Continuity via closed sets

Statement

Let $X$ and $Y$ be topological spaces and let $f \colon X \to Y$ be a function. Then $f$ is continuous if and only if for each closed set $F$ of $Y$, the preimage $f^{-1}(F)$ is closed in $X$.

Remarks

Proof: preimages commute with complements, that is, $f^{-1}(Y \setminus A) = X \setminus f^{-1}(A)$ for each $A \subseteq Y$. Then use that the complements of open sets are closed. For example, for a continuous $f \colon \mathbb{R} \to \mathbb{R}$, the set of zeros $f^{-1}(\{0\})$ is closed.

Explanations

L1 human

Continuity can be tested with closed sets as well as with open sets. This form is useful to show that a set given by an equation, such as $\{x : f(x) = 0\}$, is closed.