Theorem 1

Let $f$ be a mapping of a topological space $X$ into a topological space $X'$. Then the following statements are equivalent:

a) $f$ is continuous in $X$.

b) For every subset $A$ of $X$, $f(\overline{A}) \subset \overline{f(A)}$.

c) The inverse image under $f$ of every closed subset of $X'$ is a closed subset of $X$.

d) The inverse image under $f$ of every open subset of $X'$ is an open subset of $X$.

We have already seen that a) implies b) (Proposition 1). To show that b) implies c), let $F'$ be a closed subset of $X'$ and let $F = \overline{f^{-1}(F')}$; then by hypothesis $f(\overline{F}) \subset \overline{f(F)} \subset \overline{F'} = F'$, hence $\overline{F} \subset \overline{f^{-1}(F')} = F \subset \overline{F}$, so that $F = \overline{F}$ and $F$ is closed. By virtue of the relation $\overline{f^{-1}(A')} = \overline{f^{-1}(CA')}$ for every subset $A'$ of $X'$, c) implies d). Finally, suppose that d) is satisfied. Let $x$ be any point of $X$ and let $V'$ be any neighbourhood of $f(x)$ in $X'$; then there is an open set $A'$ in $X'$ for which

and hence $x \in \overline{f^{-1}(A')} \subset \overline{f^{-1}(V')}$. By hypothesis, $\overline{f^{-1}(A')}$ is open in $X$, so that $\overline{f^{-1}(V')}$ is a neighbourhood of $x$ in $X$. Thus d) implies a).