Hausdorff space: Difference between revisions

Latest revision as of 23:26, 16 July 2026

A Hausdorff space (or $\text{[math]}$ 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

Definition.

A topological space $\text{[math]}$ is Hausdorff, if for any two points $\text{[math]}$ , there exists two disjoint open sets $\text{[math]}$ , $\text{[math]}$ , such that $\text{[math]}$ and $\text{[math]}$ .

Equivalent Definitions

Any convergent sequence in $\text{[math]}$ has at most one limit.

Properties

Property.

Subspaces of Hausdorff spaces are Hausdorff.

Proof.

Let $\text{[math]}$ with $\text{[math]}$ Hausdorff. For $\text{[math]}$ , there exist disjoint open sets $\text{[math]}$ with $\text{[math]}$ and $\text{[math]}$ . Then $\text{[math]}$ and $\text{[math]}$ are disjoint open sets in $\text{[math]}$ containing $\text{[math]}$ and $\text{[math]}$ . □

Property.

Finite products of Hausdorff spaces are Hausdorff.

Proof.

Let $\text{[math]}$ be Hausdorff. Consider points $\text{[math]}$ . There exists an index $\text{[math]}$ with $\text{[math]}$ .

Since $\text{[math]}$ is Hausdorff, choose disjoint open sets $\text{[math]}$ containing $\text{[math]}$ and $\text{[math]}$ . Then $\text{[math]}$ and $\text{[math]}$ are disjoint open sets containing the two points. □

Property.

Compact subsets of Hausdorff spaces are closed.

Proof.

Let $\text{[math]}$ be compact and $\text{[math]}$ Hausdorff. For any $\text{[math]}$ , for each $\text{[math]}$ choose disjoint open sets $\text{[math]}$ and $\text{[math]}$ .

The collection $\text{[math]}$ covers $\text{[math]}$ . By compactness, finitely many $\text{[math]}$ cover $\text{[math]}$ . Then

$\text{[math]}$

is open, contains $\text{[math]}$ , and is disjoint from $\text{[math]}$ . Hence $\text{[math]}$ is closed. □

Examples

Example.

Every metric space is a Hausdorff space.

Terminology

The following table lists the principal terms used in this article.

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Hausdorff space	espace de Hausdorff	hausdorff-Raum	ハウスドルフ空間	Hausdorff 空间 · Hausdorff 空間	5 languages
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@@ Line 1: / Line 1: @@
+{{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.
 == 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>.
+}}
 === 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>.
+{{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 | 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 ==
-Every metric space is a Hausdorff space.
+{{Example
-{{Proof|proof=<div style="float:right; width:320px; margin:0 0 0.5em 1em;">
+ |example=Every [[metric space]] is a Hausdorff space.
-<kroki lang="tikz">
+ |proof=<div style="float:right; width:320px; margin:0 0 0.5em 1em;">
-\documentclass[tikz,border=8pt]{standalone}
+<tikz>
-\usetikzlibrary{calc}
-\usepackage{amsmath}
-\begin{document}
-\begin{tikzpicture}[scale=1.3]
 \def\r{1.5}
@@ Line 45: / Line 56: @@
 node[midway, above, sloped] {\footnotesize $r=\dfrac{d(x,y)}{2}$};
-\end{tikzpicture}
+</tikz>
-\end{document}
+</div>
-</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>.
@@ Line 55: / Line 64: @@
 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 ==
+* [[Topological space]]
+* [[Convergence]]
+{{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
+}}
+}}