Hausdorff space - Revision history

InfernalAtom683 at 15:26, 16 July 2026

2026-07-16T15:26:18Z

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	* [[Convergence]]		* [[Convergence]]

	~~==Terminology==~~
	~~{{Terminology_table\|~~
	~~{{Terminology_table/row \| Hausdorff space \| espace de Hausdorff (espace séparé) \| hausdorff-Raum (hausdorffscher Raum) \| Hausdorff 空间 \| Hausdorff 空間 \| ハウスドルフ空間 }}~~
	}}
	{{Terminology		{{Terminology
	\|rows=		\|rows=

InfernalAtom683 at 15:25, 16 July 2026

2026-07-16T15:25:41Z

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	== Definitions ==		== Definitions ==
			{{Definition\|
	A topological space <math>(X,\tau)</math> is Hausdorff, if for any two points <math>x,y\in X</math>, there exists two disjoint open sets <math>U,V\in \tau</math>, <math>U\cap V=\varnothing</math>, such that <math>x\in U</math> and <math>y\in V</math>.		A topological space <math>(X,\tau)</math> is Hausdorff, if for any two points <math>x,y\in X</math>, there exists two disjoint open sets <math>U,V\in \tau</math>, <math>U\cap V=\varnothing</math>, such that <math>x\in U</math> and <math>y\in V</math>.
			}}
	=== Equivalent Definitions ===		=== Equivalent Definitions ===
	Any convergent [[sequence]] in <math>X</math> has at most one limit.		Any convergent [[sequence]] in <math>X</math> has at most one limit.

	== Properties ==		== Properties ==
	{{Property\|property=Subspaces of Hausdorff spaces are Hausdorff.~~}}{{Proof~~\|proof=Let <math>Y \subseteq X</math> with <math>X</math> Hausdorff. For <math>y_1, y_2 \in Y, y_1 \neq y_2</math>, there exist disjoint open sets <math>U, V \subseteq X</math> with <math>y_1 \in U</math> and <math>y_2 \in V</math>. Then <math>U \cap Y</math> and <math>V \cap Y</math> are disjoint open sets in <math>Y</math> containing <math>y_1</math> and <math>y_2</math>.}}{{Property\|property=Finite products of Hausdorff spaces are Hausdorff.~~}}{{Proof~~\|proof=Let <math>X_1, \dots, X_n</math> be Hausdorff. Consider points <math>(x_1, \dots, x_n) \neq (y_1, \dots, y_n)</math>. There exists an index <math>i</math> with <math>x_i \neq y_i</math>.		{{Property
			\|property=Subspaces of Hausdorff spaces are Hausdorff.
			\|proof=Let <math>Y \subseteq X</math> with <math>X</math> Hausdorff. For <math>y_1, y_2 \in Y, y_1 \neq y_2</math>, there exist disjoint open sets <math>U, V \subseteq X</math> with <math>y_1 \in U</math> and <math>y_2 \in V</math>. Then <math>U \cap Y</math> and <math>V \cap Y</math> are disjoint open sets in <math>Y</math> containing <math>y_1</math> and <math>y_2</math>.
			}}

			{{Property
			\|property=Finite products of Hausdorff spaces are Hausdorff.
			\|proof=Let <math>X_1, \dots, X_n</math> be Hausdorff. Consider points <math>(x_1, \dots, x_n) \neq (y_1, \dots, y_n)</math>. There exists an index <math>i</math> with <math>x_i \neq y_i</math>.

			Since <math>X_i</math> is Hausdorff, choose disjoint open sets <math>U_i, V_i \subseteq X_i</math> containing <math>x_i</math> and <math>y_i</math>. Then <math>U_1 \times \dots \times U_n</math> and <math>V_1 \times \dots \times V_n</math> are disjoint open sets containing the two points.}}

	Since <math>X_i</math> is Hausdorff, choose disjoint open sets <math>U_i, V_i \subseteq X_i</math> containing <math>x_i</math> and <math>y_i</math>. Then <math>U_1 \times \dots \times U_n</math> and <math>V_1 \times \dots \times V_n</math> are disjoint open sets containing the two points.}}{{Property\|property=Compact subsets of Hausdorff spaces are closed.~~}}{{Proof~~\|proof=Let <math>K \subseteq X</math> be compact and <math>X</math> Hausdorff. For any <math>x \in X \setminus K</math>, for each <math>y \in K</math> choose disjoint open sets <math>U_y\ni x</math> and <math>V_y \ni y</math>.		{{Property
			\|property=Compact subsets of Hausdorff spaces are closed.
			\|proof=Let <math>K \subseteq X</math> be compact and <math>X</math> Hausdorff. For any <math>x \in X \setminus K</math>, for each <math>y \in K</math> choose disjoint open sets <math>U_y\ni x</math> and <math>V_y \ni y</math>.

	The collection <math>\{V_y \mid y \in K\}</math> covers <math>K</math>. By compactness, finitely many <math>V_{y_1}, \dots, V_{y_n}</math> cover <math>K</math>. Then <math display="block">U = \bigcap_{i=1}^n U_{y_i}</math> is open, contains <math>x</math>, and is disjoint from <math>K</math>. Hence <math>K</math> is closed.}}		The collection <math>\{V_y \mid y \in K\}</math> covers <math>K</math>. By compactness, finitely many <math>V_{y_1}, \dots, V_{y_n}</math> cover <math>K</math>. Then <math display="block">U = \bigcap_{i=1}^n U_{y_i}</math> is open, contains <math>x</math>, and is disjoint from <math>K</math>. Hence <math>K</math> is closed.
			}}

	== Examples ==		== Examples ==
	Every [[metric space]] is a Hausdorff space.		{{Example
	~~{{Proof~~\|proof=		\|example=Every [[metric space]] is a Hausdorff space.
	<div style="float:right; width:320px; margin:0 0 0.5em 1em;">		\|proof=<div style="float:right; width:320px; margin:0 0 0.5em 1em;">
	<tikz>		<tikz>
	\def\r{1.5}		\def\r{1.5}
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	For contradiction, assume there exists <math>z\in B(x,r)\cap B(y,r)</math>, then <math>d(x,z)<r</math> and <math>d(y,z)<r</math>. By triangular inequality, <math display="block">d(x,y)\leq d(x,y)+d(x,z)<r+r=d(x,y),</math> there exists a contradiction.		For contradiction, assume there exists <math>z\in B(x,r)\cap B(y,r)</math>, then <math>d(x,z)<r</math> and <math>d(y,z)<r</math>. By triangular inequality, <math display="block">d(x,y)\leq d(x,y)+d(x,z)<r+r=d(x,y),</math> there exists a contradiction.

	Thus, <math>B(x,r)\cap B(y,r)=\varnothing</math>; therefore <math>(X,d)</math> is Hausdorff.}}		Thus, <math>B(x,r)\cap B(y,r)=\varnothing</math>; therefore <math>(X,d)</math> is Hausdorff.
			}}

	== See also ==		== See also ==
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	{{Terminology_table\|		{{Terminology_table\|
	{{Terminology_table/row \| Hausdorff space \| espace de Hausdorff (espace séparé) \| hausdorff-Raum (hausdorffscher Raum) \| Hausdorff 空间 \| Hausdorff 空間 \| ハウスドルフ空間 }}		{{Terminology_table/row \| Hausdorff space \| espace de Hausdorff (espace séparé) \| hausdorff-Raum (hausdorffscher Raum) \| Hausdorff 空间 \| Hausdorff 空間 \| ハウスドルフ空間 }}
			}}
			{{Terminology
			\|rows=
			{{Terminology/row
			\|english=Hausdorff space
			\|french=espace de Hausdorff
			\|german=hausdorff-Raum
			\|japanese=ハウスドルフ空間
			\|chinese=Hausdorff 空间 · Hausdorff 空間
			\|highlight=yes
			}}
	}}		}}

InfernalAtom683 at 06:35, 11 July 2026

2026-07-11T06:35:34Z

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			{{Topology banner
			\| breadcrumb = {{Breadcrumb nav\|[[Mathepedia]]\|[[Topology]]\|[[General topology]]\|[[Hausdorff space]]}}
			}}
	A '''Hausdorff space''' (or '''<math>T_2</math> space''') is a type of [[topological space]] in which points can be "cleanly separated" by neighborhoods. Specifically, for any two distinct points, there exist disjoint [[Open set\|open sets]] containing each point. Consequently, Hausdorff property ensures that limits of sequences are unique when they exist.		A '''Hausdorff space''' (or '''<math>T_2</math> space''') is a type of [[topological space]] in which points can be "cleanly separated" by neighborhoods. Specifically, for any two distinct points, there exist disjoint [[Open set\|open sets]] containing each point. Consequently, Hausdorff property ensures that limits of sequences are unique when they exist.

InfernalAtom683 at 02:11, 11 July 2026

2026-07-11T02:11:49Z

← Older revision		Revision as of 10:11, 11 July 2026
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	== Examples ==		== Examples ==
	Every metric space is a Hausdorff space.		Every [[metric space]] is a Hausdorff space.
	{{Proof\|proof=		{{Proof\|proof=
	<div style="float:right; width:320px; margin:0 0 0.5em 1em;">		<div style="float:right; width:320px; margin:0 0 0.5em 1em;">

InfernalAtom683 at 08:11, 10 July 2026

2026-07-10T08:11:40Z

← Older revision		Revision as of 16:11, 10 July 2026
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	== Examples ==		== Examples ==
	Every metric space is a Hausdorff space.		Every metric space is a Hausdorff space.
	{{Proof\|proof=<div style="float:right; width:320px; margin:0 0 0.5em 1em;">		{{Proof\|proof=
			<div style="float:right; width:320px; margin:0 0 0.5em 1em;">
	<tikz>		<tikz>
	\def\r{1.5}		\def\r{1.5}
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	</tikz>		</tikz>
			</div>
	Let <math>(X,d)</math> be a metric space, take two distinct points <math>x,y</math> such that <math>d(x,y)>0</math>.		Let <math>(X,d)</math> be a metric space, take two distinct points <math>x,y</math> such that <math>d(x,y)>0</math>.

InfernalAtom683 at 08:09, 10 July 2026

2026-07-10T08:09:17Z

← Older revision		Revision as of 16:09, 10 July 2026
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	Every metric space is a Hausdorff space.		Every metric space is a Hausdorff space.
	{{Proof\|proof=<div style="float:right; width:320px; margin:0 0 0.5em 1em;">		{{Proof\|proof=<div style="float:right; width:320px; margin:0 0 0.5em 1em;">
	<~~kroki lang="~~tikz">		<tikz>
	~~\documentclass[tikz,border=8pt]{standalone}~~
	~~\usetikzlibrary{calc}~~
	~~\usepackage{amsmath}~~
	~~\begin{document}~~
	~~\begin{tikzpicture}[scale=1.3]~~

	\def\r{1.5}		\def\r{1.5}

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	node[midway, above, sloped] {\footnotesize $r=\dfrac{d(x,y)}{2}$};		node[midway, above, sloped] {\footnotesize $r=\dfrac{d(x,y)}{2}$};

	~~\end{tikzpicture}~~		</tikz>
	~~\end{document}~~
	</~~kroki></div~~>

	Let <math>(X,d)</math> be a metric space, take two distinct points <math>x,y</math> such that <math>d(x,y)>0</math>.		Let <math>(X,d)</math> be a metric space, take two distinct points <math>x,y</math> such that <math>d(x,y)>0</math>.

InfernalAtom683 at 07:16, 11 April 2026

2026-04-11T07:16:30Z

← Older revision		Revision as of 15:16, 11 April 2026
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	Since <math>X_i</math> is Hausdorff, choose disjoint open sets <math>U_i, V_i \subseteq X_i</math> containing <math>x_i</math> and <math>y_i</math>. Then <math>U_1 \times \dots \times U_n</math> and <math>V_1 \times \dots \times V_n</math> are disjoint open sets containing the two points.}}{{Property\|property=Compact subsets of Hausdorff spaces are closed.}}{{Proof\|proof=Let <math>K \subseteq X</math> be compact and <math>X</math> Hausdorff. For any <math>x \in X \setminus K</math>, for each <math>y \in K</math> choose disjoint open sets <math>U_y\ni x</math> and <math>V_y \ni y</math>.		Since <math>X_i</math> is Hausdorff, choose disjoint open sets <math>U_i, V_i \subseteq X_i</math> containing <math>x_i</math> and <math>y_i</math>. Then <math>U_1 \times \dots \times U_n</math> and <math>V_1 \times \dots \times V_n</math> are disjoint open sets containing the two points.}}{{Property\|property=Compact subsets of Hausdorff spaces are closed.}}{{Proof\|proof=Let <math>K \subseteq X</math> be compact and <math>X</math> Hausdorff. For any <math>x \in X \setminus K</math>, for each <math>y \in K</math> choose disjoint open sets <math>U_y\ni x</math> and <math>V_y \ni y</math>.

	The collection <math>\{V_y \| y \in K\}</math> covers <math>K</math>. By compactness, finitely many <math>V_{y_1}, \dots, V_{y_n}</math> cover <math>K</math>. Then <math display="block">U = \bigcap_{i=1}^n U_{y_i}</math> is open, contains <math>x</math>, and is disjoint from <math>K</math>. Hence <math>K</math> is closed.}}		The collection <math>\{V_y \mid y \in K\}</math> covers <math>K</math>. By compactness, finitely many <math>V_{y_1}, \dots, V_{y_n}</math> cover <math>K</math>. Then <math display="block">U = \bigcap_{i=1}^n U_{y_i}</math> is open, contains <math>x</math>, and is disjoint from <math>K</math>. Hence <math>K</math> is closed.}}

	== Examples ==		== Examples ==

InfernalAtom683 at 03:43, 6 April 2026

2026-04-06T03:43:50Z

← Older revision		Revision as of 11:43, 6 April 2026
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	Thus, <math>B(x,r)\cap B(y,r)=\varnothing</math>; therefore <math>(X,d)</math> is Hausdorff.}}		Thus, <math>B(x,r)\cap B(y,r)=\varnothing</math>; therefore <math>(X,d)</math> is Hausdorff.}}

			== See also ==

			* [[Topological space]]
			* [[Convergence]]

			==Terminology==
			{{Terminology_table\|
			{{Terminology_table/row \| Hausdorff space \| espace de Hausdorff (espace séparé) \| hausdorff-Raum (hausdorffscher Raum) \| Hausdorff 空间 \| Hausdorff 空間 \| ハウスドルフ空間 }}
			}}

InfernalAtom683: InfernalAtom683 moved page Hausdorffness to Hausdorff space

2026-03-24T01:55:32Z

InfernalAtom683 moved page Hausdorffness to Hausdorff space

← Older revision		Revision as of 09:55, 24 March 2026
(No difference)

InfernalAtom683: Created page with "A '''Hausdorff space''' (or ''' $T_2$ space''') is a type of topological space in which points can be "cleanly separated" by neighborhoods. Specifically, for any two distinct points, there exist disjoint open sets containing each point. Consequently, Hausdorff property ensures that limits of sequences are unique when they exist. == Definitions == A topological space $(X,\tau)$ is Hausdorff, if for any two points $x,y\in X$
2026-03-22T13:20:43Z

Created page with "A '''Hausdorff space''' (or '''<math>T_2</math> space''') is a type of topological space in which points can be "cleanly separated" by neighborhoods. Specifically, for any two distinct points, there exist disjoint open sets containing each point. Consequently, Hausdorff property ensures that limits of sequences are unique when they exist. == Definitions == A topological space <math>(X,\tau)</math> is Hausdorff, if for any two points <math>x,y\in X</math..."

New page
A '''Hausdorff space''' (or '''<math>T_2</math> space''') is a type of [[topological space]] in which points can be "cleanly separated" by neighborhoods. Specifically, for any two distinct points, there exist disjoint [[Open set|open sets]] containing each point. Consequently, Hausdorff property ensures that limits of sequences are unique when they exist.

== Definitions ==
A topological space <math>(X,\tau)</math> is Hausdorff, if for any two points <math>x,y\in X</math>, there exists two disjoint open sets <math>U,V\in \tau</math>, <math>U\cap V=\varnothing</math>, such that <math>x\in U</math> and <math>y\in V</math>.

=== Equivalent Definitions ===
Any convergent [[sequence]] in <math>X</math> has at most one limit.

== Properties ==
{{Property|property=Subspaces of Hausdorff spaces are Hausdorff.}}{{Proof|proof=Let <math>Y \subseteq X</math> with <math>X</math> Hausdorff. For <math>y_1, y_2 \in Y, y_1 \neq y_2</math>, there exist disjoint open sets <math>U, V \subseteq X</math> with <math>y_1 \in U</math> and <math>y_2 \in V</math>. Then <math>U \cap Y</math> and <math>V \cap Y</math> are disjoint open sets in <math>Y</math> containing <math>y_1</math> and <math>y_2</math>.}}{{Property|property=Finite products of Hausdorff spaces are Hausdorff.}}{{Proof|proof=Let <math>X_1, \dots, X_n</math> be Hausdorff. Consider points <math>(x_1, \dots, x_n) \neq (y_1, \dots, y_n)</math>. There exists an index <math>i</math> with <math>x_i \neq y_i</math>.

Since <math>X_i</math> is Hausdorff, choose disjoint open sets <math>U_i, V_i \subseteq X_i</math> containing <math>x_i</math> and <math>y_i</math>. Then <math>U_1 \times \dots \times U_n</math> and <math>V_1 \times \dots \times V_n</math> are disjoint open sets containing the two points.}}{{Property|property=Compact subsets of Hausdorff spaces are closed.}}{{Proof|proof=Let <math>K \subseteq X</math> be compact and <math>X</math> Hausdorff. For any <math>x \in X \setminus K</math>, for each <math>y \in K</math> choose disjoint open sets <math>U_y\ni x</math> and <math>V_y \ni y</math>.

The collection <math>\{V_y | y \in K\}</math> covers <math>K</math>. By compactness, finitely many <math>V_{y_1}, \dots, V_{y_n}</math> cover <math>K</math>. Then <math display="block">U = \bigcap_{i=1}^n U_{y_i}</math> is open, contains <math>x</math>, and is disjoint from <math>K</math>. Hence <math>K</math> is closed.}}

== Examples ==
Every metric space is a Hausdorff space.
{{Proof|proof=<div style="float:right; width:320px; margin:0 0 0.5em 1em;">
<kroki lang="tikz">
\documentclass[tikz,border=8pt]{standalone}
\usetikzlibrary{calc}
\usepackage{amsmath}
\begin{document}
\begin{tikzpicture}[scale=1.3]

\def\r{1.5}

\coordinate (x) at (0,0);

\def\ang{20}

\coordinate (y) at ({2\rcos(\ang)}, {2\rsin(\ang)});

\draw[fill=blue!25!cyan, opacity=0.2, dashed, thick] (x) circle (\r);
\draw[fill=blue!25!cyan, opacity=0.2, dashed, thick] (y) circle (\r);

\fill (x) circle (2pt);
\fill (y) circle (2pt);

\node[below left] at (x) {$x$};
\node[above right] at (y) {$y$};

\draw[black, dashed, thick]
(x) -- ({\rcos(\ang)}, {\rsin(\ang)})
node[midway, above, sloped] {\footnotesize $r=\dfrac{d(x,y)}{2}$};

\end{tikzpicture}
\end{document}
</kroki></div>

Let <math>(X,d)</math> be a metric space, take two distinct points <math>x,y</math> such that <math>d(x,y)>0</math>.

Consider open balls <math>B(x,r)</math> and <math>B(y,r)</math>where <math>r=\frac{d(x,y)}{2}>0</math>. The open balls are both open with <math>x\in B(x,r)</math> and <math>y\in B(y,r)</math>.

For contradiction, assume there exists <math>z\in B(x,r)\cap B(y,r)</math>, then <math>d(x,z)<r</math> and <math>d(y,z)<r</math>. By triangular inequality, <math display="block">d(x,y)\leq d(x,y)+d(x,z)<r+r=d(x,y),</math> there exists a contradiction.

Thus, <math>B(x,r)\cap B(y,r)=\varnothing</math>; therefore <math>(X,d)</math> is Hausdorff.}}