$f$ is continuous $\iff$ $\forall S \subseteq X, f(\text{cl}(S)) \subseteq \text{cl}(f(S))$

Asked Aug 25 '16 at 19:09

Active Aug 25 '16 at 19:09

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Let $X,Y$ be topological spaces and $f:X \to Y$. Prove that $f$ is continuous $\iff$ $\forall S \subseteq X, f(\text{cl}(S)) \subseteq \text{cl}(f(S))$

Here's my attempt:

$(\Rightarrow)$ Suppose $f$ is continous. Since $\text{cl}(f(S))$ is closed, then $f^{-1} ( \text{cl}(f(S)))$ is closed. Since $f(S) \subseteq \text{cl} (f(S))$, then $S \subseteq f^{-1} ( \text{cl}(f(S)))$.

A subset of a closed set implies the closure is also a subset. Hence $\text{cl}(S) \subseteq f^{-1} ( \text{cl}(f(S)))$, which means $f(\text{cl}(S)) \subseteq \text{cl}(f(S))$.

$(\Leftarrow)$ Suppose $f(\text{cl}(S)) \subseteq \text{cl}(f(S)) \forall S \subseteq X$

$\text{cl}(f(S))$ is closed, then if I prove that $f^{-1}\text{cl}(f(S))$ I would have $f$ continuous, but I can't do it. If the condition were $\text{cl} f^{-1}(A)\subseteq f^{-1}(\text{cl}A）$ as I found in other posts, I can do it. Maybe this condition I have is mispelled?

Thanks

asked Aug 25 '16 at 19:09

1

Let $T$ be a closed set in $Y$. Put $f^{-1}(T)=S$. Then prove that $S$ is closed. – Sungjin Kim Aug 25 '16 at 19:21
(Sorry, I only found it after posting) – Aug 25 '16 at 19:37

$f$ is continuous $\iff$ $\forall S \subseteq X, f(\text{cl}(S)) \subseteq \text{cl}(f(S))$

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