10.5 Hausdorff spaces

中文

In some topologies, for example the indiscrete topology, there are so few open sets that different points cannot be separated by non-intersecting neighbourhoods. To remedy this situation, conditions known as separation axioms are sometimes imposed on topological spaces. One of the most common of these is the Hausdorff condition: for every pair of points $x$, $y\in X$ there exist open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $U\cap V=\varnothing$ A topological space satisfying this property is known as a Hausdorff space. In an intuitive sense, no pair of distinct points of a Hausdorff space are ‘arbitrarily close to each other.

A typical ‘nice’ property of Hausdorff spaces is the fact that the limit of any convergent sequence $x_n\to x$ , defined in Problem 10.7, is unique. Suppose, for example, that $x_n\to x$ and $x_n\to x^{\prime }$ in a Hausdorff space $X$. If $x\ne x^{\prime }$ let $U$ and $U^{\prime }$ be disjoint open neighbourhoods such that $x\in U$ and $x^{\prime }\in U^{\prime }$ , and $N$ an integer such that $x_n\in U$ for all $n>N$ . Since $x_n\notin U^{\prime }$ for all $n>N$ the sequence $x_n$ cannot converge to $x^{\prime }$ . Hence $x=x^{\prime }$

In a Hausdorff space every singleton set $\{x\}$ is a closed set, for let $Y=X-\{x\}$ be its complement. Every point $y\in Y$ has an open neighbourhood $U_y$ that does not intersect some open neighbourhood of $x$. In particular $x\notin U_y$. By (Top3) the union of all these open neighbourhoods, $\begin{array}{r}Y={\bigcup }_{y\in Y}{U}_y=X-\{x\}\end{array}$ , is open. Hence $\{x\}=X-Y$ is closed since it is the complement of an open set.

Exercise

Show that on a finite set $X,$ the only Hausdorff topology is the discrete topology. For this reason, finite topologies are of limited interest.

Theorem 10.4 · Metric spaces are Hausdorff

Every metric space $(X,d)$ is a Hausdorff space.

Proof

Let $x,y\in X$ be any pair of unequal points and let $\begin{array}{r}ϵ=\frac{1}{4}d(x,y)\end{array}$ . The open balls $U=B_{ϵ}(x)$ and $V=B_{ϵ}(y)$ are open neighbourhoods of $x$ and $y$ respectively. Their intersection is empty, for if $z\in U\cap V$ then $d(x,z)<ϵ$ and $d(y,z)<ϵ$ , which contradicts the triangle inequality (Met4),

$$d(x,y)\le d(x,z)+d(z,y)\le 2ϵ<\frac{1}{2}d(x,y).$$

An immediate consequence of this theorem is that the standard topology on ${\mathbb{R}}^n$ is Hausdorff for all $n>0$

Theorem 10.5 · Continuous injections and the Hausdorff property

If $X$ and $Y$ are topological spaces and $f:X\to Y$ is a one-to-one continuous mapping, then $X$ is Hausdorff if $Y$ is Hausdorff.

Proof

Let $x$ and $x^{\prime }$ be any pair of distinct points in $X$ and set $y=f(x),y^{\prime }=f(x^{\prime })$ . Since $f$ is one-to-one these are distinct points of $Y$. If $Y$ is Hausdorff there exist non-intersecting open neighbourhoods $U_y$ and $U_{y^{\prime }}$ in $Y$ of $y$ and $y^{\prime }$ respectively. The inverse images of these sets under $f$ are open neighbourhoods of $x$ and $x^{\prime }$ respectively that are non-intersecting, since $f^{-1}(U_y)\cap f^{-1}(U_{y^{\prime }})=f^{-1}(U_y\cap U_{y^{\prime }})=f^{-1}(\varnothing )=\varnothing .$

This shows that the Hausdorff condition is a genuine topological property, invariant under topological transformations, for if $f:X\to Y$ is a homeomorphism then $f^{-1}:Y\to X$ is continuous and one-to-one.

Corollary 10.6 · Subspaces of Hausdorff spaces

Any subspace of a Hausdorff space is Hausdorff in the relative topology

Proof

Let $A$ be any subset of a topological space $X$. In the relative topology the inclusion map $i_A:A\to X$ is continuous. Since it is one-to-one, Theorem 10.5 implies that $A$ is Hausdorff.

Theorem 10.7 · Products of Hausdorff spaces

If $X$ and $Y$ are Hausdorff topological spaces then their topological product $X\times Y$ is Hausdorff.

Proof

Let $(x,y)$ and $(x^{\prime },y^{\prime })$ be any distinct pair of points in $X\times Y$ , so that either $x\ne x^{\prime }$ or $y\ne y^{\prime }$ . Suppose that $x\ne x^{\prime }.$ . There then exist open sets $U$ and $U^{\prime }$ in $X$ such that $x\in U{x}^{\prime }\in U^{\prime }$ and $U\cap U^{\prime }=\varnothing$ . The sets $U\times Y$ and $U^{\prime }\times Y$ are disjoint open neighbourhoods of $(x,y)$ and $(x^{\prime },y^{\prime })$ respectively. Similarly, if $y\ne y^{\prime }\mathbf{a}$ pair of disjoint neighbourhoods of the form $X\times V$ and $X\times V^{\prime }$ can be found that separate the two points.

Problems

Problem 10.14

If $Y$ is a Hausdorff topological space show that every continuous map $f:X\to Y$ from a topological space $X$ with indiscrete topology into $Y$ is a constant map; that is, a map of the form $f(x)=y_0$ where $y_0$ is a fixed element of $Y$.

Problem 10.15

Show that if $f:X\to Y$ and $g:X\to Y$ are continuous maps from a topological space $X$ into a Hausdorff space $Y$ then the set of points $A$ on which these maps agree, $A=\{x\in X\mid f(x)=g(x)\}$ , is closed. If $A$ is a dense subset of $X$ show that $f=g$