Past exam of the topology course of the University of Oxford / 2014 / 1 / a

The closure $\overline{A}$ is the intersection of every closed subset of $X$ containing $A$. Thus it is the smallest such closed subset. Equivalently,For equivalence, if an open neighbourhood misses $A$, its complement is a closed superset of $A$ missing $x$. Conversely, a closed superset missing $x$ has an open complement that is a neighbourhood of $x$ disjoint from $A$.

Suppose $f$ is continuous and $x\in \overline{A}$. Every open neighbourhood $V$ of $f(x)$ has an open preimage containing $x$, which meets $A$. Hence $V$ meets $f(A)$, proving $f(x)\in \overline{f(A)}$.

Conversely, suppose the indicated inclusion holds for every subset. To prove continuity, take any closed set $C\subseteq Y$ and put $A=f^{-1}(C)$. Since $f(A)\subseteq C$ and $C$ is closed,Thus $\overline{A}\subseteq f^{-1}(C)=A$, so $A$ is closed. Every closed set has closed preimage, which is the closed-set characterization of continuity.This is the closure characterization of continuity; it requires no Hausdorff or metric hypothesis.