Proof of continuity via closed sets

Proof

We first show that preimages commute with complements.

1. Let $X$ and $Y$ be topological spaces and let $f \colon X \to Y$.
2. Let $A \subseteq Y$ and $x \in X$. Then $x$ lies in the preimage $f^{-1}(Y \setminus A)$ if and only if $f(x) \notin A$.
3. This holds if and only if $x \notin f^{-1}(A)$. So $f^{-1}(Y \setminus A) = X \setminus f^{-1}(A)$, with $X \setminus f^{-1}(A)$ the complement.
4. Suppose $f$ is continuous, and let $F$ be a closed set of $Y$.
5. By the definition of a closed set, $Y \setminus F$ is open. By continuity, $f^{-1}(Y \setminus F)$ is open in $X$.
6. By step 3, $X \setminus f^{-1}(F)$ is open. By the definition of a closed set, $f^{-1}(F)$ is closed.
7. Conversely, suppose that each closed $F$ of $Y$ has a closed preimage. Let $V$ be open in $Y$.
8. By complement of an open set is closed, $Y \setminus V$ is closed in $Y$. By hypothesis, $f^{-1}(Y \setminus V)$ is closed in $X$.
9. By step 3, $X \setminus f^{-1}(V)$ is closed. By the same result, $f^{-1}(V)$ is open.
10. As $V$ was arbitrary, $f$ is continuous.