Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 21G (Linear Analysis)

Let $A,B$ be disjoint closed subsets of the metric space $(X,d)$; the cases where one is empty are immediate. The distance functionsare continuous, because each is 1-Lipschitz. Their sum is strictly positive: if both distances vanished, closedness would put $x$ in $A\cap B$. Thereforeis continuous, equals zero on $A$, and equals one on $B$. The setsare disjoint open neighbourhoods of $A$ and $B$. Thus every metric space is a normal topological space; this is the normality of every metric space.

The Urysohn lemma states that if $A$ and $B$ are disjoint closed subsets of a normal space $X$, there is a continuous $u:X\to [0,1]$ with $u=0$ on $A$ and $u=1$ on $B$.

The Tietze extension theorem states that if $A$ is a closed subset of a normal space $X$, then every continuous $f:A\to \mathbb{R}$ has a continuous extension $F:X\to \mathbb{R}$. In the bounded form, if $f(A)\subseteq [a,b]$, the extension can be chosen with $F(X)\subseteq [a,b]$.

If $X$ is compact and $f:X\to \mathbb{R}$ is continuous, then $f(X)$ is compact by the continuous image of a compact space theorem. Every compact subset of $\mathbb{R}$ is bounded, so $f$ is bounded.

Conversely, suppose the metric space $X$ is not compact. For metric spaces, compactness is equivalent to sequential compactness, so there is a sequence of distinct points $(x_n)$ having no convergent subsequence. The setis closed: an accumulation point would supply a convergent subsequence. It is also discrete for the same reason. Consequently the functionis continuous. By part (a), $X$ is normal, and the Tietze extension theorem extends $g$ to a continuous $G:X\to \mathbb{R}$. Since $G(x_n)=n$, this extension is unbounded. Therefore, if every continuous real-valued function on $X$ is bounded, $X$ must be compact. This is the bounded-continuous-function characterization of compact metric spaces.