Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 4 / 14A (Metric and Topological Spaces)

The closure $\mathrm{cl}(A)$ is the intersection of all closed subsets containing $A$; equivalently it consists of the points whose every open neighbourhood meets $A$.

Suppose $f$ is continuous and $x\in \mathrm{cl}(A)$. For any open neighbourhood $V$ of $f(x)$, continuity makes $f^{-1}(V)$ an open neighbourhood of $x$, so it meets $A$. Hence $V$ meets $f(A)$, showing $f(x)\in \mathrm{cl}(f(A))$. This proves $f(\mathrm{cl}(A))\subseteq \mathrm{cl}(f(A))$.

Conversely suppose this inclusion holds for every subset $A$. For a closed subset $C\subseteq Y$, let $A=f^{-1}(C)$. Then $f(A)\subseteq C$ and therefore $\mathrm{cl}(f(A))\subseteq C$. By the assumed inclusion, $f(\mathrm{cl}(A))\subseteq C$, so $\mathrm{cl}(A)\subseteq f^{-1}(C)=A$. Since $A\subseteq \mathrm{cl}(A)$ always, $A$ is closed. Thus every closed set has closed preimage, equivalently every open set has open preimage. This is continuity, proving the claimed equivalence.

For each integer $n\ge 1$, the open balls $\{B(x,1/n):x\in M\}$ cover the compact metric space $M$. Compactness supplies a finite subcover with centres forming a finite set $S_n$. Define $S={\bigcup }_{n\ge 1}{S}_n$, a countable union of finite sets and therefore countable.

For any $x\in M$ and $\epsilon >0$, choose $n$ with $1/n<\epsilon$. One of the covering balls contains $x$, so there is $s\in S_n\subseteq S$ with $d(x,s)<1/n<\epsilon$. Every open neighbourhood of $x$ therefore meets $S$, giving $x\in \mathrm{cl}(S)$. Thus $\mathrm{cl}(S)=M$, and $S$ is a countable dense subset. If $M$ is empty, simply take $S=\varnothing$.