub-Basis | https://proofwiki.org/wiki/Synthetic_Basis_formed_from_Synthetic_Sub-Basis | https://proofwiki.org/wiki/Synthetic_Basis_formed_from_Synthetic_Sub-Basis | [
"Synthetic Bases",
"Synthetic Sub-Bases"
] | [
"Definition:Set",
"Definition:Sub-Basis/Synthetic Sub-Basis",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Arbitrary",
"Definition:Collection",
"Definition:Subset",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Topology Generated by Synthetic Basis"
] | [
"Definition:Universal Set",
"Intersection of Empty Set",
"Set is Subset of Union/General Result",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Existential Quantifier",
"Definition:Finite Set",
"Union is Smallest Superset",
"Union of Finite Sets is Finite",
"Definition:Basis (Topology)/... |
proofwiki-1816 | Epimorphism Preserves Distributivity | Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be algebraic structures.
Let $\phi: R_1 \to R_2$ be an epimorphism.
:If $\circ_1$ is left distributive over $+_1$, then $\circ_2$ is left distributive over $+_2$.
:If $\circ_1$ is right distributive over $+_1$, then $\circ_2$ is right distributive over... | Throughout the following, we assume the morphism property holds for $\phi$ for both operations.
It remains to be shown that for non-empty $R_1$, $\paren {x +_1 y}, \paren {y +_1 z}, \paren {x \circ_1 y }, \paren {y \circ_1 z}, \paren {x \circ_1 z}, \paren { x \circ_1 \paren {y +_1 z} }, \paren { \paren {x +_1 y} \circ_... | Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be [[Definition:Algebraic Structure with Two Operations|algebraic structures]].
Let $\phi: R_1 \to R_2$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
:If $\circ_1$ is [[Definition:Left Distributive Operation|left distributive]] ove... | Throughout the following, we assume the [[Definition:Morphism Property|morphism property]] holds for $\phi$ for both operations.
It remains to be shown that for [[Definition:Non-Empty Set|non-empty]] $R_1$, $\paren {x +_1 y}, \paren {y +_1 z}, \paren {x \circ_1 y }, \paren {y \circ_1 z}, \paren {x \circ_1 z}, \paren {... | Epimorphism Preserves Distributivity | https://proofwiki.org/wiki/Epimorphism_Preserves_Distributivity | https://proofwiki.org/wiki/Epimorphism_Preserves_Distributivity | [
"Epimorphisms (Abstract Algebra)",
"Distributive Operations"
] | [
"Definition:Algebraic Structure/Two Operations",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Distributive Operation/Left",
"Definition:Distributive Operation/Left",
"Definition:Distributive Operation/Right",
"Definition:Distributive Operation/Right",
"Definition:Distributive Operation",
"... | [
"Definition:Morphism Property",
"Definition:Non-Empty Set",
"Definition:Non-Empty Set",
"Definition:Non-Empty Set"
] |
proofwiki-1817 | Epimorphism Preserves Groups | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ be a group.
Then $\struct {T, *}$ is also a group. | From Epimorphism Preserves Semigroups, if $\struct {S, \circ}$ is a semigroup then so is $\struct {T, *}$.
From Epimorphism Preserves Identity, if $\struct {S, \circ}$ has an identity $e_S$, then $\map \phi {e_S}$ is the identity for $*$.
From Epimorphism Preserves Inverses, if $x^{-1}$ is an inverse of $x$ for $\circ$... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
Let $\struct {S, \circ}$ be a [[Definition:Group|group]].
Then $\struct {T, ... | From [[Epimorphism Preserves Semigroups]], if $\struct {S, \circ}$ is a [[Definition:Semigroup|semigroup]] then so is $\struct {T, *}$.
From [[Epimorphism Preserves Identity]], if $\struct {S, \circ}$ has an [[Definition:Identity Element|identity]] $e_S$, then $\map \phi {e_S}$ is the [[Definition:Identity Element|ide... | Epimorphism Preserves Groups | https://proofwiki.org/wiki/Epimorphism_Preserves_Groups | https://proofwiki.org/wiki/Epimorphism_Preserves_Groups | [
"Group Epimorphisms"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Group",
"Definition:Group"
] | [
"Epimorphism Preserves Semigroups",
"Definition:Semigroup",
"Epimorphism Preserves Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Epimorphism Preserves Inverses",
"Definition:Inverse (Abstract Algebra)/Inverse",
"De... |
proofwiki-1818 | Totally Bounded Metric Space is Bounded | Let $M = \struct {A, d}$ be a totally bounded metric space.
Then $M$ is bounded. | Let $M = \struct {A, d}$ be totally bounded.
Then there exist $n \in \N$ and points $x_0, \dots, x_n \in A$ such that:
:$\ds \inf_{0 \mathop \le i \mathop \le n} \map d {x_i, x} \le 1$
for all $x \in A$.
Let us set:
{{begin-eqn}}
{{eqn | l = a
| o = :=
| r = x_0
}}
{{eqn | l = D
| o = :=
| r = \... | Let $M = \struct {A, d}$ be a [[Definition:Totally Bounded Metric Space|totally bounded metric space]].
Then $M$ is [[Definition:Bounded Metric Space|bounded]]. | Let $M = \struct {A, d}$ be [[Definition:Totally Bounded Metric Space|totally bounded]].
Then there exist $n \in \N$ and [[Definition:Point of Set|points]] $x_0, \dots, x_n \in A$ such that:
:$\ds \inf_{0 \mathop \le i \mathop \le n} \map d {x_i, x} \le 1$
for all $x \in A$.
Let us set:
{{begin-eqn}}
{{eqn | l = a
... | Totally Bounded Metric Space is Bounded | https://proofwiki.org/wiki/Totally_Bounded_Metric_Space_is_Bounded | https://proofwiki.org/wiki/Totally_Bounded_Metric_Space_is_Bounded | [
"Bounded Metric Spaces",
"Totally Bounded Metric Spaces"
] | [
"Definition:Totally Bounded Metric Space",
"Definition:Bounded Metric Space"
] | [
"Definition:Totally Bounded Metric Space",
"Definition:Element",
"Definition:Arbitrary",
"Definition:Bounded Metric Space"
] |
proofwiki-1819 | Existence and Uniqueness of Sigma-Algebra Generated by Collection of Subsets | Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Then $\map \sigma \GG$, the $\sigma$-algebra generated by $\GG$, exists and is unique. | === Existence ===
By Power Set is Sigma-Algebra, there is at least one $\sigma$-algebra containing $\GG$.
Next, let $\Bbb E$ be the collection of $\sigma$-algebras containing $\GG$:
:$\Bbb E := \set {\Sigma': \GG \subseteq \Sigma', \text{$\Sigma'$ is a $\sigma$-algebra} }$
By Intersection of Sigma-Algebras, $\Sigma := ... | Let $X$ be a [[Definition:Set|set]].
Let $\GG \subseteq \powerset X$ be a collection of [[Definition:Subset|subsets]] of $X$.
Then $\map \sigma \GG$, the [[Definition:Sigma-Algebra Generated by Collection of Subsets|$\sigma$-algebra generated by $\GG$]], exists and is unique. | === Existence ===
By [[Power Set is Sigma-Algebra]], there is at least one [[Definition:Sigma-Algebra|$\sigma$-algebra]] containing $\GG$.
Next, let $\Bbb E$ be the collection of [[Definition:Sigma-Algebra|$\sigma$-algebras]] containing $\GG$:
:$\Bbb E := \set {\Sigma': \GG \subseteq \Sigma', \text{$\Sigma'$ is a $\... | Existence and Uniqueness of Sigma-Algebra Generated by Collection of Subsets | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Sigma-Algebra_Generated_by_Collection_of_Subsets | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Sigma-Algebra_Generated_by_Collection_of_Subsets | [
"Sigma-Algebras Generated by Collection of Subsets"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Sigma-Algebra Generated by Collection of Subsets"
] | [
"Power Set is Sigma-Algebra",
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra",
"Intersection of Sigma-Algebras",
"Definition:Sigma-Algebra",
"Set Intersection Preserves Subsets/Families of Sets",
"Definition:Sigma-Algebra",
"Intersection is Subset/General Result",
"Definition:Sigma-Algebra"
] |
proofwiki-1820 | Basis for Topology on Cartesian Product | Let $T_1 = \struct{A_1, \tau_1}$ and $T_2 = \struct{A_2, \tau_2}$ be topological spaces.
Let $A_1 \times A_2$ be the Cartesian product of $A_1$ and $A_2$.
Let $\PP = \set {U_1 \times U_2 : U_1 \in \tau_1, U_2 \in \tau_2}$
Then $\PP$ is a synthetic basis on $A_1 \times A_2$. | We need to show that conditions $(\text B 1)$ and $(\text B 2)$ in the definition for basis hold for $\PP$. | Let $T_1 = \struct{A_1, \tau_1}$ and $T_2 = \struct{A_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
Let $A_1 \times A_2$ be the [[Definition:Cartesian Product|Cartesian product]] of $A_1$ and $A_2$.
Let $\PP = \set {U_1 \times U_2 : U_1 \in \tau_1, U_2 \in \tau_2}$
Then $\PP$ is a [[Definition... | We need to show that conditions $(\text B 1)$ and $(\text B 2)$ in the definition for [[Definition:Synthetic Basis|basis]] hold for $\PP$. | Basis for Topology on Cartesian Product | https://proofwiki.org/wiki/Basis_for_Topology_on_Cartesian_Product | https://proofwiki.org/wiki/Basis_for_Topology_on_Cartesian_Product | [
"Product Topology"
] | [
"Definition:Topological Space",
"Definition:Cartesian Product",
"Definition:Basis (Topology)/Synthetic Basis"
] | [
"Definition:Basis (Topology)/Synthetic Basis"
] |
proofwiki-1821 | Projection from Product Topology is Continuous | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its fa... | From Natural Basis of Product Topology of Finite Product, a basis for $\tau$ is:
:$\BB = \set {U \times V: U \in \tau_1, V \in \tau_2}$
Let $U$ be open in $T_1$.
Then $\map {\pr_1^{-1} } U = U \times T_2$ is one of the open sets in the basis in the definition of product topology.
Thus $\pr_1$ is continuous.
The same ar... | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
Let $T = \struct {T_1 \times T_2, \tau}$ be the [[Definition:Product Space (Topology) of Two Factor Spaces|product space]] of $T_1$ and $T_2$, where $\tau$ is the [[Definition:Product Topology on... | From [[Natural Basis of Product Topology of Finite Product]], a [[Definition:Synthetic Basis|basis]] for $\tau$ is:
:$\BB = \set {U \times V: U \in \tau_1, V \in \tau_2}$
Let $U$ be [[Definition:Open Set (Topology)|open]] in $T_1$.
Then $\map {\pr_1^{-1} } U = U \times T_2$ is one of the [[Definition:Open Set (Topolo... | Projection from Product Topology is Continuous | https://proofwiki.org/wiki/Projection_from_Product_Topology_is_Continuous | https://proofwiki.org/wiki/Projection_from_Product_Topology_is_Continuous | [
"Projection from Product Topology is Continuous",
"Projection from Product Topology is Open and Continuous",
"Product Topology",
"Continuous Mappings",
"Projections"
] | [
"Definition:Topological Space",
"Definition:Product Space (Topology)/Two Factor Spaces",
"Definition:Product Topology/Two Factor Spaces",
"Definition:Projection (Mapping Theory)",
"Definition:Product Topology/Factor Space",
"Definition:Continuous Mapping (Topology)"
] | [
"Natural Basis of Product Topology/Finite Product",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Basis (Topology)",
"Definition:Product Topology/Two Factor Spaces",
"Definition:Continuous Mapping (Topology)"
] |
proofwiki-1822 | Projection from Product Topology is Continuous | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its fa... | By definition of the product topology on $S$:
:$\tau$ is the initial topology on $S$ with respect to $\family {\pr_i}_{i \mathop \in I}$
By definition of the Initial Topoplogy: Definition 2:
:$\tau$ is the coarsest topology on $S$ such that each $\pr_i: S \to S_i$ is a $\struct{\tau, \tau_i}$-continuous. | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
Let $T = \struct {T_1 \times T_2, \tau}$ be the [[Definition:Product Space (Topology) of Two Factor Spaces|product space]] of $T_1$ and $T_2$, where $\tau$ is the [[Definition:Product Topology on... | By definition of the [[Definition:Product Topology|product topology]] on $S$:
:$\tau$ is the [[Definition:Initial Topology|initial topology]] on $S$ with respect to $\family {\pr_i}_{i \mathop \in I}$
By definition of the [[Definition:Initial Topology/Definition 2|Initial Topoplogy: Definition 2]]:
:$\tau$ is the [[De... | Projection from Product Topology is Continuous/General Result/Proof | https://proofwiki.org/wiki/Projection_from_Product_Topology_is_Continuous | https://proofwiki.org/wiki/Projection_from_Product_Topology_is_Continuous/General_Result/Proof | [
"Projection from Product Topology is Continuous",
"Projection from Product Topology is Open and Continuous",
"Product Topology",
"Continuous Mappings",
"Projections"
] | [
"Definition:Topological Space",
"Definition:Product Space (Topology)/Two Factor Spaces",
"Definition:Product Topology/Two Factor Spaces",
"Definition:Projection (Mapping Theory)",
"Definition:Product Topology/Factor Space",
"Definition:Continuous Mapping (Topology)"
] | [
"Definition:Product Topology",
"Definition:Initial Topology",
"Definition:Initial Topology/Definition 2",
"Definition:Coarser Topology",
"Definition:Continuous Mapping (Topology)"
] |
proofwiki-1823 | Continuous Mapping to Product Space | Let $T = T_1 \times T_2$ be the product space of two topological spaces $T_1$ and $T_2$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.
Let $T'$ be a topological space.
Let $f: T' \to T$ be a mapping.
Then $f$ is continuous {{iff}} $\pr_1 \circ f$ and $\pr_2... | === Necessary Condition ===
Let $f$ be continuous.
Then by:
:Projection from Product Topology is Continuous
:Composite of Continuous Mappings is Continuous
so are $\pr_1 \circ f$ and $\pr_2 \circ f$.
{{qed|lemma}} | Let $T = T_1 \times T_2$ be the [[Definition:Product Space (Topology) of Two Factor Spaces|product space]] of two [[Definition:Topological Space|topological spaces]] $T_1$ and $T_2$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the [[Definition:Projection (Mapping Theory)|first and second projections]] from $T$ on... | === Necessary Condition ===
Let $f$ be [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
Then by:
:[[Projection from Product Topology is Continuous]]
:[[Composite of Continuous Mappings is Continuous]]
so are $\pr_1 \circ f$ and $\pr_2 \circ f$.
{{qed|lemma}} | Continuous Mapping to Product Space | https://proofwiki.org/wiki/Continuous_Mapping_to_Product_Space | https://proofwiki.org/wiki/Continuous_Mapping_to_Product_Space | [
"Product Topology",
"Continuous Mappings"
] | [
"Definition:Product Space (Topology)/Two Factor Spaces",
"Definition:Topological Space",
"Definition:Projection (Mapping Theory)",
"Definition:Product Topology/Factor Space",
"Definition:Topological Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Contin... | [
"Definition:Continuous Mapping (Topology)/Everywhere",
"Projection from Product Topology is Continuous",
"Composite of Continuous Mappings is Continuous",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Continuous Mapping (Topology)... |
proofwiki-1824 | Open and Closed Sets in Topological Space | Let $T = \struct {S, \tau}$ be a topological space.
Then $S$ and $\O$ are both both open and closed in $T$. | From the definition of closed set, $U$ is open in $T$ {{iff}} $S \setminus U$ is closed in $T$.
From Underlying Set of Topological Space is Clopen, $S$ is both open and closed in $T$.
From Empty Set is Element of Topology, $\O$ is open in $T$.
From Empty Set is Closed in Topological Space, we have that $\O$ is closed i... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Then $S$ and $\O$ are both [[Definition:Clopen Set|both open and closed]] in $T$. | From the definition of [[Definition:Closed Set (Topology)|closed set]], $U$ is [[Definition:Open Set (Topology)|open]] in $T$ {{iff}} $S \setminus U$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
From [[Underlying Set of Topological Space is Clopen]], $S$ is both [[Definition:Open Set (Topology)|open]] and [... | Open and Closed Sets in Topological Space | https://proofwiki.org/wiki/Open_and_Closed_Sets_in_Topological_Space | https://proofwiki.org/wiki/Open_and_Closed_Sets_in_Topological_Space | [
"Clopen Sets"
] | [
"Definition:Topological Space",
"Definition:Clopen Set"
] | [
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Underlying Set of Topological Space is Clopen",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Empty Set is Element of Topology",
"Definition:Open Set/Topology",
"Empty Set is Clo... |
proofwiki-1825 | Absolutely Continuous Real Function is Continuous | Let $I \subseteq \R$ be a real interval.
Let $f: I \to \R$ be an absolutely continuous real function.
Then $f$ is continuous. | From Absolutely Continuous Real Function is Uniformly Continuous:
:$f$ is uniformly continuous.
Therefore, from Uniformly Continuous Real Function is Continuous:
:$f$ is continuous.
{{qed}}
Category:Absolutely Continuous Real Functions
Category:Continuous Real Functions
9uebyxkiz5v5kvnu8ztlutkxsrz1zdm | Let $I \subseteq \R$ be a [[Definition:Real Interval|real interval]].
Let $f: I \to \R$ be an [[Definition:Absolutely Continuous Real Function|absolutely continuous real function]].
Then $f$ is [[Definition:Continuous Real Function|continuous]]. | From [[Absolutely Continuous Real Function is Uniformly Continuous]]:
:$f$ is [[Definition:Uniformly Continuous Real Function|uniformly continuous]].
Therefore, from [[Uniformly Continuous Real Function is Continuous]]:
:$f$ is [[Definition:Continuous Real Function|continuous]].
{{qed}}
[[Category:Absolutely Contin... | Absolutely Continuous Real Function is Continuous | https://proofwiki.org/wiki/Absolutely_Continuous_Real_Function_is_Continuous | https://proofwiki.org/wiki/Absolutely_Continuous_Real_Function_is_Continuous | [
"Absolutely Continuous Real Functions",
"Continuous Real Functions"
] | [
"Definition:Real Interval",
"Definition:Absolute Continuity/Real Function",
"Definition:Continuous Real Function"
] | [
"Absolutely Continuous Real Function is Uniformly Continuous",
"Definition:Uniform Continuity/Real Function",
"Uniformly Continuous Function is Continuous/Real Function",
"Definition:Continuous Real Function",
"Category:Absolutely Continuous Real Functions",
"Category:Continuous Real Functions"
] |
proofwiki-1826 | Closed Set in Topological Subspace | Let $T = \struct{S, \tau}$ be a topological space.
Let $T' = \struct{H, \tau_H}$ be a subspace of $T$ where $H \subseteq S$.
Then $V \subseteq T'$ is closed in $T'$ {{iff}} $V = T' \cap W$ for some $W$ closed in $T$. | === Necessary Condition ===
Suppose $V \subseteq T'$ is closed in $T'$.
Then $T' \setminus V$ is open in $T'$ by definition.
So, by definition of subspace topology, $T' \setminus V = T' \cap U$ for some $U$ open in $T$.
Then:
{{begin-eqn}}
{{eqn | l = V
| r = T' \setminus \paren {T' \setminus V}
| c = Relat... | Let $T = \struct{S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $T' = \struct{H, \tau_H}$ be a [[Definition:Topological Subspace|subspace]] of $T$ where $H \subseteq S$.
Then $V \subseteq T'$ is [[Definition:Closed Set (Topology)|closed]] in $T'$ {{iff}} $V = T' \cap W$ for some $W$ [[Definit... | === Necessary Condition ===
Suppose $V \subseteq T'$ is [[Definition:Closed Set (Topology)|closed]] in $T'$.
Then $T' \setminus V$ is [[Definition:Open Set (Topology)|open]] in $T'$ by definition.
So, by definition of [[Definition:Topological Subspace|subspace topology]], $T' \setminus V = T' \cap U$ for some $U$ [[... | Closed Set in Topological Subspace | https://proofwiki.org/wiki/Closed_Set_in_Topological_Subspace | https://proofwiki.org/wiki/Closed_Set_in_Topological_Subspace | [
"Closed Sets",
"Topological Subspaces"
] | [
"Definition:Topological Space",
"Definition:Topological Subspace",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology"
] | [
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Definition:Topological Subspace",
"Definition:Open Set/Topology",
"Relative Complement of Relative Complement",
"Set Difference with Intersection is Difference",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"De... |
proofwiki-1827 | Continuity Defined from Closed Sets | Let $T_1$ and $T_2$ be topological spaces.
Let $f: T_1 \to T_2$ be a mapping.
Then $f$ is continuous {{iff}} for all $V$ closed in $T_2$, $f^{-1} \sqbrk V$ is closed in $T_1$. | First we show the following.
Let $W \in T_2$.
We note that $f^{-1} \sqbrk {T_2} = T_1$.
Hence, from Preimage of Set Difference under Mapping, we have:
:$f^{-1} \sqbrk {T_2 \setminus W} = T_1 \setminus f^{-1} \sqbrk W$ | Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
Let $f: T_1 \to T_2$ be a [[Definition:Mapping|mapping]].
Then $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] {{iff}} for all $V$ [[Definition:Closed Set (Topology)|closed]] in $T_2$, $f^{-1} \sqbrk V$ is [[Definit... | First we show the following.
Let $W \in T_2$.
We note that $f^{-1} \sqbrk {T_2} = T_1$.
Hence, from [[Preimage of Set Difference under Mapping]], we have:
:$f^{-1} \sqbrk {T_2 \setminus W} = T_1 \setminus f^{-1} \sqbrk W$ | Continuity Defined from Closed Sets | https://proofwiki.org/wiki/Continuity_Defined_from_Closed_Sets | https://proofwiki.org/wiki/Continuity_Defined_from_Closed_Sets | [
"Closed Sets",
"Continuous Mappings"
] | [
"Definition:Topological Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology"
] | [
"Preimage of Set Difference under Mapping"
] |
proofwiki-1828 | Real Numbers are Uncountably Infinite | The set of real numbers $\R$ is uncountably infinite. | We prove the equivalent result that every sequence $\sequence {x_k}_{k \mathop \in \N}$ omits at least one $x \in \R$.
Let $\sequence {x_k}_{k \mathop \in \N}$ be a sequence of distinct real numbers.
Let a sequence of closed real intervals $\sequence {I_n}$ be defined as follows:
Let:
{{begin-eqn}}
{{eqn | l = a_k
... | The [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] $\R$ is [[Definition:Uncountable Set|uncountably infinite]]. | We prove the equivalent result that every [[Definition:Sequence|sequence]] $\sequence {x_k}_{k \mathop \in \N}$ omits at least one $x \in \R$.
Let $\sequence {x_k}_{k \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Distinct|distinct]] [[Definition:Real Number|real numbers]].
Let a [[Definition... | Real Numbers are Uncountably Infinite/Cantor's First Proof | https://proofwiki.org/wiki/Real_Numbers_are_Uncountably_Infinite | https://proofwiki.org/wiki/Real_Numbers_are_Uncountably_Infinite/Cantor's_First_Proof | [
"Real Numbers are Uncountably Infinite",
"Real Numbers",
"Examples of Uncountable Sets"
] | [
"Definition:Set",
"Definition:Real Number",
"Definition:Uncountable/Set"
] | [
"Definition:Sequence",
"Definition:Sequence",
"Definition:Distinct",
"Definition:Real Number",
"Definition:Sequence",
"Definition:Real Interval/Closed",
"Definition:Term of Sequence",
"Definition:Distinct",
"Definition:Singleton",
"Definition:Infinite Set",
"Definition:Interior (Topology)",
"D... |
proofwiki-1829 | Real Numbers are Uncountably Infinite | The set of real numbers $\R$ is uncountably infinite. | By definition, a perfect set is a set $X$ such that every element $x \in X$ is the limit of a sequence of elements of $X$ distinct from $x$.
From Real Numbers form Perfect Set, $\R$ is perfect.
Therefore it is sufficient to show that a perfect subset of $X \subseteq \R^k$ is uncountable.
We prove the equivalent result ... | The [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] $\R$ is [[Definition:Uncountable Set|uncountably infinite]]. | By definition, a [[Definition:Perfect Set|perfect set]] is a [[Definition:Set|set]] $X$ such that every [[Definition:Element|element]] $x \in X$ is the [[Definition:Limit of Sequence (Topology)|limit of a sequence]] of [[Definition:Element|elements]] of $X$ [[Definition:Distinct|distinct]] from $x$.
From [[Real Number... | Real Numbers are Uncountably Infinite/Cantor's Second Proof | https://proofwiki.org/wiki/Real_Numbers_are_Uncountably_Infinite | https://proofwiki.org/wiki/Real_Numbers_are_Uncountably_Infinite/Cantor's_Second_Proof | [
"Real Numbers are Uncountably Infinite",
"Real Numbers",
"Examples of Uncountable Sets"
] | [
"Definition:Set",
"Definition:Real Number",
"Definition:Uncountable/Set"
] | [
"Definition:Perfect Set",
"Definition:Set",
"Definition:Element",
"Definition:Limit of Sequence/Topological Space",
"Definition:Element",
"Definition:Distinct",
"Real Numbers form Perfect Set",
"Definition:Perfect Set",
"Definition:Perfect Set",
"Definition:Subset",
"Definition:Uncountable/Set",... |
proofwiki-1830 | Real Numbers are Uncountably Infinite | The set of real numbers $\R$ is uncountably infinite. | It is sufficient to show that the real interval $I = \set {x \in \R: 0 < x \le 1}$ is uncountable.
Let $x \in I$.
From Existence of Base-N Representation, $x$ has a unique representation of the form:
:$x = \dfrac {\epsilon_1} 3 + \dfrac {\epsilon_2} {3^2} + \dfrac {\epsilon_3} {3^3} + \cdots$
where $\epsilon_k = 0, 1$... | The [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] $\R$ is [[Definition:Uncountable Set|uncountably infinite]]. | It is sufficient to show that the [[Definition:Real Interval|real interval]] $I = \set {x \in \R: 0 < x \le 1}$ is [[Definition:Uncountable Set|uncountable]].
Let $x \in I$.
From [[Existence of Base-N Representation]], $x$ has a [[Definition:Unique|unique]] representation of the form:
:$x = \dfrac {\epsilon_1} 3 + \d... | Real Numbers are Uncountably Infinite/Proof 1 using Ternary Notation | https://proofwiki.org/wiki/Real_Numbers_are_Uncountably_Infinite | https://proofwiki.org/wiki/Real_Numbers_are_Uncountably_Infinite/Proof_1_using_Ternary_Notation | [
"Real Numbers are Uncountably Infinite",
"Real Numbers",
"Examples of Uncountable Sets"
] | [
"Definition:Set",
"Definition:Real Number",
"Definition:Uncountable/Set"
] | [
"Definition:Real Interval",
"Definition:Uncountable/Set",
"Existence of Base-N Representation",
"Definition:Unique",
"Definition:Infinite Set",
"Definition:Countably Infinite/Set",
"Definition:Ternary Notation",
"Definition:Real Number",
"Definition:Element",
"Definition:Element",
"Definition:Pr... |
proofwiki-1831 | Real Numbers are Uncountably Infinite | The set of real numbers $\R$ is uncountably infinite. | Define a mapping $f: \powerset {\N_{>0} } \to \R$ thus:
:$\map f S = 0.d_1 d_2 \ldots$, interpreted as a ternary expansion where $\sequence {d_n}$ is the characteristic function of $S$.
That is:
:$\ds \map f S = \sum_{i \mathop \in S} 3^{-i}$
By the lemma, $f$ is an injection.
{{AimForCont}} that $\R$ is countable.
The... | The [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] $\R$ is [[Definition:Uncountable Set|uncountably infinite]]. | Define a [[Definition:Mapping|mapping]] $f: \powerset {\N_{>0} } \to \R$ thus:
:$\map f S = 0.d_1 d_2 \ldots$, interpreted as a [[Definition:Ternary Notation|ternary expansion]] where $\sequence {d_n}$ is the characteristic function of $S$.
That is:
:$\ds \map f S = \sum_{i \mathop \in S} 3^{-i}$
By the [[Real Numbe... | Real Numbers are Uncountably Infinite/Proof 2 using Ternary Notation | https://proofwiki.org/wiki/Real_Numbers_are_Uncountably_Infinite | https://proofwiki.org/wiki/Real_Numbers_are_Uncountably_Infinite/Proof_2_using_Ternary_Notation | [
"Real Numbers are Uncountably Infinite",
"Real Numbers",
"Examples of Uncountable Sets"
] | [
"Definition:Set",
"Definition:Real Number",
"Definition:Uncountable/Set"
] | [
"Definition:Mapping",
"Definition:Ternary Notation",
"Real Numbers are Uncountably Infinite/Proof 2 using Ternary Notation/Lemma",
"Definition:Injection",
"Definition:Countable Set",
"Definition:Injection",
"Composite of Injections is Injection",
"Definition:Injection",
"Definition:Contradiction",
... |
proofwiki-1832 | Real Numbers are Uncountably Infinite | The set of real numbers $\R$ is uncountably infinite. | By Surjection from Natural Numbers iff Countable, a set $A$ is countable {{iff}} there exists a surjection $f: \N \to A$.
Suppose there exists a surjection $f: \N \to \R$.
Then $\forall x \in \R: \exists n \in \N: \map f n = x$ as $f$ is surjective.
Let $d_{n, 0}$ be the integer before the decimal point of $\map f n$.
... | The [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] $\R$ is [[Definition:Uncountable Set|uncountably infinite]]. | By [[Surjection from Natural Numbers iff Countable]], a set $A$ is [[Definition:Countable Set|countable]] {{iff}} there exists a [[Definition:Surjection|surjection]] $f: \N \to A$.
Suppose there exists a [[Definition:Surjection|surjection]] $f: \N \to \R$.
Then $\forall x \in \R: \exists n \in \N: \map f n = x$ as $f... | Real Numbers are Uncountably Infinite/Set-Theoretical Approach: Proof 1 | https://proofwiki.org/wiki/Real_Numbers_are_Uncountably_Infinite | https://proofwiki.org/wiki/Real_Numbers_are_Uncountably_Infinite/Set-Theoretical_Approach:_Proof_1 | [
"Real Numbers are Uncountably Infinite",
"Real Numbers",
"Examples of Uncountable Sets"
] | [
"Definition:Set",
"Definition:Real Number",
"Definition:Uncountable/Set"
] | [
"Surjection from Natural Numbers iff Countable",
"Definition:Countable Set",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Integer",
"Definition:Decimal Expansion/Decimal Point",
"Definition:Digit",
"Definition:Decimal Expansion",
"Definition:Integer",
"D... |
proofwiki-1833 | Real Numbers are Uncountably Infinite | The set of real numbers $\R$ is uncountably infinite. | By Cantor's Theorem there is no surjection:
:$\N \twoheadrightarrow \powerset \N$
Additionally, we have Power Set of Natural Numbers is not Countable.
Therefore, if we can show that $\powerset \N$ injects into $\R$ then there is no injection $\R \hookrightarrow \N$ and $\R$ is uncountable.
To prove the theorem we const... | The [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] $\R$ is [[Definition:Uncountable Set|uncountably infinite]]. | By [[Cantor's Theorem]] there is no [[Definition:Surjection|surjection]]:
:$\N \twoheadrightarrow \powerset \N$
Additionally, we have [[Power Set of Natural Numbers is not Countable]].
Therefore, if we can show that $\powerset \N$ [[Definition:Injection|injects]] into $\R$ then there is no [[Definition:Injection|inje... | Real Numbers are Uncountably Infinite/Set-Theoretical Approach: Proof 2 | https://proofwiki.org/wiki/Real_Numbers_are_Uncountably_Infinite | https://proofwiki.org/wiki/Real_Numbers_are_Uncountably_Infinite/Set-Theoretical_Approach:_Proof_2 | [
"Real Numbers are Uncountably Infinite",
"Real Numbers",
"Examples of Uncountable Sets"
] | [
"Definition:Set",
"Definition:Real Number",
"Definition:Uncountable/Set"
] | [
"Cantor's Theorem",
"Definition:Surjection",
"Power Set of Natural Numbers is Uncountable",
"Definition:Injection",
"Definition:Injection",
"Definition:Injection",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Sequence/Infin... |
proofwiki-1834 | Condition for Point being in Closure | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $H^-$ denote the closure of $H$ in $T$.
Let $x \in S$.
Then $x \in H^-$ {{iff}} every open neighborhood of $x$ contains a point in $H$. | From the definition of closure, we have that $H^-$ is the union of $H$ and all the limit points of $H$ in $T$.
By definition, an open neighborhood of $x$ in $T$ is an open set of $T$ which contains $x$.
=== Necessary Condition ===
Let $x \in H^-$.
Then either:
:$(1): \quad x \in H$, in which case every open neighborhoo... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$.
Let $H^-$ denote the [[Definition:Closure (Topology)|closure]] of $H$ in $T$.
Let $x \in S$.
Then $x \in H^-$ {{iff}} every [[Definition:Open Neighborhood of Point|open neighborhood]] of $x$ contains a point i... | From the definition of [[Definition:Closure (Topology)|closure]], we have that $H^-$ is the [[Definition:Set Union|union]] of $H$ and all the [[Definition:Limit Point of Set|limit points]] of $H$ in $T$.
By definition, an [[Definition:Open Neighborhood of Point|open neighborhood]] of $x$ in $T$ is an [[Definition:Open... | Condition for Point being in Closure/Proof 1 | https://proofwiki.org/wiki/Condition_for_Point_being_in_Closure | https://proofwiki.org/wiki/Condition_for_Point_being_in_Closure/Proof_1 | [
"Set Closures",
"Condition for Point being in Closure"
] | [
"Definition:Topological Space",
"Definition:Closure (Topology)",
"Definition:Open Neighborhood/Point"
] | [
"Definition:Closure (Topology)",
"Definition:Set Union",
"Definition:Limit Point/Topology/Set",
"Definition:Open Neighborhood/Point",
"Definition:Open Set/Topology",
"Definition:Open Neighborhood/Point",
"Definition:Limit Point/Topology/Set",
"Definition:Limit Point/Topology/Set",
"Definition:Open N... |
proofwiki-1835 | Condition for Point being in Closure | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $H^-$ denote the closure of $H$ in $T$.
Let $x \in S$.
Then $x \in H^-$ {{iff}} every open neighborhood of $x$ contains a point in $H$. | The condition to be proved is equivalent to showing that $x \in H^-$ {{iff}}, for every open neighborhood $U$ of $x$, the intersection $H \cap U$ is non-empty.
For $U \in \tau$, let $U^{\complement}$ denote the relative complement of $U$ in $S$.
By definition, $U^{\complement}$ is closed in $T$
We have that:
{{begin-eq... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$.
Let $H^-$ denote the [[Definition:Closure (Topology)|closure]] of $H$ in $T$.
Let $x \in S$.
Then $x \in H^-$ {{iff}} every [[Definition:Open Neighborhood of Point|open neighborhood]] of $x$ contains a point i... | The condition to be proved is equivalent to showing that $x \in H^-$ {{iff}}, for every [[Definition:Open Neighborhood of Point|open neighborhood]] $U$ of $x$, the [[Definition:Set Intersection|intersection]] $H \cap U$ is [[Definition:Non-Empty Set|non-empty]].
For $U \in \tau$, let $U^{\complement}$ denote the [[De... | Condition for Point being in Closure/Proof 2 | https://proofwiki.org/wiki/Condition_for_Point_being_in_Closure | https://proofwiki.org/wiki/Condition_for_Point_being_in_Closure/Proof_2 | [
"Set Closures",
"Condition for Point being in Closure"
] | [
"Definition:Topological Space",
"Definition:Closure (Topology)",
"Definition:Open Neighborhood/Point"
] | [
"Definition:Open Neighborhood/Point",
"Definition:Set Intersection",
"Definition:Non-Empty Set",
"Definition:Relative Complement",
"Definition:Closed Set/Topology",
"Empty Intersection iff Subset of Complement",
"Set Closure is Smallest Closed Set/Topology",
"Relative Complement inverts Subsets",
"R... |
proofwiki-1836 | Topological Closure of Subset is Subset of Topological Closure | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq K$ and $K \subseteq S$.
Then:
:$\map \cl H \subseteq \map \cl K$
where $\map \cl H$ denotes the closure of $H$. | From Topological Closure is Closed, $\map \cl K$ is closed.
From Set is Subset of its Topological Closure:
:$K \subseteq \map \cl K$
By Subset Relation is Transitive, it follows that:
:$H \subseteq \map \cl K$
Hence, by definition of closure as the smallest closed set that contains $H$:
:$\map \cl H \subseteq \map \cl ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq K$ and $K \subseteq S$.
Then:
:$\map \cl H \subseteq \map \cl K$
where $\map \cl H$ denotes the [[Definition:Closure (Topology)|closure]] of $H$. | From [[Topological Closure is Closed]], $\map \cl K$ is [[Definition:Closed Set (Topology)|closed]].
From [[Set is Subset of its Topological Closure]]:
:$K \subseteq \map \cl K$
By [[Subset Relation is Transitive]], it follows that:
:$H \subseteq \map \cl K$
Hence, by definition of [[Definition:Closure (Topology)... | Topological Closure of Subset is Subset of Topological Closure/Proof 1 | https://proofwiki.org/wiki/Topological_Closure_of_Subset_is_Subset_of_Topological_Closure | https://proofwiki.org/wiki/Topological_Closure_of_Subset_is_Subset_of_Topological_Closure/Proof_1 | [
"Subsets",
"Set Closures",
"Topological Closure of Subset is Subset of Topological Closure"
] | [
"Definition:Topological Space",
"Definition:Closure (Topology)"
] | [
"Topological Closure is Closed",
"Definition:Closed Set/Topology",
"Set is Subset of its Topological Closure",
"Subset Relation is Transitive",
"Definition:Closure (Topology)/Definition 3",
"Definition:Smallest Set by Set Inclusion",
"Definition:Closed Set/Topology",
"Definition:Subset"
] |
proofwiki-1837 | Topological Closure of Subset is Subset of Topological Closure | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq K$ and $K \subseteq S$.
Then:
:$\map \cl H \subseteq \map \cl K$
where $\map \cl H$ denotes the closure of $H$. | From the definition of closure:
:$\map \cl H$ is the union of $H$ and its limit points.
Let $x \in \map \cl H$.
If $x \in H$ then $x \in K \implies x \in \map \cl K$.
Otherwise $x$ is a limit point of $H$.
That is, every open set $U$ of $T$ such that $x \in U$ contains $y \in H$ such that $y \ne x$.
But as $y \in H$ it... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq K$ and $K \subseteq S$.
Then:
:$\map \cl H \subseteq \map \cl K$
where $\map \cl H$ denotes the [[Definition:Closure (Topology)|closure]] of $H$. | From the definition of [[Definition:Closure (Topology)|closure]]:
:$\map \cl H$ is the [[Definition:Set Union|union]] of $H$ and its [[Definition:Limit Point of Set|limit points]].
Let $x \in \map \cl H$.
If $x \in H$ then $x \in K \implies x \in \map \cl K$.
Otherwise $x$ is a [[Definition:Limit Point of Set|limi... | Topological Closure of Subset is Subset of Topological Closure/Proof 2 | https://proofwiki.org/wiki/Topological_Closure_of_Subset_is_Subset_of_Topological_Closure | https://proofwiki.org/wiki/Topological_Closure_of_Subset_is_Subset_of_Topological_Closure/Proof_2 | [
"Subsets",
"Set Closures",
"Topological Closure of Subset is Subset of Topological Closure"
] | [
"Definition:Topological Space",
"Definition:Closure (Topology)"
] | [
"Definition:Closure (Topology)",
"Definition:Set Union",
"Definition:Limit Point/Topology/Set",
"Definition:Limit Point/Topology/Set",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Limit Point/Topology/Set",
"Set Union Preserves Subsets"
] |
proofwiki-1838 | Closure of Topological Closure equals Closure | Let $T$ be a topological space.
Let $H \subseteq T$.
Then:
:$\map \cl {\map \cl H} = \map \cl H$
where $\cl$ denotes the closure of $H$. | It follows directly from Set is Subset of its Topological Closure that:
:$\map \cl H \subseteq \map \cl {\map \cl H}$
{{qed|lemma}}
Let $x \in \map \cl {\map \cl H}$.
Then from Condition for Point being in Closure, any $U$ which is open in $T$ such that $x \in U$ contains some $y \in \map \cl H$.
If we consider $U$ as ... | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq T$.
Then:
:$\map \cl {\map \cl H} = \map \cl H$
where $\cl$ denotes the [[Definition:Closure (Topology)|closure]] of $H$. | It follows directly from [[Set is Subset of its Topological Closure]] that:
:$\map \cl H \subseteq \map \cl {\map \cl H}$
{{qed|lemma}}
Let $x \in \map \cl {\map \cl H}$.
Then from [[Condition for Point being in Closure]], any $U$ which is [[Definition:Open Set (Topology)|open]] in $T$ such that $x \in U$ contains s... | Closure of Topological Closure equals Closure | https://proofwiki.org/wiki/Closure_of_Topological_Closure_equals_Closure | https://proofwiki.org/wiki/Closure_of_Topological_Closure_equals_Closure | [
"Set Closures"
] | [
"Definition:Topological Space",
"Definition:Closure (Topology)"
] | [
"Set is Subset of its Topological Closure",
"Condition for Point being in Closure",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Subset",
"Definition:Set Equality"
] |
proofwiki-1839 | Topological Closure is Closed | Let $T$ be a topological space.
Let $H \subseteq T$.
Then the closure $\map \cl H$ of $H$ is closed in $T$. | From Closure of Topological Closure equals Closure:
:$\map \cl {\map \cl H} = \map \cl H$
From Set is Closed iff Equals Topological Closure, it follows that $\map \cl H$ is closed.
{{qed}} | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq T$.
Then the [[Definition:Closure (Topology)|closure]] $\map \cl H$ of $H$ is [[Definition:Closed Set (Topology)|closed]] in $T$. | From [[Closure of Topological Closure equals Closure]]:
:$\map \cl {\map \cl H} = \map \cl H$
From [[Set is Closed iff Equals Topological Closure]], it follows that $\map \cl H$ is [[Definition:Closed Set (Topology)|closed]].
{{qed}} | Topological Closure is Closed | https://proofwiki.org/wiki/Topological_Closure_is_Closed | https://proofwiki.org/wiki/Topological_Closure_is_Closed | [
"Set Closures",
"Closed Sets"
] | [
"Definition:Topological Space",
"Definition:Closure (Topology)",
"Definition:Closed Set/Topology"
] | [
"Closure of Topological Closure equals Closure",
"Set is Closed iff Equals Topological Closure",
"Definition:Closed Set/Topology"
] |
proofwiki-1840 | Equivalence of Definitions of Closure of Topological Subspace | {{TFAE|def = Closure (Topology)|view = closure|context = Topology (Mathematical Branch)|contextview = topology}}
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$. | === Definition $1$ is equivalent to Definition $2$ ===
This is proved in Set Closure as Intersection of Closed Sets.
{{qed|lemma}} | {{TFAE|def = Closure (Topology)|view = closure|context = Topology (Mathematical Branch)|contextview = topology}}
Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$. | === Definition $1$ is equivalent to Definition $2$ ===
This is proved in [[Set Closure as Intersection of Closed Sets]].
{{qed|lemma}} | Equivalence of Definitions of Closure of Topological Subspace | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Closure_of_Topological_Subspace | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Closure_of_Topological_Subspace | [
"Set Closures"
] | [
"Definition:Topological Space"
] | [
"Set Closure as Intersection of Closed Sets"
] |
proofwiki-1841 | Closure of Intersection is Subset of Intersection of Closures | Let $T$ be a topological space.
Let $I$ be an indexing set.
Let $\forall i \in I: H_i \subseteq T$.
Then:
:$\ds \map \cl {\bigcap_I H_i} \subseteq \bigcap_I \map \cl {H_i}$
where $\map \cl {H_i}$ denotes the closure of $H_i$. | Since $\ds \bigcap_I \map \cl {H_i}$ is an intersection of closed sets, it is closed, from Topology Defined by Closed Sets.
Also, it contains $\ds \bigcap_I H_i$ and so by the main definition of closure also contains $\ds \map \cl {\bigcap_I H_i}$.
{{qed}} | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\forall i \in I: H_i \subseteq T$.
Then:
:$\ds \map \cl {\bigcap_I H_i} \subseteq \bigcap_I \map \cl {H_i}$
where $\map \cl {H_i}$ denotes the [[Definition:Closure (Topology)|closure]] of $H... | Since $\ds \bigcap_I \map \cl {H_i}$ is an intersection of [[Definition:Closed Set (Topology)|closed sets]], it is closed, from [[Topology Defined by Closed Sets]].
Also, it contains $\ds \bigcap_I H_i$ and so by the main definition of [[Definition:Closure (Topology)|closure]] also contains $\ds \map \cl {\bigcap_I H_... | Closure of Intersection is Subset of Intersection of Closures | https://proofwiki.org/wiki/Closure_of_Intersection_is_Subset_of_Intersection_of_Closures | https://proofwiki.org/wiki/Closure_of_Intersection_is_Subset_of_Intersection_of_Closures | [
"Set Intersection",
"Set Closures"
] | [
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Closure (Topology)"
] | [
"Definition:Closed Set/Topology",
"Topology Defined by Closed Sets",
"Definition:Closure (Topology)"
] |
proofwiki-1842 | Closure of Finite Union equals Union of Closures | Let $T$ be a topological space.
Let $n \in \N$.
Let:
:$\forall i \in \set {1, 2, \ldots, n}: H_i \subseteq T$
Then:
:$\ds \bigcup_{i \mathop = 1}^n \map \cl {H_i} = \map \cl {\bigcup_{i \mathop = 1}^n H_i}$
where $\map \cl {H_i}$ denotes the closure of $H_i$. | From Closure of Union contains Union of Closures we have that:
:$\ds \bigcup_{i \mathop = 1}^n \map \cl {H_i} \subseteq \map \cl {\bigcup_{i \mathop = 1}^n H_i}$
We need now to show that:
:$\ds \bigcup_{i \mathop = 1}^n \map \cl {H_i} \supseteq \map \cl {\bigcup_{i \mathop = 1}^n H_i}$
Let $\ds K = \bigcup_{i \mathop ... | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $n \in \N$.
Let:
:$\forall i \in \set {1, 2, \ldots, n}: H_i \subseteq T$
Then:
:$\ds \bigcup_{i \mathop = 1}^n \map \cl {H_i} = \map \cl {\bigcup_{i \mathop = 1}^n H_i}$
where $\map \cl {H_i}$ denotes the [[Definition:Closure (Topology)|closure]... | From [[Closure of Union contains Union of Closures]] we have that:
:$\ds \bigcup_{i \mathop = 1}^n \map \cl {H_i} \subseteq \map \cl {\bigcup_{i \mathop = 1}^n H_i}$
We need now to show that:
:$\ds \bigcup_{i \mathop = 1}^n \map \cl {H_i} \supseteq \map \cl {\bigcup_{i \mathop = 1}^n H_i}$
Let $\ds K = \bigcup_{... | Closure of Finite Union equals Union of Closures | https://proofwiki.org/wiki/Closure_of_Finite_Union_equals_Union_of_Closures | https://proofwiki.org/wiki/Closure_of_Finite_Union_equals_Union_of_Closures | [
"Set Union",
"Set Closures"
] | [
"Definition:Topological Space",
"Definition:Closure (Topology)"
] | [
"Closure of Union contains Union of Closures",
"Topological Closure is Closed",
"Definition:Closed Set/Topology",
"Definition:Set Union",
"Definition:Finite Set",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Topology Defined by Closed Sets",
"Set is Subset of its Topological ... |
proofwiki-1843 | Compact First-Countable Space is Sequentially Compact | Let $T = \struct {S, \tau}$ be a compact first-countable topological space.
Then every infinite sequence in $S$ has a convergent subsequence; that is, $T$ is sequentially compact. | Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $S$.
{{AimForCont}} that $\sequence {x_n}_{n \mathop \in \N}$ does not have a convergent subsequence.
By Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit, it follows that $\sequence {x_n}_{n \mathop \in \N}$ does ... | Let $T = \struct {S, \tau}$ be a [[Definition:Compact Topological Space|compact]] [[Definition:First-Countable Space|first-countable]] [[Definition:Topological Space|topological space]].
Then every [[Definition:Infinite Sequence|infinite sequence]] in $S$ has a [[Definition:Convergent Sequence (Topology)|convergent]] ... | Let $\sequence {x_n}_{n \mathop \in \N}$ be an [[Definition:Infinite Sequence|infinite sequence]] in $S$.
{{AimForCont}} that $\sequence {x_n}_{n \mathop \in \N}$ does not have a [[Definition:Convergent Sequence (Topology)|convergent]] [[Definition:Subsequence|subsequence]].
By [[Accumulation Point of Infinite Seque... | Compact First-Countable Space is Sequentially Compact | https://proofwiki.org/wiki/Compact_First-Countable_Space_is_Sequentially_Compact | https://proofwiki.org/wiki/Compact_First-Countable_Space_is_Sequentially_Compact | [
"First-Countable Spaces",
"Compact Topological Spaces",
"Sequentially Compact Spaces"
] | [
"Definition:Compact Topological Space",
"Definition:First-Countable Space",
"Definition:Topological Space",
"Definition:Sequence/Infinite Sequence",
"Definition:Convergent Sequence/Topology",
"Definition:Subsequence",
"Definition:Sequentially Compact Space"
] | [
"Definition:Sequence/Infinite Sequence",
"Definition:Convergent Sequence/Topology",
"Definition:Subsequence",
"Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit",
"Definition:Accumulation Point/Sequence",
"Definition:Open Set/Topology",
"Definition:Finite Set",
"D... |
proofwiki-1844 | Compact First-Countable Space is Sequentially Compact | Let $T = \struct {S, \tau}$ be a compact first-countable topological space.
Then every infinite sequence in $S$ has a convergent subsequence; that is, $T$ is sequentially compact. | Follows directly from:
: Infinite Sequence in Countably Compact Space has Accumulation Point
: Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Compact Topological Space|compact]] [[Definition:First-Countable Space|first-countable]] [[Definition:Topological Space|topological space]].
Then every [[Definition:Infinite Sequence|infinite sequence]] in $S$ has a [[Definition:Convergent Sequence (Topology)|convergent]] ... | Follows directly from:
: [[Infinite Sequence in Countably Compact Space has Accumulation Point]]
: [[Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit]]
{{qed}} | Countably Compact First-Countable Space is Sequentially Compact/Proof 1 | https://proofwiki.org/wiki/Compact_First-Countable_Space_is_Sequentially_Compact | https://proofwiki.org/wiki/Countably_Compact_First-Countable_Space_is_Sequentially_Compact/Proof_1 | [
"First-Countable Spaces",
"Compact Topological Spaces",
"Sequentially Compact Spaces"
] | [
"Definition:Compact Topological Space",
"Definition:First-Countable Space",
"Definition:Topological Space",
"Definition:Sequence/Infinite Sequence",
"Definition:Convergent Sequence/Topology",
"Definition:Subsequence",
"Definition:Sequentially Compact Space"
] | [
"Infinite Sequence in Countably Compact Space has Accumulation Point",
"Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit"
] |
proofwiki-1845 | Compact First-Countable Space is Sequentially Compact | Let $T = \struct {S, \tau}$ be a compact first-countable topological space.
Then every infinite sequence in $S$ has a convergent subsequence; that is, $T$ is sequentially compact. | Let $T = \struct {S, \tau}$ be a countably compact first-countable topological space.
By definition of sequentially compact, it is sufficient to show that every infinite sequence in $S$ has a convergent subsequence.
Let $\sequence {s_n}$ be any sequence in $S$.
By Infinite Sequence in Countably Compact Space has Accumu... | Let $T = \struct {S, \tau}$ be a [[Definition:Compact Topological Space|compact]] [[Definition:First-Countable Space|first-countable]] [[Definition:Topological Space|topological space]].
Then every [[Definition:Infinite Sequence|infinite sequence]] in $S$ has a [[Definition:Convergent Sequence (Topology)|convergent]] ... | Let $T = \struct {S, \tau}$ be a [[Definition:Countably Compact Space|countably compact]] [[Definition:First-Countable Space|first-countable]] [[Definition:Topological Space|topological space]].
By definition of [[Definition:Sequentially Compact Space|sequentially compact]], it is sufficient to show that every [[Defin... | Countably Compact First-Countable Space is Sequentially Compact/Proof 2 | https://proofwiki.org/wiki/Compact_First-Countable_Space_is_Sequentially_Compact | https://proofwiki.org/wiki/Countably_Compact_First-Countable_Space_is_Sequentially_Compact/Proof_2 | [
"First-Countable Spaces",
"Compact Topological Spaces",
"Sequentially Compact Spaces"
] | [
"Definition:Compact Topological Space",
"Definition:First-Countable Space",
"Definition:Topological Space",
"Definition:Sequence/Infinite Sequence",
"Definition:Convergent Sequence/Topology",
"Definition:Subsequence",
"Definition:Sequentially Compact Space"
] | [
"Definition:Countably Compact Space",
"Definition:First-Countable Space",
"Definition:Topological Space",
"Definition:Sequentially Compact Space",
"Definition:Sequence/Infinite Sequence",
"Definition:Convergent Sequence/Topology",
"Definition:Subsequence",
"Definition:Sequence/Infinite Sequence",
"I... |
proofwiki-1846 | Closure of Open Ball in Metric Space | Let $M = \struct {A, d}$ be a metric space.
Let $\map {B_\epsilon} x$ be an open $\epsilon$-ball in $M$.
Let $y \in \map \cl {\map {B_\epsilon} x}$, where $\cl$ denotes the closure of $\map {B_\epsilon} x$.
Then $\map d {x, y} \le \epsilon$. | Suppose $\map d {x, y} > \epsilon$.
Then $\map {B_{\map d {x, y} - \epsilon} } y$ is an open set containing $y$ and not meeting $\map {B_\epsilon} x$.
Hence:
:$y \notin \map \cl {\map {B_\epsilon} x}$
{{qed}} | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Let $\map {B_\epsilon} x$ be an [[Definition:Open Ball of Metric Space|open $\epsilon$-ball]] in $M$.
Let $y \in \map \cl {\map {B_\epsilon} x}$, where $\cl$ denotes the [[Definition:Closure (Metric Space)|closure]] of $\map {B_\epsilon} x$.
Th... | Suppose $\map d {x, y} > \epsilon$.
Then $\map {B_{\map d {x, y} - \epsilon} } y$ is an [[Definition:Open Set (Metric Space)|open set]] containing $y$ and not meeting $\map {B_\epsilon} x$.
Hence:
:$y \notin \map \cl {\map {B_\epsilon} x}$
{{qed}} | Closure of Open Ball in Metric Space | https://proofwiki.org/wiki/Closure_of_Open_Ball_in_Metric_Space | https://proofwiki.org/wiki/Closure_of_Open_Ball_in_Metric_Space | [
"Open Balls"
] | [
"Definition:Metric Space",
"Definition:Open Ball",
"Definition:Closure (Topology)/Metric Space"
] | [
"Definition:Open Set/Metric Space"
] |
proofwiki-1847 | Nowhere Dense iff Complement of Closure is Everywhere Dense | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Then $H$ is nowhere dense in $T$ {{iff}} $S \setminus H^-$ is everywhere dense in $T$, where $H^-$ denotes the closure of $H$.
=== Corollary ===
{{:Nowhere Dense iff Complement of Closure is Everywhere Dense/Corollary}} | Let:
:$H^\circ$ denote the interior of any $H \subseteq S$
:$H^-$ denote the closure of any $H \subseteq S$.
From the definition, $H$ is nowhere dense in $T$ {{iff}} $\paren {H^-}^\circ = \O$.
From the definition of interior, it follows that $\paren {H^-}^\circ = \O$ {{iff}} every open set of $T$ contains a point of $S... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$.
Then $H$ is [[Definition:Nowhere Dense|nowhere dense]] in $T$ {{iff}} $S \setminus H^-$ is [[Definition:Everywhere Dense|everywhere dense]] in $T$, where $H^-$ denotes the [[Definition:Closure (Topology)|closure... | Let:
:$H^\circ$ denote the [[Definition:Interior (Topology)|interior]] of any $H \subseteq S$
:$H^-$ denote the [[Definition:Closure (Topology)|closure]] of any $H \subseteq S$.
From the definition, $H$ is [[Definition:Nowhere Dense|nowhere dense]] in $T$ {{iff}} $\paren {H^-}^\circ = \O$.
From the definition of [[D... | Nowhere Dense iff Complement of Closure is Everywhere Dense | https://proofwiki.org/wiki/Nowhere_Dense_iff_Complement_of_Closure_is_Everywhere_Dense | https://proofwiki.org/wiki/Nowhere_Dense_iff_Complement_of_Closure_is_Everywhere_Dense | [
"Nowhere Dense iff Complement of Closure is Everywhere Dense",
"Set Closures",
"Nowhere Dense"
] | [
"Definition:Topological Space",
"Definition:Nowhere Dense",
"Definition:Everywhere Dense",
"Definition:Closure (Topology)",
"Nowhere Dense iff Complement of Closure is Everywhere Dense/Corollary"
] | [
"Definition:Interior (Topology)",
"Definition:Closure (Topology)",
"Definition:Nowhere Dense",
"Definition:Interior (Topology)",
"Definition:Open Set/Topology",
"Definition:Everywhere Dense"
] |
proofwiki-1848 | Convergence in Indiscrete Space | Let $\struct {S, \set {S, \O} }$ be an indiscrete space.
Let $\sequence {x_n}$ be any sequence in $S$.
Then $\sequence {x_n}$ converges to any point $x$ of $S$. | For any open set $U \subseteq S$ such that $x \in U$, we must have $U = S$.
Hence:
:$\forall n \ge 1: x_n \in U$
The result follows from the definition of a convergent sequence in a topological space.
{{qed}} | Let $\struct {S, \set {S, \O} }$ be an [[Definition:Indiscrete Topology|indiscrete space]].
Let $\sequence {x_n}$ be any [[Definition:Sequence|sequence in $S$]].
Then $\sequence {x_n}$ [[Definition:Convergent Sequence (Topology)|converges]] to any point $x$ of $S$. | For any [[Definition:Open Set (Topology)|open set]] $U \subseteq S$ such that $x \in U$, we must have $U = S$.
Hence:
:$\forall n \ge 1: x_n \in U$
The result follows from the definition of a [[Definition:Convergent Sequence (Topology)|convergent sequence in a topological space]].
{{qed}} | Convergence in Indiscrete Space | https://proofwiki.org/wiki/Convergence_in_Indiscrete_Space | https://proofwiki.org/wiki/Convergence_in_Indiscrete_Space | [
"Indiscrete Topology",
"Convergence"
] | [
"Definition:Indiscrete Topology",
"Definition:Sequence",
"Definition:Convergent Sequence/Topology"
] | [
"Definition:Open Set/Topology",
"Definition:Convergent Sequence/Topology"
] |
proofwiki-1849 | Totally Bounded Metric Space is Separable | A totally bounded metric space is separable. | Let $M = \struct {A, d}$ be a totally bounded metric space.
By the definition of total boundedness, we can use the axiom of countable choice to construct a sequence $\sequence {F_n}_{n \mathop \ge 1}$ such that:
:For all natural numbers $n \ge 1$, $F_n$ is a finite $\paren {1/n}$-net for $M$.
Let $\ds S = \bigcup_{n \m... | A [[Definition:Totally Bounded Metric Space|totally bounded metric space]] is [[Definition:Separable Space|separable]]. | Let $M = \struct {A, d}$ be a [[Definition:Totally Bounded Metric Space|totally bounded metric space]].
By the definition of [[Definition:Totally Bounded Metric Space|total boundedness]], we can use the [[Axiom:Axiom of Countable Choice|axiom of countable choice]] to construct a [[Definition:Sequence|sequence]] $\seq... | Totally Bounded Metric Space is Separable | https://proofwiki.org/wiki/Totally_Bounded_Metric_Space_is_Separable | https://proofwiki.org/wiki/Totally_Bounded_Metric_Space_is_Separable | [
"Separable Spaces",
"Totally Bounded Metric Spaces"
] | [
"Definition:Totally Bounded Metric Space",
"Definition:Separable Space"
] | [
"Definition:Totally Bounded Metric Space",
"Definition:Totally Bounded Metric Space",
"Axiom:Axiom of Countable Choice",
"Definition:Sequence",
"Definition:Natural Numbers",
"Definition:Epsilon-Net/Finite Net",
"Countable Union of Countable Sets is Countable",
"Definition:Countable Set",
"Definition... |
proofwiki-1850 | Convergent Sequence in T2 Space has Unique Limit | Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Let $\sequence {x_n}$ be a convergent sequence in $T$.
Then $\sequence {x_n}$ has exactly one limit. | From the definition of convergent sequence, we have that $\sequence {x_n}$ converges to at least one limit.
{{AimForCont}} $l$ and $m$ are both limits of $\sequence{x_n}$.
As $T$ is a $T_2$ (Hausdorff) space, there exist $U, V \in \tau$ such that:
:$l \in U, m \in V$ and $U \cap V = \O$
Then, from the definition of con... | Let $T = \struct {S, \tau}$ be a [[Definition:T2 Space|$T_2$ (Hausdorff) space]].
Let $\sequence {x_n}$ be a [[Definition:Convergent Sequence (Topology)|convergent sequence]] in $T$.
Then $\sequence {x_n}$ has exactly one [[Definition:Limit of Sequence (Topology)|limit]]. | From the definition of [[Definition:Convergent Sequence (Topology)|convergent sequence]], we have that $\sequence {x_n}$ converges to at least one [[Definition:Limit of Sequence (Topology)|limit]].
{{AimForCont}} $l$ and $m$ are both [[Definition:Limit of Sequence (Topology)|limits]] of $\sequence{x_n}$.
As $T$ is a... | Convergent Sequence in T2 Space has Unique Limit | https://proofwiki.org/wiki/Convergent_Sequence_in_T2_Space_has_Unique_Limit | https://proofwiki.org/wiki/Convergent_Sequence_in_T2_Space_has_Unique_Limit | [
"Limits of Sequences",
"Hausdorff Spaces"
] | [
"Definition:T2 Space",
"Definition:Convergent Sequence/Topology",
"Definition:Limit of Sequence/Topological Space"
] | [
"Definition:Convergent Sequence/Topology",
"Definition:Limit of Sequence/Topological Space",
"Definition:Limit of Sequence/Topological Space",
"Definition:T2 Space",
"Definition:Convergent Sequence/Topology",
"Definition:Contradiction"
] |
proofwiki-1851 | Metric Space is T2 | A metric space $M = \struct {A, d}$ is a $T_2$ (Hausdorff) space. | Let $x, y \in A: x \ne y$.
Then from Distinct Points in Metric Space have Disjoint Open Balls, there exist open $\epsilon$-balls $\map {B_\epsilon} x$ and $\map {B_\epsilon} y$ which are disjoint open sets containing $x$ and $y$ respectively.
Hence the result by the definition of $T_2$ (Hausdorff) space.
{{qed}} | A [[Definition:Metric Space|metric space]] $M = \struct {A, d}$ is a [[Definition:T2 Space|$T_2$ (Hausdorff) space]]. | Let $x, y \in A: x \ne y$.
Then from [[Distinct Points in Metric Space have Disjoint Open Balls]], there exist [[Definition:Open Ball of Metric Space|open $\epsilon$-balls]] $\map {B_\epsilon} x$ and $\map {B_\epsilon} y$ which are [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set of Metric Space|open sets]]... | Metric Space is T2/Proof 1 | https://proofwiki.org/wiki/Metric_Space_is_T2 | https://proofwiki.org/wiki/Metric_Space_is_T2/Proof_1 | [
"Metric Space is T2",
"Metric Space fulfils all Separation Axioms",
"Examples of Hausdorff Spaces"
] | [
"Definition:Metric Space",
"Definition:T2 Space"
] | [
"Distinct Points in Metric Space have Disjoint Open Balls",
"Definition:Open Ball",
"Definition:Disjoint Sets",
"Definition:Open Set/Metric Space",
"Definition:T2 Space"
] |
proofwiki-1852 | Metric Space is T2 | A metric space $M = \struct {A, d}$ is a $T_2$ (Hausdorff) space. | {{AimForCont}} $M$ is not Hausdorff.
That is, there are $x, y \in A: x \ne y$ such that:
:$\forall \epsilon \in \R_{>0}: \exists z \in \map {B_\epsilon} x \cap \map {B_\epsilon} y$
where $\map {B_\epsilon} x$ denote the open $\epsilon$-ball of $x$ in $M$.
Let $r = \dfrac {\map d {x, y} } 2$.
Let $z \in \map {B_r} x \ca... | A [[Definition:Metric Space|metric space]] $M = \struct {A, d}$ is a [[Definition:T2 Space|$T_2$ (Hausdorff) space]]. | {{AimForCont}} $M$ is not [[Definition:T2 Space|Hausdorff]].
That is, there are $x, y \in A: x \ne y$ such that:
:$\forall \epsilon \in \R_{>0}: \exists z \in \map {B_\epsilon} x \cap \map {B_\epsilon} y$
where $\map {B_\epsilon} x$ denote the [[Definition:Open Ball|open $\epsilon$-ball]] of $x$ in $M$.
Let $r = \... | Metric Space is T2/Proof 2 | https://proofwiki.org/wiki/Metric_Space_is_T2 | https://proofwiki.org/wiki/Metric_Space_is_T2/Proof_2 | [
"Metric Space is T2",
"Metric Space fulfils all Separation Axioms",
"Examples of Hausdorff Spaces"
] | [
"Definition:Metric Space",
"Definition:T2 Space"
] | [
"Definition:T2 Space",
"Definition:Open Ball",
"Definition:Contradiction",
"Definition:T2 Space"
] |
proofwiki-1853 | Convergent Real Sequence has Unique Limit | Let $\sequence {s_n}$ be a real sequence.
Then $\sequence {s_n}$ can have at most one limit. | {{AimForCont}} that $\sequence {s_n}$ converges to $l$ and also to $m$.
That is, suppose that:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
and:
:$\ds \lim_{n \mathop \to \infty} x_n = m$
{{WLOG}}, assume that $l \ne m$.
Let:
:$\epsilon = \dfrac {\size {l - m} } 2$
As $l \ne m$, it follows that $\epsilon > 0$.
As $\seque... | Let $\sequence {s_n}$ be a [[Definition:Real Sequence|real sequence]].
Then $\sequence {s_n}$ can have at most one [[Definition:Limit of Real Sequence|limit]]. | {{AimForCont}} that $\sequence {s_n}$ [[Definition:Convergent Real Sequence|converges]] to $l$ and also to $m$.
That is, suppose that:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
and:
:$\ds \lim_{n \mathop \to \infty} x_n = m$
{{WLOG}}, assume that $l \ne m$.
Let:
:$\epsilon = \dfrac {\size {l - m} } 2$
As $l \ne m... | Convergent Real Sequence has Unique Limit/Proof 1 | https://proofwiki.org/wiki/Convergent_Real_Sequence_has_Unique_Limit | https://proofwiki.org/wiki/Convergent_Real_Sequence_has_Unique_Limit/Proof_1 | [
"Limits of Sequences",
"Real Numbers",
"Convergent Real Sequence has Unique Limit"
] | [
"Definition:Real Sequence",
"Definition:Limit of Sequence/Real Numbers"
] | [
"Definition:Convergent Sequence/Real Numbers",
"Triangle Inequality/Real Numbers",
"Definition:Contradiction",
"Proof by Contradiction"
] |
proofwiki-1854 | Convergent Real Sequence has Unique Limit | Let $\sequence {s_n}$ be a real sequence.
Then $\sequence {s_n}$ can have at most one limit. | We have that the real number line is a metric space
The result then follows from Convergent Sequence in Metric Space has Unique Limit.
{{qed}} | Let $\sequence {s_n}$ be a [[Definition:Real Sequence|real sequence]].
Then $\sequence {s_n}$ can have at most one [[Definition:Limit of Real Sequence|limit]]. | We have that the [[Real Number Line is Metric Space|real number line is a metric space]]
The result then follows from [[Convergent Sequence in Metric Space has Unique Limit]].
{{qed}} | Convergent Real Sequence has Unique Limit/Proof 2 | https://proofwiki.org/wiki/Convergent_Real_Sequence_has_Unique_Limit | https://proofwiki.org/wiki/Convergent_Real_Sequence_has_Unique_Limit/Proof_2 | [
"Limits of Sequences",
"Real Numbers",
"Convergent Real Sequence has Unique Limit"
] | [
"Definition:Real Sequence",
"Definition:Limit of Sequence/Real Numbers"
] | [
"Real Number Line is Metric Space",
"Convergent Sequence in Metric Space has Unique Limit"
] |
proofwiki-1855 | Distinct Points in Metric Space have Disjoint Open Balls | Let $M = \struct {A, d}$ be a metric space.
Let $x, y \in M: x \ne y$.
Then there exist disjoint open $\epsilon$-balls $\map {B_\epsilon} x$ and $\map {B_\epsilon} y$ containing $x$ and $y$ respectively. | Let $x, y \in A: x \ne y$.
Then $\map d {x, y} > 0$.
Put $\epsilon = \dfrac {\map d {x, y} } 2$.
Let $\map {B_\epsilon} x$ and $\map {B_\epsilon} y$ be the open $\epsilon$-balls of $x$ and $y$ respectively.
Suppose $\map {B_\epsilon} x$ and $\map {B_\epsilon} y$ are not disjoint.
Then $\exists z \in M$ such that $z \in... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Let $x, y \in M: x \ne y$.
Then there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Ball of Metric Space|open $\epsilon$-balls]] $\map {B_\epsilon} x$ and $\map {B_\epsilon} y$ containing $x$ and $y$ respectively. | Let $x, y \in A: x \ne y$.
Then $\map d {x, y} > 0$.
Put $\epsilon = \dfrac {\map d {x, y} } 2$.
Let $\map {B_\epsilon} x$ and $\map {B_\epsilon} y$ be the [[Definition:Open Ball of Metric Space|open $\epsilon$-balls]] of $x$ and $y$ respectively.
Suppose $\map {B_\epsilon} x$ and $\map {B_\epsilon} y$ are not [[D... | Distinct Points in Metric Space have Disjoint Open Balls/Proof 2 | https://proofwiki.org/wiki/Distinct_Points_in_Metric_Space_have_Disjoint_Open_Balls | https://proofwiki.org/wiki/Distinct_Points_in_Metric_Space_have_Disjoint_Open_Balls/Proof_2 | [
"Open Balls",
"Distinct Points in Metric Space have Disjoint Open Balls"
] | [
"Definition:Metric Space",
"Definition:Disjoint Sets",
"Definition:Open Ball"
] | [
"Definition:Open Ball",
"Definition:Disjoint Sets",
"Definition:Metric Space/Metric",
"Definition:Disjoint Sets"
] |
proofwiki-1856 | Sequentially Compact Metric Space is Lindelöf | Let $M = \struct {X, d}$ be a sequentially compact metric space.
Then $M$ is also a Lindelöf space.
That is, from every open cover of $M$, it is possible to extract a countable subcover. | Take any open cover $C$ of $M$.
We need to find a countable subset of $C$ which still covers $X$.
We have that a Sequentially Compact Metric Space is Second-Countable.
Thus, by definition, the topology of $M$ has a countable basis.
Let $\BB$ be a countable basis for the topology induced by $d$ of $M$.
Let $x \in X$.
As... | Let $M = \struct {X, d}$ be a [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Metric Space|metric space]].
Then $M$ is also a [[Definition:Lindelöf Space|Lindelöf space]].
That is, from every [[Definition:Open Cover|open cover]] of $M$, it is possible to extract a [[Definition:Countable ... | Take any [[Definition:Open Cover|open cover]] $C$ of $M$.
We need to find a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $C$ which still [[Definition:Cover of Set|covers]] $X$.
We have that a [[Sequentially Compact Metric Space is Second-Countable]].
Thus, by definition, the [[Definition:T... | Sequentially Compact Metric Space is Lindelöf | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Lindelöf | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Lindelöf | [
"Lindelöf Spaces",
"Metric Spaces",
"Sequentially Compact Spaces"
] | [
"Definition:Sequentially Compact Space",
"Definition:Metric Space",
"Definition:Lindelöf Space",
"Definition:Open Cover",
"Definition:Subcover/Countable"
] | [
"Definition:Open Cover",
"Definition:Countable Set",
"Definition:Subset",
"Definition:Cover of Set",
"Sequentially Compact Metric Space is Second-Countable",
"Definition:Topology Induced by Metric",
"Definition:Countable Basis",
"Definition:Countable Basis",
"Definition:Topology Induced by Metric",
... |
proofwiki-1857 | Sequentially Compact Metric Space is Compact | A sequentially compact metric space is compact. | Let $\struct {A, d}$ be a sequentially compact metric space.
From Sequentially Compact Metric Space is Lindelöf, every open cover of $A$ has a countable subcover.
Let $C$ be an open cover $A$.
Extract from it a countable subcover $\set {U_1, U_2, \ldots}$.
{{AimForCont}} there exists no finite subcover of $C$.
Then, fo... | A [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Metric Space|metric space]] is [[Definition:Compact Metric Space|compact]]. | Let $\struct {A, d}$ be a [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Metric Space|metric space]].
From [[Sequentially Compact Metric Space is Lindelöf]], every [[Definition:Open Cover|open cover]] of $A$ has a [[Definition:Countable Subcover|countable subcover]].
Let $C$ be an [[Defin... | Sequentially Compact Metric Space is Compact/Proof 1 | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Compact | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Compact/Proof_1 | [
"Sequentially Compact Metric Space is Compact",
"Sequentially Compact Spaces",
"Compact Metric Spaces",
"Metric Spaces"
] | [
"Definition:Sequentially Compact Space",
"Definition:Metric Space",
"Definition:Compact Space/Metric Space"
] | [
"Definition:Sequentially Compact Space",
"Definition:Metric Space",
"Sequentially Compact Metric Space is Lindelöf",
"Definition:Open Cover",
"Definition:Subcover/Countable",
"Definition:Open Cover",
"Definition:Subcover/Countable",
"Definition:Subcover/Finite",
"Definition:Set",
"Definition:Cover... |
proofwiki-1858 | Sequentially Compact Metric Space is Compact | A sequentially compact metric space is compact. | Let $M = \struct {A, d}$ be a sequentially compact metric space.
Let $\UU$ be any open cover of $M$.
By Lebesgue's Number Lemma, there exists a Lebesgue number for $\UU$.
Let $\set {x_1, x_2, \ldots, x_n}$ be a finite $\epsilon$-net for $M$, where $\epsilon$ is this same Lebesgue number.
This exists by Sequentially Com... | A [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Metric Space|metric space]] is [[Definition:Compact Metric Space|compact]]. | Let $M = \struct {A, d}$ be a [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Metric Space|metric space]].
Let $\UU$ be any [[Definition:Open Cover|open cover]] of $M$.
By [[Lebesgue's Number Lemma/Sequentially Compact Space|Lebesgue's Number Lemma]], there exists a [[Definition:Lebesgue N... | Sequentially Compact Metric Space is Compact/Proof 2 | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Compact | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Compact/Proof_2 | [
"Sequentially Compact Metric Space is Compact",
"Sequentially Compact Spaces",
"Compact Metric Spaces",
"Metric Spaces"
] | [
"Definition:Sequentially Compact Space",
"Definition:Metric Space",
"Definition:Compact Space/Metric Space"
] | [
"Definition:Sequentially Compact Space",
"Definition:Metric Space",
"Definition:Open Cover",
"Lebesgue's Number Lemma/Sequentially Compact Space",
"Definition:Lebesgue Number",
"Definition:Epsilon-Net/Finite Net",
"Definition:Lebesgue Number",
"Sequentially Compact Metric Space is Totally Bounded",
... |
proofwiki-1859 | Sequentially Compact Metric Space is Compact | A sequentially compact metric space is compact. | Follows directly from:
:Sequentially Compact Space is Countably Compact
:Countably Compact Metric Space is Compact
{{qed}} | A [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Metric Space|metric space]] is [[Definition:Compact Metric Space|compact]]. | Follows directly from:
:[[Sequentially Compact Space is Countably Compact]]
:[[Countably Compact Metric Space is Compact]]
{{qed}} | Sequentially Compact Metric Space is Compact/Proof 3 | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Compact | https://proofwiki.org/wiki/Sequentially_Compact_Metric_Space_is_Compact/Proof_3 | [
"Sequentially Compact Metric Space is Compact",
"Sequentially Compact Spaces",
"Compact Metric Spaces",
"Metric Spaces"
] | [
"Definition:Sequentially Compact Space",
"Definition:Metric Space",
"Definition:Compact Space/Metric Space"
] | [
"Sequentially Compact Space is Countably Compact",
"Countably Compact Metric Space is Compact"
] |
proofwiki-1860 | Compact Subspace of Hausdorff Space is Closed | Let $H = \struct {A, \tau}$ be a $T_2$ (Hausdorff) space.
Let $C$ be a compact subspace of $H$.
Then $C$ is closed in $H$. | From Subspace of Hausdorff Space is Hausdorff, a subspace of a $T_2$ (Hausdorff) space is itself $T_2$ (Hausdorff).
Let $a \in A \setminus C$.
We are going to prove that there exists an open set $U_a$ such that $a \in U_a \subseteq A \setminus C$.
For any single point $x \in C$, the Hausdorff condition ensures the exis... | Let $H = \struct {A, \tau}$ be a [[Definition:T2 Space|$T_2$ (Hausdorff) space]].
Let $C$ be a [[Definition:Compact Topological Subspace|compact subspace]] of $H$.
Then $C$ is [[Definition:Closed Set (Topology)|closed]] in $H$. | From [[Subspace of Hausdorff Space is Hausdorff]], a [[Definition:Topological Subspace|subspace]] of a [[Definition:T2 Space|$T_2$ (Hausdorff) space]] is itself [[Definition:T2 Space|$T_2$ (Hausdorff)]].
Let $a \in A \setminus C$.
We are going to prove that there exists an [[Definition:Open Set (Topology)|open set]] ... | Compact Subspace of Hausdorff Space is Closed/Proof 1 | https://proofwiki.org/wiki/Compact_Subspace_of_Hausdorff_Space_is_Closed | https://proofwiki.org/wiki/Compact_Subspace_of_Hausdorff_Space_is_Closed/Proof_1 | [
"Compact Subspace of Hausdorff Space is Closed",
"Hausdorff Spaces",
"Compact Topological Spaces",
"Closed Sets",
"Topological Subspaces"
] | [
"Definition:T2 Space",
"Definition:Compact Topological Space/Subspace",
"Definition:Closed Set/Topology"
] | [
"T2 Property is Hereditary",
"Definition:Topological Subspace",
"Definition:T2 Space",
"Definition:T2 Space",
"Definition:Open Set/Topology",
"Definition:T2 Space",
"Definition:Disjoint Sets",
"Definition:Open Set/Topology",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Open Cove... |
proofwiki-1861 | Compact Subspace of Hausdorff Space is Closed | Let $H = \struct {A, \tau}$ be a $T_2$ (Hausdorff) space.
Let $C$ be a compact subspace of $H$.
Then $C$ is closed in $H$. | For a subset $S \subseteq A$, let $S^{\complement}$ denote the relative complement of $S$ in $A$.
Consider an arbitrary point $x \in C^{\complement}$.
Define the set:
:$\ds \OO = \set {V \in \tau: \exists U \in \tau: x \in U \subseteq V^\complement}$
By Empty Intersection iff Subset of Complement:
:$U \subseteq V^\comp... | Let $H = \struct {A, \tau}$ be a [[Definition:T2 Space|$T_2$ (Hausdorff) space]].
Let $C$ be a [[Definition:Compact Topological Subspace|compact subspace]] of $H$.
Then $C$ is [[Definition:Closed Set (Topology)|closed]] in $H$. | For a [[Definition:Subset|subset]] $S \subseteq A$, let $S^{\complement}$ denote the [[Definition:Relative Complement|relative complement]] of $S$ in $A$.
Consider an arbitrary [[Definition:Element|point]] $x \in C^{\complement}$.
Define the [[Definition:Set|set]]:
:$\ds \OO = \set {V \in \tau: \exists U \in \tau: x... | Compact Subspace of Hausdorff Space is Closed/Proof 2 | https://proofwiki.org/wiki/Compact_Subspace_of_Hausdorff_Space_is_Closed | https://proofwiki.org/wiki/Compact_Subspace_of_Hausdorff_Space_is_Closed/Proof_2 | [
"Compact Subspace of Hausdorff Space is Closed",
"Hausdorff Spaces",
"Compact Topological Spaces",
"Closed Sets",
"Topological Subspaces"
] | [
"Definition:T2 Space",
"Definition:Compact Topological Space/Subspace",
"Definition:Closed Set/Topology"
] | [
"Definition:Subset",
"Definition:Relative Complement",
"Definition:Element",
"Definition:Set",
"Empty Intersection iff Subset of Complement",
"Definition:T2 Space",
"Definition:Open Cover",
"Definition:Compact Topological Space/Subspace",
"Definition:Subcover/Finite",
"Principle of Finite Choice",... |
proofwiki-1862 | Continuous Image of Compact Space is Compact | Let $T_1$ and $T_2$ be topological spaces.
Let $f: T_1 \to T_2$ be a continuous mapping.
If $T_1$ is compact then so is its image $f \sqbrk {T_1}$ under $f$.
That is, compactness is a continuous invariant. | By {{Corollary|Continuous Image of Compact Space is Compact|2}}, $f \sqbrk S$ is bounded.
By Supremum of Bounded Above Set of Reals is in Closure:
:$\map \sup {f \sqbrk S} \in \map \cl {f \sqbrk S}$
and by Infimum of Bounded Below Set of Reals is in Closure:
:$\map \inf {f \sqbrk S} \in \map \cl {f \sqbrk S}$
From Cont... | Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
Let $f: T_1 \to T_2$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
If $T_1$ is [[Definition:Compact Topological Space|compact]] then so is its [[Definition:Image of Mapping|image]] $f \sqbrk {T_1}$ under $f... | By {{Corollary|Continuous Image of Compact Space is Compact|2}}, $f \sqbrk S$ is [[Definition:Bounded Mapping to Metric Space|bounded]].
By [[Supremum of Bounded Above Set of Reals is in Closure]]:
:$\map \sup {f \sqbrk S} \in \map \cl {f \sqbrk S}$
and by [[Infimum of Bounded Below Set of Reals is in Closure]]:
:$\ma... | Continuous Image of Compact Space is Compact/Corollary 3/Proof 1 | https://proofwiki.org/wiki/Continuous_Image_of_Compact_Space_is_Compact | https://proofwiki.org/wiki/Continuous_Image_of_Compact_Space_is_Compact/Corollary_3/Proof_1 | [
"Continuous Image of Compact Space is Compact",
"Compact Topological Spaces",
"Examples of Continuous Invariants"
] | [
"Definition:Topological Space",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Compact Topological Space",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Compact Topological Space",
"Definition:Continuous Invariant"
] | [
"Definition:Bounded Mapping/Metric Space",
"Supremum of Bounded Above Set of Reals is in Closure",
"Infimum of Bounded Below Set of Reals is in Closure",
"Continuous Image of Compact Space is Compact",
"Definition:Compact Topological Space/Subspace",
"Non-Closed Set of Real Numbers is not Compact",
"Rul... |
proofwiki-1863 | Continuous Image of Compact Space is Compact | Let $T_1$ and $T_2$ be topological spaces.
Let $f: T_1 \to T_2$ be a continuous mapping.
If $T_1$ is compact then so is its image $f \sqbrk {T_1}$ under $f$.
That is, compactness is a continuous invariant. | By Continuous Image of Compact Space is Compact, $f \sqbrk S$ is compact.
From Compact Metric Space is Complete and Compact Metric Space is Totally Bounded, $f \sqbrk S$ is complete and totally bounded.
A Totally Bounded Metric Space is Bounded.
Hence both the supremum and the infimum of $f \sqbrk S$ exist in $\R$.
Bec... | Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
Let $f: T_1 \to T_2$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
If $T_1$ is [[Definition:Compact Topological Space|compact]] then so is its [[Definition:Image of Mapping|image]] $f \sqbrk {T_1}$ under $f... | By [[Continuous Image of Compact Space is Compact]], $f \sqbrk S$ is [[Definition:Compact Metric Space|compact]].
From [[Compact Metric Space is Complete]] and [[Compact Metric Space is Totally Bounded]], $f \sqbrk S$ is [[Definition:Complete Metric Space|complete]] and [[Definition:Totally Bounded Metric Space|totall... | Continuous Image of Compact Space is Compact/Corollary 3/Proof 2 | https://proofwiki.org/wiki/Continuous_Image_of_Compact_Space_is_Compact | https://proofwiki.org/wiki/Continuous_Image_of_Compact_Space_is_Compact/Corollary_3/Proof_2 | [
"Continuous Image of Compact Space is Compact",
"Compact Topological Spaces",
"Examples of Continuous Invariants"
] | [
"Definition:Topological Space",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Compact Topological Space",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Compact Topological Space",
"Definition:Continuous Invariant"
] | [
"Continuous Image of Compact Space is Compact",
"Definition:Compact Space/Metric Space",
"Compact Metric Space is Complete",
"Compact Metric Space is Totally Bounded",
"Definition:Complete Metric Space",
"Definition:Totally Bounded Metric Space",
"Totally Bounded Metric Space is Bounded",
"Definition:... |
proofwiki-1864 | Continuous Image of Compact Space is Compact | Let $T_1$ and $T_2$ be topological spaces.
Let $f: T_1 \to T_2$ be a continuous mapping.
If $T_1$ is compact then so is its image $f \sqbrk {T_1}$ under $f$.
That is, compactness is a continuous invariant. | From Restriction of Mapping to Image is Surjection, the restriction of $f$ to $\Img f$ is a surjection.
The result follows from Compactness is Preserved under Continuous Surjection.
{{qed}} | Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
Let $f: T_1 \to T_2$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
If $T_1$ is [[Definition:Compact Topological Space|compact]] then so is its [[Definition:Image of Mapping|image]] $f \sqbrk {T_1}$ under $f... | From [[Restriction of Mapping to Image is Surjection]], the [[Definition:Restriction of Mapping|restriction]] of $f$ to $\Img f$ is a [[Definition:Surjection|surjection]].
The result follows from [[Compactness is Preserved under Continuous Surjection]].
{{qed}} | Continuous Image of Compact Space is Compact/Proof 1 | https://proofwiki.org/wiki/Continuous_Image_of_Compact_Space_is_Compact | https://proofwiki.org/wiki/Continuous_Image_of_Compact_Space_is_Compact/Proof_1 | [
"Continuous Image of Compact Space is Compact",
"Compact Topological Spaces",
"Examples of Continuous Invariants"
] | [
"Definition:Topological Space",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Compact Topological Space",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Compact Topological Space",
"Definition:Continuous Invariant"
] | [
"Restriction of Mapping to Image is Surjection",
"Definition:Restriction/Mapping",
"Definition:Surjection",
"Compactness is Preserved under Continuous Surjection"
] |
proofwiki-1865 | Continuous Image of Compact Space is Compact | Let $T_1$ and $T_2$ be topological spaces.
Let $f: T_1 \to T_2$ be a continuous mapping.
If $T_1$ is compact then so is its image $f \sqbrk {T_1}$ under $f$.
That is, compactness is a continuous invariant. | Suppose $\UU$ is an open cover of $f \sqbrk {T_1}$ by sets open in $T_2$.
Because $f$ is continuous, it follows that $f^{-1} \sqbrk U$ is open in $T_1$ for all $U \in \UU$.
The set $\set {f^{-1} \sqbrk U: U \in \UU}$ is an open cover of $T_1$, because for any $x \in T_1$, it follows that $\map f x$ must be in some $U \... | Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
Let $f: T_1 \to T_2$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
If $T_1$ is [[Definition:Compact Topological Space|compact]] then so is its [[Definition:Image of Mapping|image]] $f \sqbrk {T_1}$ under $f... | Suppose $\UU$ is an [[Definition:Open Cover|open cover]] of $f \sqbrk {T_1}$ by sets [[Definition:Open Set (Topology)|open in $T_2$]].
Because $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]], it follows that $f^{-1} \sqbrk U$ is [[Definition:Open Set (Topology)|open in $T_1$]] for all $U \in ... | Continuous Image of Compact Space is Compact/Proof 2 | https://proofwiki.org/wiki/Continuous_Image_of_Compact_Space_is_Compact | https://proofwiki.org/wiki/Continuous_Image_of_Compact_Space_is_Compact/Proof_2 | [
"Continuous Image of Compact Space is Compact",
"Compact Topological Spaces",
"Examples of Continuous Invariants"
] | [
"Definition:Topological Space",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Compact Topological Space",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Compact Topological Space",
"Definition:Continuous Invariant"
] | [
"Definition:Open Cover",
"Definition:Open Set/Topology",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Open Set/Topology",
"Definition:Open Cover",
"Definition:Compact Topological Space",
"Definition:Subcover/Finite",
"Definition:Subcover/Finite"
] |
proofwiki-1866 | Closed Subspace of Compact Space is Compact | Let $T = \struct {S, \tau}$ be a compact topological space.
Let $C = \struct {H, \tau_H}$ be a subspace of $T$.
Let $C$ be closed in $T$.
Then $\struct {H, \tau}$ is compact.
That is, the property of being compact is weakly hereditary. | Let $T$ be a compact space.
Let $C$ be a closed subspace of $T$.
Let $\UU$ be an open cover of $C$.
We have that $C$ is closed in $T$.
It follows by definition of closed that $T \setminus C$ is open in $T$.
So if we add $T \setminus C$ to $\UU$, we see that $\UU \cup \set {T \setminus C}$ is also an open cover of $T$.
... | Let $T = \struct {S, \tau}$ be a [[Definition:Compact Topological Space|compact topological space]].
Let $C = \struct {H, \tau_H}$ be a [[Definition:Topological Subspace|subspace]] of $T$.
Let $C$ be [[Definition:Closed Set (Topology)|closed]] in $T$.
Then $\struct {H, \tau}$ is [[Definition:Compact Topological Spa... | Let $T$ be a [[Definition:Compact Topological Space|compact space]].
Let $C$ be a [[Definition:Closed Set (Topology)|closed]] [[Definition:Topological Subspace|subspace]] of $T$.
Let $\UU$ be an [[Definition:Open Cover|open cover]] of $C$.
We have that $C$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
It ... | Closed Subspace of Compact Space is Compact | https://proofwiki.org/wiki/Closed_Subspace_of_Compact_Space_is_Compact | https://proofwiki.org/wiki/Closed_Subspace_of_Compact_Space_is_Compact | [
"Closed Sets",
"Examples of Compact Topological Spaces",
"Examples of Weakly Hereditary Properties"
] | [
"Definition:Compact Topological Space",
"Definition:Topological Subspace",
"Definition:Closed Set/Topology",
"Definition:Compact Topological Space",
"Definition:Property",
"Definition:Compact Topological Space",
"Definition:Weakly Hereditary Property"
] | [
"Definition:Compact Topological Space",
"Definition:Closed Set/Topology",
"Definition:Topological Subspace",
"Definition:Open Cover",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Definition:Open Cover",
"Definition:Compact Topological Space",
... |
proofwiki-1867 | Closure of Real Interval is Closed Real Interval | Let $I$ be a non-empty real interval such that one of these holds:
:$I = \openint a b$
:$I = \hointr a b$
:$I = \hointl a b$
:$I = \closedint a b$
Let $I^-$ denote the closure of $I$.
Then $I^-$ is the closed real interval $\closedint a b$. | There are four cases to cover:
:$(1): \quad$ Let $I = \openint a b$.
From Closure of Open Real Interval is Closed Real Interval:
:$I^- = \closedint a b$
{{qed|lemma}}
:$(2): \quad$ Let $I = \hointr a b$.
From Closure of Half-Open Real Interval is Closed Real Interval:
:$I^- = \closedint a b$
{{qed|lemma}}
:$(3): \quad$... | Let $I$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Real Interval|real interval]] such that one of these holds:
:$I = \openint a b$
:$I = \hointr a b$
:$I = \hointl a b$
:$I = \closedint a b$
Let $I^-$ denote the [[Definition:Closure (Topology)|closure]] of $I$.
Then $I^-$ is the [[Definition:Closed Rea... | There are four cases to cover:
:$(1): \quad$ Let $I = \openint a b$.
From [[Closure of Open Real Interval is Closed Real Interval]]:
:$I^- = \closedint a b$
{{qed|lemma}}
:$(2): \quad$ Let $I = \hointr a b$.
From [[Closure of Half-Open Real Interval is Closed Real Interval]]:
:$I^- = \closedint a b$
{{qed|lemma}}
... | Closure of Real Interval is Closed Real Interval/Proof 1 | https://proofwiki.org/wiki/Closure_of_Real_Interval_is_Closed_Real_Interval | https://proofwiki.org/wiki/Closure_of_Real_Interval_is_Closed_Real_Interval/Proof_1 | [
"Closure of Real Interval is Closed Real Interval",
"Real Intervals",
"Set Closures"
] | [
"Definition:Non-Empty Set",
"Definition:Real Interval",
"Definition:Closure (Topology)",
"Definition:Real Interval/Closed"
] | [
"Closure of Open Real Interval is Closed Real Interval",
"Closure of Half-Open Real Interval is Closed Real Interval",
"Closure of Half-Open Real Interval is Closed Real Interval",
"Closed Real Interval is Closed in Real Number Line",
"Definition:Closed Set/Topology",
"Set is Closed iff Equals Topological... |
proofwiki-1868 | Closure of Real Interval is Closed Real Interval | Let $I$ be a non-empty real interval such that one of these holds:
:$I = \openint a b$
:$I = \hointr a b$
:$I = \hointl a b$
:$I = \closedint a b$
Let $I^-$ denote the closure of $I$.
Then $I^-$ is the closed real interval $\closedint a b$. | Let $I$ be one of the intervals as specified in the exposition.
Note that:
:$(1): \quad$ By Condition for Point being in Closure, $x \in I^-$ {{iff}} every open set in $\R$ containing $x$ contains a point in $I$.
:$(2): \quad$ From Union of Open Sets of Metric Space is Open, every open set in $\R$ is a union of open in... | Let $I$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Real Interval|real interval]] such that one of these holds:
:$I = \openint a b$
:$I = \hointr a b$
:$I = \hointl a b$
:$I = \closedint a b$
Let $I^-$ denote the [[Definition:Closure (Topology)|closure]] of $I$.
Then $I^-$ is the [[Definition:Closed Rea... | Let $I$ be one of the intervals as specified in the exposition.
Note that:
:$(1): \quad$ By [[Condition for Point being in Closure]], $x \in I^-$ {{iff}} every [[Definition:Open Set (Real Analysis)|open set]] in $\R$ containing $x$ contains a point in $I$.
:$(2): \quad$ From [[Union of Open Sets of Metric Space is O... | Closure of Real Interval is Closed Real Interval/Proof 2 | https://proofwiki.org/wiki/Closure_of_Real_Interval_is_Closed_Real_Interval | https://proofwiki.org/wiki/Closure_of_Real_Interval_is_Closed_Real_Interval/Proof_2 | [
"Closure of Real Interval is Closed Real Interval",
"Real Intervals",
"Set Closures"
] | [
"Definition:Non-Empty Set",
"Definition:Real Interval",
"Definition:Closure (Topology)",
"Definition:Real Interval/Closed"
] | [
"Condition for Point being in Closure",
"Definition:Open Set/Real Analysis",
"Union of Open Sets of Metric Space is Open",
"Definition:Open Set/Real Analysis",
"Definition:Set Union",
"Definition:Real Interval/Open",
"Definition:Real Interval/Open",
"Closure of Real Interval is Closed Real Interval/Pr... |
proofwiki-1869 | Topological Product of Compact Spaces | Let $T_1$ and $T_2$ be topological spaces.
Let $T_1 \times T_2$ be the product space of $T_1$ and $T_2$.
Then $T_1 \times T_2$ is compact {{iff}} both $T_1$ and $T_2$ are compact. | === Necessary Condition ===
Let $T_1 \times T_2$ be compact.
From Projection from Product Topology is Continuous, both $T_1$ and $T_2$ are the continuous images of $T_1 \times T_2$ under the projections $\pr_1$ and $\pr_2$ respectively.
So both $T_1$ and $T_2$ are compact by Continuous Image of Compact Space is Compact... | Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
Let $T_1 \times T_2$ be the [[Definition:Product Space (Topology) of Two Factor Spaces|product space]] of $T_1$ and $T_2$.
Then $T_1 \times T_2$ is [[Definition:Compact Topological Space|compact]] {{iff}} both $T_1$ and $T_2$ are [[Definitio... | === Necessary Condition ===
Let $T_1 \times T_2$ be [[Definition:Compact Topological Space|compact]].
From [[Projection from Product Topology is Continuous]], both $T_1$ and $T_2$ are the [[Definition:Continuous Mapping (Topology)|continuous]] [[Definition:Image of Subset under Mapping|images]] of $T_1 \times T_2$ un... | Topological Product of Compact Spaces | https://proofwiki.org/wiki/Topological_Product_of_Compact_Spaces | https://proofwiki.org/wiki/Topological_Product_of_Compact_Spaces | [
"Topology"
] | [
"Definition:Topological Space",
"Definition:Product Space (Topology)/Two Factor Spaces",
"Definition:Compact Topological Space",
"Definition:Compact Topological Space"
] | [
"Definition:Compact Topological Space",
"Projection from Product Topology is Continuous",
"Definition:Continuous Mapping (Topology)",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Projection (Mapping Theory)",
"Definition:Compact Topological Space",
"Continuous Image of Compact Space is Co... |
proofwiki-1870 | Heine-Cantor Theorem | Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $M_1$ be compact.
Let $f: A_1 \to A_2$ be a continuous mapping.
Then $f$ is uniformly continuous. | Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
For all $x \in A_1$, define:
:$\map \Delta x = \set {\delta \in \R_{>0}: \forall y \in A_1: \map {d_1} {x, y} < 2 \delta \implies \map {d_2} {\map f x, \map f y} < \dfrac \epsilon 2}$
Define:
:$\CC = \set {\map {B_{\delta} } x: x \in A_1, \, \delta \in \map... | Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
Let $M_1$ be [[Definition:Compact Metric Space|compact]].
Let $f: A_1 \to A_2$ be a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]].
Then $f$ is [[Definition:Uniformly Continuous Mapping ... | Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
For all $x \in A_1$, define:
:$\map \Delta x = \set {\delta \in \R_{>0}: \forall y \in A_1: \map {d_1} {x, y} < 2 \delta \implies \map {d_2} {\map f x, \map f y} < \dfrac \epsilon 2}$
Define:
:$\CC = \set {\ma... | Heine-Cantor Theorem/Proof 1 | https://proofwiki.org/wiki/Heine-Cantor_Theorem | https://proofwiki.org/wiki/Heine-Cantor_Theorem/Proof_1 | [
"Heine-Cantor Theorem",
"Compact Metric Spaces",
"Uniformly Continuous Mappings",
"Continuous Mappings",
"Metric Spaces"
] | [
"Definition:Metric Space",
"Definition:Compact Space/Metric Space",
"Definition:Continuous Mapping (Metric Space)",
"Definition:Uniform Continuity/Metric Space"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:Open Ball",
"Definition:Continuous Mapping (Metric Space)",
"Definition:Cover of Set",
"Open Ball is Open Set/Pseudometric Space",
"Definition:Open Cover",
"Definition:Compact Space/Metric Space",
"Definition:Subcover/Finite",
"Definition:Cover... |
proofwiki-1871 | Heine-Cantor Theorem | Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $M_1$ be compact.
Let $f: A_1 \to A_2$ be a continuous mapping.
Then $f$ is uniformly continuous. | Let $A_1 \times A_1$ and $A_2 \times A_2$ be considered with the product topology.
Let $F: A_1 \times A_1 \to A_2 \times A_2$ be the mapping defined as:
:$\map F {x, y} = \tuple {\map f x, \map f y}$
By Projection from Product Topology is Continuous, we have that the (first and second) projections on $A_1 \times A_1$ a... | Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
Let $M_1$ be [[Definition:Compact Metric Space|compact]].
Let $f: A_1 \to A_2$ be a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]].
Then $f$ is [[Definition:Uniformly Continuous Mapping ... | Let $A_1 \times A_1$ and $A_2 \times A_2$ be considered with the [[Definition:Product Topology|product topology]].
Let $F: A_1 \times A_1 \to A_2 \times A_2$ be the [[Definition:Mapping|mapping]] defined as:
:$\map F {x, y} = \tuple {\map f x, \map f y}$
By [[Projection from Product Topology is Continuous]], we have ... | Heine-Cantor Theorem/Proof 2 | https://proofwiki.org/wiki/Heine-Cantor_Theorem | https://proofwiki.org/wiki/Heine-Cantor_Theorem/Proof_2 | [
"Heine-Cantor Theorem",
"Compact Metric Spaces",
"Uniformly Continuous Mappings",
"Continuous Mappings",
"Metric Spaces"
] | [
"Definition:Metric Space",
"Definition:Compact Space/Metric Space",
"Definition:Continuous Mapping (Metric Space)",
"Definition:Uniform Continuity/Metric Space"
] | [
"Definition:Product Topology",
"Definition:Mapping",
"Projection from Product Topology is Continuous",
"Definition:Projection (Mapping Theory)",
"Definition:Continuous Mapping (Topology)",
"Composite of Continuous Mappings is Continuous",
"Continuous Mapping to Product Space",
"Definition:Continuous M... |
proofwiki-1872 | Continuous Function on Closed Real Interval is Uniformly Continuous | Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \R$ be a continuous function.
Then $f$ is uniformly continuous on $\closedint a b$. | We have that $\R$ is a metric space under the usual (Euclidean) metric.
We also have from the Heine-Borel Theorem that $\closedint a b$ is compact.
So the result Heine-Cantor Theorem applies.
{{qed}} | Let $\closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]].
Let $f: \closedint a b \to \R$ be a [[Definition:Continuous Real Function|continuous function]].
Then $f$ is [[Definition:Uniformly Continuous Real Function|uniformly continuous]] on $\closedint a b$. | We have that [[Real Number Line is Metric Space|$\R$ is a metric space]] under the [[Definition:Euclidean Metric on Real Number Line|usual (Euclidean) metric]].
We also have from the [[Heine-Borel Theorem]] that $\closedint a b$ is [[Definition:Compact Topological Space|compact]].
So the result [[Heine-Cantor Theorem... | Continuous Function on Closed Real Interval is Uniformly Continuous/Proof 1 | https://proofwiki.org/wiki/Continuous_Function_on_Closed_Real_Interval_is_Uniformly_Continuous | https://proofwiki.org/wiki/Continuous_Function_on_Closed_Real_Interval_is_Uniformly_Continuous/Proof_1 | [
"Continuous Function on Closed Real Interval is Uniformly Continuous",
"Continuous Real Functions",
"Uniformly Continuous Real Functions"
] | [
"Definition:Real Interval/Closed",
"Definition:Continuous Real Function",
"Definition:Uniform Continuity/Real Function"
] | [
"Real Number Line is Metric Space",
"Definition:Euclidean Metric/Real Number Line",
"Heine-Borel Theorem",
"Definition:Compact Topological Space",
"Heine-Cantor Theorem"
] |
proofwiki-1873 | Continuous Function on Closed Real Interval is Uniformly Continuous | Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \R$ be a continuous function.
Then $f$ is uniformly continuous on $\closedint a b$. | {{AimForCont}} $f$ is not uniformly continuous.
Then there exists an $\epsilon \in \R_{>0}$ such that for all $\delta \in \R_{>0}$ there are points $x ,y \in \closedint a b$ for which:
:$\size {x - y} < \delta$
and:
:$\size {\map f x - \map f y} \ge \epsilon$
We will now choose for each $k \in \N$ numbers $x_k, y_k \i... | Let $\closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]].
Let $f: \closedint a b \to \R$ be a [[Definition:Continuous Real Function|continuous function]].
Then $f$ is [[Definition:Uniformly Continuous Real Function|uniformly continuous]] on $\closedint a b$. | {{AimForCont}} $f$ is not [[Definition:Uniformly Continuous Real Function|uniformly continuous]].
Then there exists an $\epsilon \in \R_{>0}$ such that for all $\delta \in \R_{>0}$ there are points $x ,y \in \closedint a b$ for which:
:$\size {x - y} < \delta$
and:
:$\size {\map f x - \map f y} \ge \epsilon$
We will... | Continuous Function on Closed Real Interval is Uniformly Continuous/Proof 2 | https://proofwiki.org/wiki/Continuous_Function_on_Closed_Real_Interval_is_Uniformly_Continuous | https://proofwiki.org/wiki/Continuous_Function_on_Closed_Real_Interval_is_Uniformly_Continuous/Proof_2 | [
"Continuous Function on Closed Real Interval is Uniformly Continuous",
"Continuous Real Functions",
"Uniformly Continuous Real Functions"
] | [
"Definition:Real Interval/Closed",
"Definition:Continuous Real Function",
"Definition:Uniform Continuity/Real Function"
] | [
"Definition:Uniform Continuity/Real Function",
"Definition:Real Sequence",
"Definition:Bounded Sequence/Real",
"Bolzano-Weierstrass Theorem",
"Definition:Convergent Sequence/Real Numbers",
"Definition:Subsequence",
"Definition:Limit of Sequence/Real Numbers",
"Definition:Limit of Sequence/Real Numbers... |
proofwiki-1874 | Sine of Complement equals Cosine | :$\map \sin {\dfrac \pi 2 - \theta} = \cos \theta$ | {{begin-eqn}}
{{eqn | l = \map \sin {\frac \pi 2 - \theta}
| r = \sin \frac \pi 2 \cos \theta - \cos \frac \pi 2 \sin \theta
| c = Sine of Difference
}}
{{eqn | r = 1 \times \cos \theta - 0 \times \sin \theta
| c = Sine of Right Angle and Cosine of Right Angle
}}
{{eqn | r = \cos \theta
}}
{{end-eqn}}... | :$\map \sin {\dfrac \pi 2 - \theta} = \cos \theta$ | {{begin-eqn}}
{{eqn | l = \map \sin {\frac \pi 2 - \theta}
| r = \sin \frac \pi 2 \cos \theta - \cos \frac \pi 2 \sin \theta
| c = [[Sine of Difference]]
}}
{{eqn | r = 1 \times \cos \theta - 0 \times \sin \theta
| c = [[Sine of Right Angle]] and [[Cosine of Right Angle]]
}}
{{eqn | r = \cos \theta
}}... | Sine of Complement equals Cosine/Proof 1 | https://proofwiki.org/wiki/Sine_of_Complement_equals_Cosine | https://proofwiki.org/wiki/Sine_of_Complement_equals_Cosine/Proof_1 | [
"Sine of Complement equals Cosine",
"Sine Function",
"Cosine Function",
"Complementary Angles"
] | [] | [
"Sine of Difference",
"Sine of Right Angle",
"Cosine of Right Angle"
] |
proofwiki-1875 | Sine of Complement equals Cosine | :$\map \sin {\dfrac \pi 2 - \theta} = \cos \theta$ | {{begin-eqn}}
{{eqn | l = \map \sin {\frac \pi 2 - \theta}
| r = -\map \sin {\theta - \frac \pi 2}
| c = Sine Function is Odd
}}
{{eqn | r = \map \cos {\theta - \frac \pi 2 + \frac \pi 2}
| c = Cosine of Angle plus Right Angle
}}
{{eqn | r = \cos \theta
}}
{{end-eqn}}
{{qed}} | :$\map \sin {\dfrac \pi 2 - \theta} = \cos \theta$ | {{begin-eqn}}
{{eqn | l = \map \sin {\frac \pi 2 - \theta}
| r = -\map \sin {\theta - \frac \pi 2}
| c = [[Sine Function is Odd]]
}}
{{eqn | r = \map \cos {\theta - \frac \pi 2 + \frac \pi 2}
| c = [[Cosine of Angle plus Right Angle]]
}}
{{eqn | r = \cos \theta
}}
{{end-eqn}}
{{qed}} | Sine of Complement equals Cosine/Proof 2 | https://proofwiki.org/wiki/Sine_of_Complement_equals_Cosine | https://proofwiki.org/wiki/Sine_of_Complement_equals_Cosine/Proof_2 | [
"Sine of Complement equals Cosine",
"Sine Function",
"Cosine Function",
"Complementary Angles"
] | [] | [
"Sine Function is Odd",
"Cosine of Angle plus Right Angle"
] |
proofwiki-1876 | Sine of Complement equals Cosine | :$\map \sin {\dfrac \pi 2 - \theta} = \cos \theta$ | {{begin-eqn}}
{{eqn | l = \map \sin {\dfrac \pi 2 - \theta}
| r = \map \Im {e^{i \paren {\frac \pi 2 - \theta} } }
| c = Euler's Formula
}}
{{eqn | r = \map \Im {e^{i \frac \pi 2} e^{-i \theta} }
| c = Exponent Combination Laws
}}
{{eqn | r = \map \Im {\paren {\cos \dfrac \pi 2+i \sin \dfrac \pi 2} e^... | :$\map \sin {\dfrac \pi 2 - \theta} = \cos \theta$ | {{begin-eqn}}
{{eqn | l = \map \sin {\dfrac \pi 2 - \theta}
| r = \map \Im {e^{i \paren {\frac \pi 2 - \theta} } }
| c = [[Euler's Formula]]
}}
{{eqn | r = \map \Im {e^{i \frac \pi 2} e^{-i \theta} }
| c = [[Exponent Combination Laws]]
}}
{{eqn | r = \map \Im {\paren {\cos \dfrac \pi 2+i \sin \dfrac \... | Sine of Complement equals Cosine/Proof 3 | https://proofwiki.org/wiki/Sine_of_Complement_equals_Cosine | https://proofwiki.org/wiki/Sine_of_Complement_equals_Cosine/Proof_3 | [
"Sine of Complement equals Cosine",
"Sine Function",
"Cosine Function",
"Complementary Angles"
] | [] | [
"Euler's Formula",
"Exponent Combination Laws",
"Euler's Formula",
"Cosine of Right Angle",
"Sine of Right Angle",
"Euler's Formula",
"Cosine Function is Even"
] |
proofwiki-1877 | Sine of Complement equals Cosine | :$\map \sin {\dfrac \pi 2 - \theta} = \cos \theta$ | :300px
Let $\angle xOP$ and $\angle QOy$ be complementary.
Then:
:$\angle xOP = \angle QOy$
Hence:
:the projection of $OP$ on the $x$-axis
equals:
:the projection of $OQ$ on the $y$-axis.
Hence the result.
{{qed}} | :$\map \sin {\dfrac \pi 2 - \theta} = \cos \theta$ | :[[File:ComplementaryAngles.png|300px]]
Let $\angle xOP$ and $\angle QOy$ be [[Definition:Complementary Angles|complementary]].
Then:
:$\angle xOP = \angle QOy$
Hence:
:the [[Definition:Geometric Projection|projection]] of $OP$ on the [[Definition:X-Axis|$x$-axis]]
equals:
:the [[Definition:Geometric Projection|proj... | Sine of Complement equals Cosine/Proof 4 | https://proofwiki.org/wiki/Sine_of_Complement_equals_Cosine | https://proofwiki.org/wiki/Sine_of_Complement_equals_Cosine/Proof_4 | [
"Sine of Complement equals Cosine",
"Sine Function",
"Cosine Function",
"Complementary Angles"
] | [] | [
"File:ComplementaryAngles.png",
"Definition:Complementary Angles",
"Definition:Projection (Geometry)",
"Definition:Axis/X-Axis",
"Definition:Projection (Geometry)",
"Definition:Axis/Y-Axis"
] |
proofwiki-1878 | Cosine of Complement equals Sine | :$\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$ | {{begin-eqn}}
{{eqn | l = \map \cos {\frac \pi 2 - \theta}
| r = \cos \frac \pi 2 \cos \theta + \sin \frac \pi 2 \sin \theta
| c = Cosine of Difference
}}
{{eqn | r = 0 \times \cos \theta + 1 \times \sin \theta
| c = Cosine of Right Angle and Sine of Right Angle
}}
{{eqn | r = \sin \theta
}}
{{end-eqn... | :$\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$ | {{begin-eqn}}
{{eqn | l = \map \cos {\frac \pi 2 - \theta}
| r = \cos \frac \pi 2 \cos \theta + \sin \frac \pi 2 \sin \theta
| c = [[Cosine of Difference]]
}}
{{eqn | r = 0 \times \cos \theta + 1 \times \sin \theta
| c = [[Cosine of Right Angle]] and [[Sine of Right Angle]]
}}
{{eqn | r = \sin \theta
... | Cosine of Complement equals Sine/Proof 1 | https://proofwiki.org/wiki/Cosine_of_Complement_equals_Sine | https://proofwiki.org/wiki/Cosine_of_Complement_equals_Sine/Proof_1 | [
"Cosine of Complement equals Sine",
"Sine Function",
"Cosine Function",
"Complementary Angles"
] | [] | [
"Cosine of Difference",
"Cosine of Right Angle",
"Sine of Right Angle"
] |
proofwiki-1879 | Cosine of Complement equals Sine | :$\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$ | {{begin-eqn}}
{{eqn | l = \map \cos {\frac \pi 2 - \theta}
| r = \map \cos {\theta - \frac \pi 2}
| c = Cosine Function is Even
}}
{{eqn | r = \map \sin {\theta - \frac \pi 2 + \frac \pi 2}
| c = Sine of Angle plus Right Angle
}}
{{eqn | r = \sin \theta
}}
{{end-eqn}}
{{qed}} | :$\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$ | {{begin-eqn}}
{{eqn | l = \map \cos {\frac \pi 2 - \theta}
| r = \map \cos {\theta - \frac \pi 2}
| c = [[Cosine Function is Even]]
}}
{{eqn | r = \map \sin {\theta - \frac \pi 2 + \frac \pi 2}
| c = [[Sine of Angle plus Right Angle]]
}}
{{eqn | r = \sin \theta
}}
{{end-eqn}}
{{qed}} | Cosine of Complement equals Sine/Proof 2 | https://proofwiki.org/wiki/Cosine_of_Complement_equals_Sine | https://proofwiki.org/wiki/Cosine_of_Complement_equals_Sine/Proof_2 | [
"Cosine of Complement equals Sine",
"Sine Function",
"Cosine Function",
"Complementary Angles"
] | [] | [
"Cosine Function is Even",
"Sine of Angle plus Right Angle"
] |
proofwiki-1880 | Cosine of Complement equals Sine | :$\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$ | {{begin-eqn}}
{{eqn | l = \map \cos {\dfrac \pi 2 - \theta}
| r = \frac 1 2 \paren {e^{i \paren {\frac \pi 2 - \theta} } + e^{-i \paren {\frac \pi 2 - \theta} } }
| c = Euler's Cosine Identity
}}
{{eqn | r = \frac 1 2 \paren {e^{i \frac \pi 2} e^{-i \theta} + e^{-i \frac \pi 2} e^{i \theta} }
| c = Ex... | :$\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$ | {{begin-eqn}}
{{eqn | l = \map \cos {\dfrac \pi 2 - \theta}
| r = \frac 1 2 \paren {e^{i \paren {\frac \pi 2 - \theta} } + e^{-i \paren {\frac \pi 2 - \theta} } }
| c = [[Euler's Cosine Identity]]
}}
{{eqn | r = \frac 1 2 \paren {e^{i \frac \pi 2} e^{-i \theta} + e^{-i \frac \pi 2} e^{i \theta} }
| c ... | Cosine of Complement equals Sine/Proof 3 | https://proofwiki.org/wiki/Cosine_of_Complement_equals_Sine | https://proofwiki.org/wiki/Cosine_of_Complement_equals_Sine/Proof_3 | [
"Cosine of Complement equals Sine",
"Sine Function",
"Cosine Function",
"Complementary Angles"
] | [] | [
"Euler's Cosine Identity",
"Exponential of Sum/Complex Numbers",
"Euler's Formula",
"Cosine of Right Angle",
"Sine of Right Angle",
"Cosine Function is Even",
"Sine Function is Odd",
"Euler's Sine Identity"
] |
proofwiki-1881 | Cosine of Complement equals Sine | :$\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$ | :300px
Let $\angle xOP$ and $\angle QOy$ be complementary.
Then:
:$\angle xOP = \angle QOy$
Hence:
:the projection of $OP$ on the $x$-axis
equals:
:the projection of $OQ$ on the $y$-axis.
Hence the result.
{{qed}} | :$\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$ | :[[File:ComplementaryAngles.png|300px]]
Let $\angle xOP$ and $\angle QOy$ be [[Definition:Complementary Angles|complementary]].
Then:
:$\angle xOP = \angle QOy$
Hence:
:the [[Definition:Geometric Projection|projection]] of $OP$ on the [[Definition:X-Axis|$x$-axis]]
equals:
:the [[Definition:Geometric Projection|proj... | Cosine of Complement equals Sine/Proof 4 | https://proofwiki.org/wiki/Cosine_of_Complement_equals_Sine | https://proofwiki.org/wiki/Cosine_of_Complement_equals_Sine/Proof_4 | [
"Cosine of Complement equals Sine",
"Sine Function",
"Cosine Function",
"Complementary Angles"
] | [] | [
"File:ComplementaryAngles.png",
"Definition:Complementary Angles",
"Definition:Projection (Geometry)",
"Definition:Axis/X-Axis",
"Definition:Projection (Geometry)",
"Definition:Axis/Y-Axis"
] |
proofwiki-1882 | Continuous Bijection from Compact to Hausdorff is Homeomorphism | Let $T_1$ be a compact space.
Let $T_2$ be a Hausdorff space.
Let $f: T_1 \to T_2$ be a continuous bijection.
Then $f$ is a homeomorphism. | Let $g = f^{-1}$.
We need to show that $g: T_2 \to T_1$ is continuous.
For any $V \subseteq T_1$, we have $g^{-1} \sqbrk V = \map f V$.
We are to show that if $V$ is closed in $T_1$, then $g^{-1} \sqbrk V$ is closed in $T_2$.
Suppose $V$ is closed in $T_1$.
Since $T_1$ is compact, $V$ is compact by Closed Subspace of C... | Let $T_1$ be a [[Definition:Compact Topological Space|compact space]].
Let $T_2$ be a [[Definition:Hausdorff Space|Hausdorff space]].
Let $f: T_1 \to T_2$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Bijection|bijection]].
Then $f$ is a [[Definition:Homeomorphism (Topological... | Let $g = f^{-1}$.
We need to show that $g: T_2 \to T_1$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
For any $V \subseteq T_1$, we have $g^{-1} \sqbrk V = \map f V$.
We are to show that if $V$ is [[Definition:Closed Set (Topology)|closed]] in $T_1$, then $g^{-1} \sqbrk V$ is [[Definition:Cl... | Continuous Bijection from Compact to Hausdorff is Homeomorphism | https://proofwiki.org/wiki/Continuous_Bijection_from_Compact_to_Hausdorff_is_Homeomorphism | https://proofwiki.org/wiki/Continuous_Bijection_from_Compact_to_Hausdorff_is_Homeomorphism | [
"Continuous Bijection from Compact to Hausdorff is Homeomorphism",
"Homeomorphisms (Topological Spaces)",
"Compact Topological Spaces",
"Hausdorff Spaces",
"Continuous Mappings"
] | [
"Definition:Compact Topological Space",
"Definition:T2 Space",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Bijection",
"Definition:Homeomorphism/Topological Spaces"
] | [
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Compact Topological Space",
"Definition:Compact Topological Space",
"Closed Subspace of Compact Space is Compact",
"Definition:Compac... |
proofwiki-1883 | Indirect Proof | Let $P$ be a proposition whose truth value is to be proved (either true or false).
An '''indirect proof''' has one of the following two argument forms:
=== Reductio ad Absurdum ===
A '''reductio ad absurdum''' argument for the truth of $P$ is a valid argument which takes as a premise the negation of $P$, and from it de... | Let $x$ be an even integer.
Let $y = 2 n + 5$.
Assume $y = x + 5$ is not an odd integer.
Then:
:$y = x + 5 = 2 n$
where $n \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = x
| r = 2 n - 5
| c =
}}
{{eqn | r = \paren {2 n - 6} + 1
| c =
}}
{{eqn | r = 2 \paren {n - 3} + 1
| c =
}}
{{eqn | r = 2 r + 1
... | Let $P$ be a [[Definition:Proposition|proposition]] whose [[Definition:Truth Value|truth value]] is to be [[Definition:Proof|proved]] (either [[Definition:True|true]] or [[Definition:False|false]]).
An '''[[Definition:Indirect Proof|indirect proof]]''' has one of the following two [[Definition:Argument Form|argument f... | Let $x$ be an [[Definition:Even Integer|even integer]].
Let $y = 2 n + 5$.
Assume $y = x + 5$ is not an [[Definition:Odd Integer|odd integer]].
Then:
:$y = x + 5 = 2 n$
where $n \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = x
| r = 2 n - 5
| c =
}}
{{eqn | r = \paren {2 n - 6} + 1
| c =
}}
{{eqn | ... | Even Integer Plus 5 is Odd/Indirect Proof | https://proofwiki.org/wiki/Indirect_Proof | https://proofwiki.org/wiki/Even_Integer_Plus_5_is_Odd/Indirect_Proof | [
"Indirect Proof",
"Logic",
"Proof Techniques"
] | [
"Definition:Proposition",
"Definition:Truth Value",
"Definition:Proof",
"Definition:True",
"Definition:False",
"Indirect Proof",
"Definition:Argument Form",
"Reductio ad Absurdum",
"Reductio ad Absurdum",
"Definition:True",
"Definition:Valid Argument",
"Definition:Premise",
"Definition:Logic... | [
"Definition:Even Integer",
"Definition:Odd Integer",
"Definition:Odd Integer",
"Definition:False",
"Definition:Even Integer",
"Rule of Transposition",
"Definition:Even Integer",
"Definition:Odd Integer"
] |
proofwiki-1884 | Indirect Proof | Let $P$ be a proposition whose truth value is to be proved (either true or false).
An '''indirect proof''' has one of the following two argument forms:
=== Reductio ad Absurdum ===
A '''reductio ad absurdum''' argument for the truth of $P$ is a valid argument which takes as a premise the negation of $P$, and from it de... | For proofs, see:
* Reductio ad Absurdum/Sequent Form
* Proof by Contradiction/Sequent Form | Let $P$ be a [[Definition:Proposition|proposition]] whose [[Definition:Truth Value|truth value]] is to be [[Definition:Proof|proved]] (either [[Definition:True|true]] or [[Definition:False|false]]).
An '''[[Definition:Indirect Proof|indirect proof]]''' has one of the following two [[Definition:Argument Form|argument f... | For proofs, see:
* [[Reductio ad Absurdum/Sequent Form]]
* [[Proof by Contradiction/Sequent Form]] | Indirect Proof | https://proofwiki.org/wiki/Indirect_Proof | https://proofwiki.org/wiki/Indirect_Proof | [
"Indirect Proof",
"Logic",
"Proof Techniques"
] | [
"Definition:Proposition",
"Definition:Truth Value",
"Definition:Proof",
"Definition:True",
"Definition:False",
"Indirect Proof",
"Definition:Argument Form",
"Reductio ad Absurdum",
"Reductio ad Absurdum",
"Definition:True",
"Definition:Valid Argument",
"Definition:Premise",
"Definition:Logic... | [
"Reductio ad Absurdum/Sequent Form",
"Proof by Contradiction/Sequent Form"
] |
proofwiki-1885 | Continuous Mapping of Separation | Let $T$ and $T'$ be topological spaces.
Let $A \mid B$ be a separation of $T$.
Let $f: T \to T'$ be a mapping such that the restrictions $f \restriction_A$ and $f \restriction_B$ are both continuous.
Then $f$ is continuous on the whole of $T$. | Follows directly from Pasting Lemma for Union of Open Sets.
{{Qed}} | Let $T$ and $T'$ be [[Definition:Topological Space|topological spaces]].
Let $A \mid B$ be a [[Definition:Separation (Topology)|separation]] of $T$.
Let $f: T \to T'$ be a [[Definition:Mapping|mapping]] such that the [[Definition:Restriction of Mapping|restrictions]] $f \restriction_A$ and $f \restriction_B$ are both... | Follows directly from [[Pasting Lemma for Union of Open Sets]].
{{Qed}} | Continuous Mapping of Separation | https://proofwiki.org/wiki/Continuous_Mapping_of_Separation | https://proofwiki.org/wiki/Continuous_Mapping_of_Separation | [
"Separations",
"Continuous Mappings"
] | [
"Definition:Topological Space",
"Definition:Separation (Topology)",
"Definition:Mapping",
"Definition:Restriction/Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Continuous Mapping (Topology)/Everywhere"
] | [
"Pasting Lemma/Union of Open Sets"
] |
proofwiki-1886 | Fermat's Last Theorem | :$\forall a, b, c, n \in \Z_{>0}: n > 2$, the equation $a^n + b^n = c^n$ has no solutions. | The proof of this theorem is beyond the current scope of {{ProofWiki}}, and indeed, is beyond the understanding of many high level mathematicians.
For the curious reader, the proof can be found [https://www.jstor.org/stable/2118559 here], in a paper published by {{AuthorRef|Andrew John Wiles|Andrew Wiles}}, entitled ''... | :$\forall a, b, c, n \in \Z_{>0}: n > 2$, the [[Definition:Equation|equation]] $a^n + b^n = c^n$ has no solutions. | The proof of this theorem is beyond the current scope of {{ProofWiki}}, and indeed, is beyond the understanding of many high level mathematicians.
For the curious reader, the proof can be found [https://www.jstor.org/stable/2118559 here], in a paper published by {{AuthorRef|Andrew John Wiles|Andrew Wiles}}, entitled '... | Fermat's Last Theorem | https://proofwiki.org/wiki/Fermat's_Last_Theorem | https://proofwiki.org/wiki/Fermat's_Last_Theorem | [
"Number Theory",
"Classic Problems",
"Fermat's Last Theorem"
] | [
"Definition:Equation"
] | [
"Taniyama-Shimura Conjecture",
"Epsilon Conjecture"
] |
proofwiki-1887 | Subset of Real Numbers is Interval iff Connected | Let the real number line $\R$ be considered as a topological space.
Let $S$ be a subspace of $\R$.
Then $S$ is connected {{iff}} $S$ is an interval of $\R$.
That is, the only subspaces of $\R$ that are connected are intervals. | From Rule of Transposition, we may replace the ''only if'' statement by its contrapositive.
Therefore, the following suffices: | Let the [[Definition:Real Number Line|real number line]] $\R$ be considered as a [[Definition:Topological Space|topological space]].
Let $S$ be a [[Definition:Topological Subspace|subspace]] of $\R$.
Then $S$ is [[Definition:Connected Topological Space|connected]] {{iff}} $S$ is an [[Definition:Real Interval|interva... | From [[Rule of Transposition]], we may replace the ''only if'' statement by its [[Definition:Contrapositive Statement|contrapositive]].
Therefore, the following suffices: | Subset of Real Numbers is Interval iff Connected | https://proofwiki.org/wiki/Subset_of_Real_Numbers_is_Interval_iff_Connected | https://proofwiki.org/wiki/Subset_of_Real_Numbers_is_Interval_iff_Connected | [
"Real Intervals",
"Connected Topological Spaces"
] | [
"Definition:Real Number/Real Number Line",
"Definition:Topological Space",
"Definition:Topological Subspace",
"Definition:Connected Topological Space",
"Definition:Real Interval",
"Definition:Topological Subspace",
"Definition:Connected Topological Space",
"Definition:Real Interval"
] | [
"Rule of Transposition",
"Definition:Contrapositive Statement"
] |
proofwiki-1888 | Continuous Image of Connected Space is Connected | Let $T_1$ and $T_2$ be topological spaces.
Let $S_1 \subseteq T_1$ be connected.
Let $f: T_1 \to T_2$ be a continuous mapping.
Then the image $f \sqbrk {S_1}$ is connected. | Let $i: f \sqbrk {T_1} \to T_2$ be the inclusion mapping.
Let $g: T_1 \to f \sqbrk {T_1}$ be the surjective restriction of $f$.
Then $f = i \circ g$.
Hence, by Continuity of Composite with Inclusion: Inclusion on Mapping, it follows that $g$ is continuous.
;Proof by Contradiction
{{Recall|Connected Topological Space|co... | Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
Let $S_1 \subseteq T_1$ be [[Definition:Connected Set (Topology)|connected]].
Let $f: T_1 \to T_2$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
Then the [[Definition:Image of Mapping|image]] $f \sqbrk {S_... | Let $i: f \sqbrk {T_1} \to T_2$ be the [[Definition:Inclusion Mapping|inclusion mapping]].
Let $g: T_1 \to f \sqbrk {T_1}$ be the [[Surjection by Restriction of Codomain|surjective restriction]] of $f$.
Then $f = i \circ g$.
Hence, by [[Continuity of Composite with Inclusion/Inclusion on Mapping|Continuity of Compos... | Continuous Image of Connected Space is Connected/Proof 1 | https://proofwiki.org/wiki/Continuous_Image_of_Connected_Space_is_Connected | https://proofwiki.org/wiki/Continuous_Image_of_Connected_Space_is_Connected/Proof_1 | [
"Continuous Image of Connected Space is Connected",
"Connected Topological Spaces",
"Examples of Continuous Invariants"
] | [
"Definition:Topological Space",
"Definition:Connected Set (Topology)",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Connected Set (Topology)"
] | [
"Definition:Inclusion Mapping",
"Restriction of Mapping to Image is Surjection",
"Continuity of Composite with Inclusion/Inclusion on Mapping",
"Definition:Continuous Mapping (Topology)",
"Proof by Contradiction",
"Definition:Separation (Topology)",
"Preimage of Separation under Continuous Mapping is Se... |
proofwiki-1889 | Continuous Image of Connected Space is Connected | Let $T_1$ and $T_2$ be topological spaces.
Let $S_1 \subseteq T_1$ be connected.
Let $f: T_1 \to T_2$ be a continuous mapping.
Then the image $f \sqbrk {S_1}$ is connected. | Suppose that $S_2 = f \sqbrk {S_1}$ is not connected in $T_2$.
Then by definition there exist open sets $U_2$ and $V_2$ in $T_2$ such that:
:$S_2 \subseteq U_2 \cup V_2$
:$U_2 \cap V_2 \cap S_2 = \O$
:$U_2 \cap S_2 \ne \O$
:$V_2 \cap S_2 \ne \O$
We have {{hypothesis}} that $f: T_1 \to T_2$ is continuous.
Thus $U_1 = f^... | Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
Let $S_1 \subseteq T_1$ be [[Definition:Connected Set (Topology)|connected]].
Let $f: T_1 \to T_2$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
Then the [[Definition:Image of Mapping|image]] $f \sqbrk {S_... | Suppose that $S_2 = f \sqbrk {S_1}$ is not [[Definition:Connected Set (Topology)|connected]] in $T_2$.
Then [[Definition:Connected Set (Topology)/Definition 2|by definition]] there exist [[Definition:Open Set (Topology)|open sets]] $U_2$ and $V_2$ in $T_2$ such that:
:$S_2 \subseteq U_2 \cup V_2$
:$U_2 \cap V_2 \cap S... | Continuous Image of Connected Space is Connected/Proof 2 | https://proofwiki.org/wiki/Continuous_Image_of_Connected_Space_is_Connected | https://proofwiki.org/wiki/Continuous_Image_of_Connected_Space_is_Connected/Proof_2 | [
"Continuous Image of Connected Space is Connected",
"Connected Topological Spaces",
"Examples of Continuous Invariants"
] | [
"Definition:Topological Space",
"Definition:Connected Set (Topology)",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Connected Set (Topology)"
] | [
"Definition:Connected Set (Topology)",
"Definition:Connected Set (Topology)/Definition 2",
"Definition:Open Set/Topology",
"Definition:Continuous Mapping (Topology)",
"Definition:Open Set/Topology",
"Definition:Contradiction",
"Definition:Connected Set (Topology)",
"Rule of Transposition"
] |
proofwiki-1890 | Union of Connected Sets with Non-Empty Intersections is Connected | Let $T = \struct {S, \tau}$ be a topological space.
Let $I$ be an indexing set.
Let $\AA = \family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $S$, all connected in $T$.
Let $\AA$ be such that no two of its elements are disjoint:
:$\forall B, C \in \AA: B \cap C \ne \O$
Then $\ds \bigcup \AA$ ... | Let $A := \ds \bigcup \AA$.
Let $D = \set {0, 1}$, with the discrete topology.
Let $f: A \to D$ be a continuous mapping.
To show that $A$ is connected, we need to show that $f$ is not a surjection.
From Connected Set in Subspace, each $C \in \AA$ is connected in $A$.
From Restriction of Continuous Mapping is Continuous... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\AA = \family {A_\alpha}_{\alpha \mathop \in I}$ be an [[Definition:Indexed Family of Subsets|indexed family]] of [[Definition:Subset|subsets]] of $S$, all [[Definition:Con... | Let $A := \ds \bigcup \AA$.
Let $D = \set {0, 1}$, with the [[Definition:Discrete Topology|discrete topology]].
Let $f: A \to D$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
To show that $A$ is [[Definition:Connected Set (Topology)|connected]], we need to show that $f$ is not a [[... | Union of Connected Sets with Non-Empty Intersections is Connected | https://proofwiki.org/wiki/Union_of_Connected_Sets_with_Non-Empty_Intersections_is_Connected | https://proofwiki.org/wiki/Union_of_Connected_Sets_with_Non-Empty_Intersections_is_Connected | [
"Union of Connected Sets with Non-Empty Intersections is Connected",
"Connected Sets (Topology)"
] | [
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Indexing Set/Family of Subsets",
"Definition:Subset",
"Definition:Connected Set (Topology)",
"Definition:Element",
"Definition:Disjoint Sets",
"Definition:Connected Set (Topology)"
] | [
"Definition:Discrete Topology",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Connected Set (Topology)",
"Definition:Surjection",
"Connected Set in Subspace",
"Definition:Connected Set (Topology)",
"Restriction of Continuous Mapping is Continuous/Topological Spaces",
"Definition:R... |
proofwiki-1891 | Union of Connected Sets with Non-Empty Intersections is Connected | Let $T = \struct {S, \tau}$ be a topological space.
Let $I$ be an indexing set.
Let $\AA = \family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $S$, all connected in $T$.
Let $\AA$ be such that no two of its elements are disjoint:
:$\forall B, C \in \AA: B \cap C \ne \O$
Then $\ds \bigcup \AA$ ... | Let $C \in \AA$.
From Union of Connected Sets with Non-Empty Intersections is Connected applied to $B$ and $C$, the union $B \cup C$ is connected.
Thus the set $\tilde \AA = \set {B \cup C: C \in \AA}$ satisfies the conditions of the theorem.
Hence the result.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\AA = \family {A_\alpha}_{\alpha \mathop \in I}$ be an [[Definition:Indexed Family of Subsets|indexed family]] of [[Definition:Subset|subsets]] of $S$, all [[Definition:Con... | Let $C \in \AA$.
From [[Union of Connected Sets with Non-Empty Intersections is Connected]] applied to $B$ and $C$, the [[Definition:Set Union|union]] $B \cup C$ is [[Definition:Connected Set (Topology)|connected]].
Thus the set $\tilde \AA = \set {B \cup C: C \in \AA}$ satisfies the conditions of the theorem.
Hence... | Union of Connected Sets with Non-Empty Intersections is Connected/Corollary/Proof 1 | https://proofwiki.org/wiki/Union_of_Connected_Sets_with_Non-Empty_Intersections_is_Connected | https://proofwiki.org/wiki/Union_of_Connected_Sets_with_Non-Empty_Intersections_is_Connected/Corollary/Proof_1 | [
"Union of Connected Sets with Non-Empty Intersections is Connected",
"Connected Sets (Topology)"
] | [
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Indexing Set/Family of Subsets",
"Definition:Subset",
"Definition:Connected Set (Topology)",
"Definition:Element",
"Definition:Disjoint Sets",
"Definition:Connected Set (Topology)"
] | [
"Union of Connected Sets with Non-Empty Intersections is Connected",
"Definition:Set Union",
"Definition:Connected Set (Topology)"
] |
proofwiki-1892 | Union of Connected Sets with Non-Empty Intersections is Connected | Let $T = \struct {S, \tau}$ be a topological space.
Let $I$ be an indexing set.
Let $\AA = \family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $S$, all connected in $T$.
Let $\AA$ be such that no two of its elements are disjoint:
:$\forall B, C \in \AA: B \cap C \ne \O$
Then $\ds \bigcup \AA$ ... | Let $\ds H = B \cup \bigcup \AA$
{{AimForCont}} that $H$ is not connected.
That is, that $H$ is disconnected.
We have that:
:$\forall C \in \AA: B \cap C \ne \O$
Thus:
:$\exists x \in H: x \in B \cap C$
From the definition of disconnected, there exist separated sets $U, V$ whose union is $H$.
So:
:$\forall s \in S: s \... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\AA = \family {A_\alpha}_{\alpha \mathop \in I}$ be an [[Definition:Indexed Family of Subsets|indexed family]] of [[Definition:Subset|subsets]] of $S$, all [[Definition:Con... | Let $\ds H = B \cup \bigcup \AA$
{{AimForCont}} that $H$ is not [[Definition:Connected Set (Topology)|connected]].
That is, that $H$ is [[Definition:Disconnected Set|disconnected]].
We have that:
:$\forall C \in \AA: B \cap C \ne \O$
Thus:
:$\exists x \in H: x \in B \cap C$
From the definition of [[Definition:Dis... | Union of Connected Sets with Non-Empty Intersections is Connected/Corollary/Proof 2 | https://proofwiki.org/wiki/Union_of_Connected_Sets_with_Non-Empty_Intersections_is_Connected | https://proofwiki.org/wiki/Union_of_Connected_Sets_with_Non-Empty_Intersections_is_Connected/Corollary/Proof_2 | [
"Union of Connected Sets with Non-Empty Intersections is Connected",
"Connected Sets (Topology)"
] | [
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Indexing Set/Family of Subsets",
"Definition:Subset",
"Definition:Connected Set (Topology)",
"Definition:Element",
"Definition:Disjoint Sets",
"Definition:Connected Set (Topology)"
] | [
"Definition:Connected Set (Topology)",
"Definition:Disconnected (Topology)/Set",
"Definition:Disconnected (Topology)/Set",
"Definition:Separated Sets",
"Definition:Set Union",
"Definition:Separated Sets",
"Definition:Set Union",
"Definition:Cover of Set",
"Definition:Disconnected (Topology)/Set",
... |
proofwiki-1893 | Product Space of 2 Factors is Connected iff Factors are Connected | Let $T_1$ and $T_2$ be topological spaces.
Then the product space $T_1 \times T_2$ is connected {{iff}} $T_1$ and $T_2$ are connected. | ==== Necessary Condition ====
Let $T_1 \times T_2$ be connected.
By Projection from Product Topology is Continuous, $T_1$ and $T_2$ are continuous images under the projections $\pr_1$ and $\pr_2$.
Hence by Continuous Image of Connected Space is Connected, $T_1$ and $T_2$ are connected.
{{qed|lemma}}
==== Sufficient Con... | Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
Then the [[Definition:Product Space (Topology)|product space]] $T_1 \times T_2$ is [[Definition:Connected Topological Space|connected]] {{iff}} $T_1$ and $T_2$ are [[Definition:Connected Topological Space|connected]]. | ==== Necessary Condition ====
Let $T_1 \times T_2$ be [[Definition:Connected Topological Space|connected]].
By [[Projection from Product Topology is Continuous]], $T_1$ and $T_2$ are [[Definition:Everywhere Continuous Mapping (Topology)|continuous images]] under the [[Definition:Projection (Mapping Theory)|projection... | Product Space of 2 Factors is Connected iff Factors are Connected | https://proofwiki.org/wiki/Product_Space_of_2_Factors_is_Connected_iff_Factors_are_Connected | https://proofwiki.org/wiki/Product_Space_of_2_Factors_is_Connected_iff_Factors_are_Connected | [
"Product Spaces",
"Connected Topological Spaces"
] | [
"Definition:Topological Space",
"Definition:Product Space (Topology)",
"Definition:Connected Topological Space",
"Definition:Connected Topological Space"
] | [
"Definition:Connected Topological Space",
"Projection from Product Topology is Continuous",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Projection (Mapping Theory)",
"Continuous Image of Connected Space is Connected",
"Definition:Connected Topological Space",
"Definition:Connected... |
proofwiki-1894 | Set between Connected Set and Closure is Connected | Let $T$ be a topological space.
Let $H$ be a connected set of $T$.
Let $H \subseteq K \subseteq H^-$, where $H^-$ denotes the closure of $H$.
Then $K$ is connected. | Let $D$ be the discrete space $\set {0, 1}$.
Let $f: K \to D$ be an arbitrary continuous mapping.
From Continuity of Composite with Inclusion, the restriction $f \restriction_H$ is continuous.
We have that:
:$H$ is connected
:$f \restriction_H$ is continuous
Thus by definition of connected set:
:$f \sqbrk H = \set 0$ o... | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $H$ be a [[Definition:Connected Set (Topology)|connected set]] of $T$.
Let $H \subseteq K \subseteq H^-$, where $H^-$ denotes the [[Definition:Closure (Topology)|closure]] of $H$.
Then $K$ is [[Definition:Connected Set (Topology)|connected]]. | Let $D$ be the [[Definition:Discrete Topology|discrete space]] $\set {0, 1}$.
Let $f: K \to D$ be an arbitrary [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
From [[Continuity of Composite with Inclusion]], the [[Definition:Restriction of Mapping|restriction]] $f \restriction_H$ is [[Defi... | Set between Connected Set and Closure is Connected/Proof 1 | https://proofwiki.org/wiki/Set_between_Connected_Set_and_Closure_is_Connected | https://proofwiki.org/wiki/Set_between_Connected_Set_and_Closure_is_Connected/Proof_1 | [
"Set between Connected Set and Closure is Connected",
"Set Closures",
"Connected Sets (Topology)"
] | [
"Definition:Topological Space",
"Definition:Connected Set (Topology)",
"Definition:Closure (Topology)",
"Definition:Connected Set (Topology)"
] | [
"Definition:Discrete Topology",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Continuity of Composite with Inclusion",
"Definition:Restriction/Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Connected Set (Topology)",
"Definition:Continuous Mapping (Topology)/Every... |
proofwiki-1895 | Set between Connected Set and Closure is Connected | Let $T$ be a topological space.
Let $H$ be a connected set of $T$.
Let $H \subseteq K \subseteq H^-$, where $H^-$ denotes the closure of $H$.
Then $K$ is connected. | Let $T_K = \struct {K, \tau_K}$ be the topological subspace of $T$ whose underlying set is $K$.
Let $\map {\cl_K} H$ denote the closure of $H$ in $K$.
From Closure of Subset in Subspace:
:$\map {\cl_K} H = K \cap H^-$
By hypothesis:
:$K \subseteq H^-$
and so by Intersection with Subset is Subset:
:$\map {\cl_K} H = K$
... | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $H$ be a [[Definition:Connected Set (Topology)|connected set]] of $T$.
Let $H \subseteq K \subseteq H^-$, where $H^-$ denotes the [[Definition:Closure (Topology)|closure]] of $H$.
Then $K$ is [[Definition:Connected Set (Topology)|connected]]. | Let $T_K = \struct {K, \tau_K}$ be the [[Definition:Topological Subspace|topological subspace]] of $T$ whose [[Definition:Underlying Set of Topological Space|underlying set]] is $K$.
Let $\map {\cl_K} H$ denote the [[Definition:Closure (Topology)|closure]] of $H$ in $K$.
From [[Closure of Subset in Subspace]]:
:$\map... | Set between Connected Set and Closure is Connected/Proof 2 | https://proofwiki.org/wiki/Set_between_Connected_Set_and_Closure_is_Connected | https://proofwiki.org/wiki/Set_between_Connected_Set_and_Closure_is_Connected/Proof_2 | [
"Set between Connected Set and Closure is Connected",
"Set Closures",
"Connected Sets (Topology)"
] | [
"Definition:Topological Space",
"Definition:Connected Set (Topology)",
"Definition:Closure (Topology)",
"Definition:Connected Set (Topology)"
] | [
"Definition:Topological Subspace",
"Definition:Underlying Set/Topological Space",
"Definition:Closure (Topology)",
"Closure of Subset in Subspace",
"Definition:By Hypothesis",
"Intersection with Subset is Subset",
"Definition:Discrete Topology",
"Definition:Continuous Mapping (Topology)/Everywhere",
... |
proofwiki-1896 | Path-Connected Space is Connected | Let $T$ be a topological space which is path-connected.
Then $T$ is connected. | {{AimForCont}} $X \mid Y$ is a separation of $T$.
As $X$ and $Y$ are non-empty, we can find $x \in X$ and $y \in Y$.
As $T$ is path-connected, there exists a path $f : \closedint 0 1 \to T$ with initial point $x$ and final point $y$.
Subset of Real Numbers is Interval iff Connected shows that $\closedint 0 1$ is connec... | Let $T$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Path-Connected Space|path-connected]].
Then $T$ is [[Definition:Connected Topological Space|connected]]. | {{AimForCont}} $X \mid Y$ is a [[Definition:Separation (Topology)|separation]] of $T$.
As $X$ and $Y$ are [[Definition:Non-Empty Set|non-empty]], we can find $x \in X$ and $y \in Y$.
As $T$ is [[Definition:Path-Connected Space|path-connected]], there exists a [[Definition:Path (Topology)|path]] $f : \closedint 0 1 \t... | Path-Connected Space is Connected/Proof 1 | https://proofwiki.org/wiki/Path-Connected_Space_is_Connected | https://proofwiki.org/wiki/Path-Connected_Space_is_Connected/Proof_1 | [
"Path-Connected Space is Connected",
"Path-Connected Spaces",
"Connected Topological Spaces",
"Sequence of Implications of Connectedness Properties"
] | [
"Definition:Topological Space",
"Definition:Path-Connected/Topological Space",
"Definition:Connected Topological Space"
] | [
"Definition:Separation (Topology)",
"Definition:Non-Empty Set",
"Definition:Path-Connected/Topological Space",
"Definition:Path (Topology)",
"Definition:Path (Topology)/Initial Point",
"Definition:Path (Topology)/Final Point",
"Subset of Real Numbers is Interval iff Connected",
"Definition:Connected S... |
proofwiki-1897 | Path-Connected Space is Connected | Let $T$ be a topological space which is path-connected.
Then $T$ is connected. | Let $D$ be the discrete space $\set {0, 1}$.
Let $T$ be path-connected.
Let $f: T \to D$ be a continuous surjection.
Let $x, y \in T: \map f x = 0, \map f y = 1$.
Let $I \subset \R$ be the closed real interval $\closedint 0 1$.
Let $g: I \to T$ be a path from $x$ to $y$.
Then by Composite of Continuous Mappings is Cont... | Let $T$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Path-Connected Space|path-connected]].
Then $T$ is [[Definition:Connected Topological Space|connected]]. | Let $D$ be the [[Definition:Discrete Topology|discrete space]] $\set {0, 1}$.
Let $T$ be [[Definition:Path-Connected Space|path-connected]].
Let $f: T \to D$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Surjection|surjection]].
Let $x, y \in T: \map f x = 0, \map f y = 1$.
Le... | Path-Connected Space is Connected/Proof 2 | https://proofwiki.org/wiki/Path-Connected_Space_is_Connected | https://proofwiki.org/wiki/Path-Connected_Space_is_Connected/Proof_2 | [
"Path-Connected Space is Connected",
"Path-Connected Spaces",
"Connected Topological Spaces",
"Sequence of Implications of Connectedness Properties"
] | [
"Definition:Topological Space",
"Definition:Path-Connected/Topological Space",
"Definition:Connected Topological Space"
] | [
"Definition:Discrete Topology",
"Definition:Path-Connected/Topological Space",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Surjection",
"Definition:Real Interval/Closed",
"Definition:Path (Topology)",
"Composite of Continuous Mappings is Continuous",
"Definition:Continuous Mappi... |
proofwiki-1898 | Hausdorff Paradox | There exists a disjoint decomposition of the sphere $\mathbb S^2$ into four sets $A, B, C, Q$ such that $A, B, C, B \cup C$ are all congruent and $Q$ is countable. | Let $R \subset \map {\mathbb {SO} } 3$ be the group generated by the $\pi$ and $\dfrac {2 \pi} 3$ rotations around different axes.
The elements:
:$\psi = \begin {pmatrix} -\tfrac 1 2 & \tfrac {\sqrt 3} 2 & 0 \\ -\tfrac {\sqrt 3} 2 & -\tfrac 1 2 & 0 \\ 0 & 0 & 1 \\ \end {pmatrix}$
:$\phi = \begin {pmatrix} -\map \cos \v... | There exists a disjoint [[Definition:Decomposition (Topology)|decomposition]] of the [[Definition:Sphere (Topology)|sphere]] $\mathbb S^2$ into four sets $A, B, C, Q$ such that $A, B, C, B \cup C$ are all [[Definition:Congruence (Metric Spaces)|congruent]] and $Q$ is [[Definition:Countable|countable]]. | Let $R \subset \map {\mathbb {SO} } 3$ be the [[Definition:Group|group]] [[Definition:Generator of Group|generated]] by the $\pi$ and $\dfrac {2 \pi} 3$ rotations around different axes.
The elements:
:$\psi = \begin {pmatrix} -\tfrac 1 2 & \tfrac {\sqrt 3} 2 & 0 \\ -\tfrac {\sqrt 3} 2 & -\tfrac 1 2 & 0 \\ 0 & 0 & 1 \\... | Hausdorff Paradox/Proof 1 | https://proofwiki.org/wiki/Hausdorff_Paradox | https://proofwiki.org/wiki/Hausdorff_Paradox/Proof_1 | [
"Hausdorff Paradox",
"Topology",
"Veridical Paradoxes",
"Antinomies"
] | [
"Definition:Decomposable Set",
"Definition:Sphere/Topology",
"Definition:Congruence (Metric Spaces)",
"Definition:Countable Set"
] | [
"Definition:Group",
"Definition:Generator of Group",
"Definition:Basis (Topology)",
"Definition:Identity Mapping",
"Definition:Countable Set",
"Set of Orbits forms Partition",
"Definition:Set Partition",
"Definition:Orbit (Group Theory)",
"Definition:Orbit (Group Theory)",
"Definition:Group Action... |
proofwiki-1899 | Hausdorff Paradox | There exists a disjoint decomposition of the sphere $\mathbb S^2$ into four sets $A, B, C, Q$ such that $A, B, C, B \cup C$ are all congruent and $Q$ is countable. | === Lemma $1$ ===
{{:Hausdorff Paradox/Lemma 1}}{{qed|lemma}}
=== Lemma $2$ ===
{{:Hausdorff Paradox/Lemma 2}}{{qed|lemma}}
Let $Q$ be the set of all fixed points of $\Bbb S^2$ of all rotations $\alpha \in G$.
Each such $\alpha$ has $2$ fixed points.
Hence $Q$ is countable.
The set $\Bbb S^2 \setminus Q$ is a disjoint ... | There exists a disjoint [[Definition:Decomposition (Topology)|decomposition]] of the [[Definition:Sphere (Topology)|sphere]] $\mathbb S^2$ into four sets $A, B, C, Q$ such that $A, B, C, B \cup C$ are all [[Definition:Congruence (Metric Spaces)|congruent]] and $Q$ is [[Definition:Countable|countable]]. | === [[Hausdorff Paradox/Lemma 1|Lemma $1$]] ===
{{:Hausdorff Paradox/Lemma 1}}{{qed|lemma}}
=== [[Hausdorff Paradox/Lemma 2|Lemma $2$]] ===
{{:Hausdorff Paradox/Lemma 2}}{{qed|lemma}}
Let $Q$ be the [[Definition:Set|set]] of all [[Definition:Fixed Point|fixed points]] of $\Bbb S^2$ of all [[Definition:Space Rotatio... | Hausdorff Paradox/Proof 2 | https://proofwiki.org/wiki/Hausdorff_Paradox | https://proofwiki.org/wiki/Hausdorff_Paradox/Proof_2 | [
"Hausdorff Paradox",
"Topology",
"Veridical Paradoxes",
"Antinomies"
] | [
"Definition:Decomposable Set",
"Definition:Sphere/Topology",
"Definition:Congruence (Metric Spaces)",
"Definition:Countable Set"
] | [
"Hausdorff Paradox/Lemma 1",
"Hausdorff Paradox/Lemma 2",
"Definition:Set",
"Definition:Fixed Point",
"Definition:Rotation (Geometry)/Space",
"Definition:Fixed Point",
"Definition:Countable Set",
"Definition:Set",
"Definition:Disjoint Union",
"Definition:Orbit (Group Theory)",
"Definition:Set",
... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.