Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 2 / 10F (Analysis and Topology)

A topological space is compact if every open cover has a finite subcover. It is Hausdorff if every pair of distinct points has disjoint open neighbourhoods.

Let $Y$ be closed in compact $X$, and let $\{U_i\}$ be an open cover of $Y$ by sets open in $X$. Adding the open set $X\setminus Y$ gives a cover of $X$, which has a finite subcover. Removing $X\setminus Y$ leaves a finite subcover of $Y$. Thus every closed subspace of a compact space is compact.

Now let $A,B$ be disjoint closed subsets of a compact Hausdorff space. They are compact. For each $a\in A$ and $b\in B$, choose disjoint open sets $U_{a,b}\ni a$ and $V_{a,b}\ni b$. Fixing $a$, finitely many $V_{a,b_j}$ cover $B$. PutThen $U_a$ contains $a$, $V_a$ contains $B$, and they are disjoint. Finitely many $U_{a_i}$ cover $A$. Thereforeare disjoint open neighbourhoods of $A$ and $B$. This proves the normality of a compact Hausdorff space.

Let $X$ be compact Hausdorff, let $x\in X$, and let $U$ be a neighbourhood of $x$. Choose an open $O$ with $x\in O\subseteq U$. Apply the separation result from part (a) to the disjoint closed sets $\{x\}$ and $X\setminus O$. There is an open $V\ni x$ whose closure lies in $O$. Thenis compact, contains the neighbourhood $V$ of $x$, and lies in $U$. Thus $X$ is locally compact.

Conversely, suppose $X$ is locally compact Hausdorff and $A\cap K$ is closed in every compact $K\subseteq X$. For $x\notin A$, choose a compact neighbourhood $K$ of $x$ and an open set $U$ withSince $A\cap K$ is closed in $K$, there is an open set $W\subseteq X$ such thatThen $U\cap W$ is an open neighbourhood of $x$ disjoint from $A$. Thus $X\setminus A$ is open andThis is the compactly detected closed-set theorem in a locally compact Hausdorff space.