A Function is Continuous iff it takes Closed Sets back to Closed Sets

Last updated Nov 1, 2022

Let $(X, \tau), (Y, \rho)$ be Topological Spaces and let $f: X \to Y$ be a Function. Then $f$ is a Continuous Function If and Only If for all Closed $K \subset Y$, $f^{-1}(Y)$ is Closed in $X$.

$(\Rightarrow)$: Suppose $f$ is a Continuous Function. Suppose $K \subset Y$ is Closed. Then, by definition $V:= K^{C}$ is Open. So $f^{-1}(V) = f^{-1}(K^{C}) = f^{-1}(K)^{C}$ is Open, where the last equality follows from Function Preimage preserves Elementary Set Operations. Thus $(f^{-1}(K)^{C})^{C} = f^{-1}(K)$ is Closed. $\checkmark$

($\Leftarrow$): Let $V \subset Y$ be Open. Then, by definition, $V^{C}$ is Closed in $Y$. By assumption $f^{-1}(V^{C})$ is Closed. Because Function Preimage preserves Elementary Set Operations, $$f^{-1}(V^{C}) = f^{-1}(V)^{C}$$ so $(f^{-1}(V)^{C})^{C} = f^{-1}(V)$ is Open. Thus $f$ is a Continuous Function.

• Function Preimage