Hausdorff space

A Hausdorff space separates any two distinct points by disjoint open neighborhoods.

A Hausdorff space is completely regular if a point $x$ outside a closed set $F$ can be separated from $F$ by a continuous function $u:X\to [0,1]$ with $u(x)=1$ and $u{|}_F=0$. These functions determine the original topology: for any neighborhood $O$ of $x$, choose $F=X\setminus O$ to obtain $x\in \{u>1/2\}\subseteq O$. This is exactly the point-separation property needed for the evaluation embedding into the weak-star dual of bounded continuous functions.

Let $K\subseteq X$, where the ambient topological space $X$ is Hausdorff and $K$ is compact in its subspace topology. Then $K$ is a closed set in $X$: the complement $X\setminus K$ is an open set in $X$. The ambient space need not be compact. For example, $[0,1]$ is a compact closed interval in the non-compact Hausdorff real line.^[ref]

The ambient Hausdorff hypothesis matters even when $K$ itself is Hausdorff. In the Sierpiński space $S=\{0,1\}$, the open sets are $\varnothing ,\{1\},S$, so the closed sets are $\varnothing ,\{0\},S$. Thus $K=\{1\}$ is compact and Hausdorff in its one-point subspace topology, but is not closed in $S$: its complement $\{0\}$ is not an open set. Every open neighborhood of $0$ contains $1$, so separation from $K$ fails. In contrast, the discrete topology on the same underlying set is Hausdorff and makes both singletons closed.

The diagram lists every open set and closed set, so the distinction can be checked directly. All subsets in both panels are compact: for any open cover of a finite set, choose one covering open set per point, obtaining a finite subcover. Compactness alone therefore cannot distinguish the two panels.

Here is the finite-selection mechanism in the general proof. For $x\in X\setminus K$ and each $k\in K$, the Hausdorff property gives disjoint open neighborhoods $U_k$ of $k$ and $V_k$ of $x$. The $U_k$ give an ambient open cover of $K$. Select a finite subcover $U_{k_1},\dots ,U_{k_m}$ and setThen $K\subseteq U$, $x\in V$, and $U\cap V=\varnothing$. The finite intersection makes $V$ an open set; an infinite intersection would not guarantee this. Thus every point of $X\setminus K$ has an open neighborhood contained in that complement, proving that $K$ is closed. The empty set is already closed.

This theorem combines with the continuous image of a compact space theorem to make every continuous map from a compact space to a Hausdorff space a closed map: a closed subset of its domain is compact, and its image is compact and hence closed in the target. This is the reason the inverse in the compact-to-Hausdorff continuous bijection theorem is continuous. The converse implication, from closed to compact, needs additional hypotheses: the real line is closed in itself but not compact; in a compact space, every closed subset is compact.

In a Hausdorff space, a point $x$ outside a compact subset $K$ can be separated from all of $K$ by disjoint open sets. Choose disjoint open neighbourhoods $U_k$ of $k$ and $V_k$ of $x$ for each $k\in K$. Finitely many $U_k$ cover $K$ by compactness. Their union and the matching finite intersection of the $V_k$ give the displayed separation. If $K$ is empty, choose $U=\varnothing$, $V=X$. Thus $X\setminus K$ is open, proving that $K$ is closed.

Two disjoint compact subsets of a Hausdorff space have disjoint open neighbourhoods. For every $p\in P$, separation of a point and a compact set gives disjoint open sets $G_p\ni p$ and $H_p\supseteq Q$. Choose finitely many $G_p$ covering $P$, then take their union for $U$ and the matching finite intersection of the $H_p$ for $V$. The compactness of both subsets is used, once in the point-versus-compact set result and once in the outer finite selection. The ambient Hausdorff space need not be compact or a normal space.

A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. It maps closed sets to compact sets, which are closed in the codomain, so it is a closed map and its inverse is continuous.

A space is Hausdorff exactly when its diagonal is closed in its square.