# Continuous function between two topological spaces: an ELEMENTARY property. [duplicate]

I'm reading the first chapter of a book on general topology. It has a lot of small, simple exercises on almost all pages and I try to do them all to fully understand the subject.

This one I did not manage to solve, though it's probably really simple, cause the previous ones were:

Suppose $X$ and $Y$ are topological spaces. $f:X\to Y$ is any map.

Prove that f is continuous if and only if: $$f(\overline{A}) \subseteq \overline{f(A)}\quad for\quad all\quad A\subseteq X$$

I think I'll probably have to only use the open subset criterion for continuity and the definition of A-bar and that'll probably be it, but please, help me a bit. Thanks!

## marked as duplicate by copper.hat, Johanna, Community♦May 22 '15 at 16:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• – m_gnacik May 22 '15 at 16:32
• Thank you, I'll check it out. – I want to make games May 22 '15 at 16:34

## 1 Answer

You aren't dealing with pre-images, so we'll have to use the definition directly.

Suppose that $f$ is continuous, but that $f(\overline{A})\not\subset\overline{f(A)}$. Then exists $y \in f(\overline{A})$ such that $y \not\in \overline{f(A)}$. Then exists an open neighborhood of $y$, $V$, such that $V \cap A = \varnothing$. But $y = f(x)$ with $x \in \overline{A}$. By continuity, there is a neighborhood of $x$, $U$, such that $f(U)\subset V$. Then: $$U \cap A \neq \varnothing \implies \varnothing \neq f(U \cap A) = f(U)\cap f(A) \subset V \cap A = \varnothing,$$ contradiction. So $f(\overline{A})\subset \overline{f(A)}$.

For the other direction, you'll use that $f$ is continuous if and only if pre-images of closed sets are closed. Can you go on?