Hausdorffness

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 $x,y\in X$, there exists two disjoint open sets $U,V\in \tau$, $U\cap V=\varnothing$, such that $x\in U$ and $y\in V$.

Equivalent Definitions

Any convergent sequence in $X$ has at most one limit.

Properties

Property

Subspaces of Hausdorff spaces are Hausdorff.

Proof

Let $Y\subseteq X$ with $X$ Hausdorff. For $y_1,y_2\in Y,y_1\ne y_2$, there exist disjoint open sets $U,V\subseteq X$ with $y_1\in U$ and $y_2\in V$. Then $U\cap Y$ and $V\cap Y$ are disjoint open sets in $Y$ containing $y_1$ and $y_2$.

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Property

Finite products of Hausdorff spaces are Hausdorff.

Proof

Let $X_1,\dots ,X_n$ be Hausdorff. Consider points $(x_1,\dots ,x_n)\ne (y_1,\dots ,y_n)$. There exists an index $i$ with $x_i\ne y_i$.

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

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Property

Compact subsets of Hausdorff spaces are closed.

Proof

Let $K\subseteq X$ be compact and $X$ Hausdorff. For any $x\in X\setminus K$, for each $y\in K$ choose disjoint open sets $U_y\ni x$ and $V_y\ni y$.

The collection $\{V_y|y\in K\}$ covers $K$. By compactness, finitely many $V_{y_1},\dots ,V_{y_n}$ cover $K$. Then $$U=\bigcap _{i=1}^n{U}_{y_i}$$ is open, contains $x$, and is disjoint from $K$. Hence $K$ is closed.

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Examples

Every metric space is a Hausdorff space.

Proof

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

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

For contradiction, assume there exists $z\in B(x,r)\cap B(y,r)$, then $d(x,z)<r$ and $d(y,z)<r$. By triangular inequality, $$d(x,y)\le d(x,y)+d(x,z)<r+r=d(x,y),$$ there exists a contradiction.

Thus, $B(x,r)\cap B(y,r)=\varnothing$; therefore $(X,d)$ is Hausdorff.

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