# proof $f(\overline{A})\subseteq \overline{f(A)} \Leftrightarrow f$ continuous [duplicate]

It'd be great if someone checked the proof I did for the following problem:

$f:X\longrightarrow Y$,

$f(\overline{A})\subseteq \overline{f(A)},\forall A\subseteq X \Leftrightarrow f$ continuous

proof:

Suppose $f(\overline{A})\subseteq \overline{f(A)}$ for any $A\subseteq X$ . Let $C\subseteq Y$ be closed and define $A=f^{-1}(C)$. $f(\overline{A})\subseteq \overline{f(A)}\subseteq\overline{C}=C$, therefore $\overline{A}\subseteq f^{-1}(C)=A$, so $A$ is closed.

*Conversely, suppose $f$ is continuous.

Let $p\in\overline{A}$ and let $V$ be an open neighborhood of $f(p)$. Since $f$ is continuous, $f^{-1}(V)$ is open. On the other hand $p\in f^{-1}(V)$, but since $p\in\overline{A}$ we have that $f^{-1}(V)$ contains points of $A$. Therefore $V$ contains points of $f(A)$. Since $V$ is an arbitrary neighborhood of $f(p)$, we have $f(p)\in \overline{f(A)}$. So finally, $f(\overline{A})\subseteq\overline{f(A)}$.

Thank you very much!

## marked as duplicate by José Carlos Santos general-topology StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 27 '18 at 22:47

• Since you don't need that $f^{-1}(V)$ intersects $X-A$, you may not need to treat $p\in A$ and $p\in\partial A$ as separate cases. Butg as the case $p\in A$ is so trivial, it wonÄt change much. – Hagen von Eitzen Feb 7 '17 at 12:36
• You assume that $f(f^{-1}(C))=C$, which is not in general true. I think you can use $f(f^{-1}(C))\subseteq C$ instead, but as it stands the proof is incomplete. – TonyK Feb 7 '17 at 12:37
• True!thanks, I think it is ok now. – Gerard Gracia Feb 7 '17 at 16:10

Suppose $$A \subseteq X$$. Then $$A \subseteq f^{-1}[f[A]] \subseteq f^{-1}[\overline{f[A]}]$$ where the latter set is closed by continuity, so $$\overline{A} \subseteq f^{-1}[\overline{f[A]}$$ as well. This implies that $$f[\overline{A}] \subseteq \overline{f[A]}$$ directly.
I found this proof very awkward when I first came across it so for everyone else like me I'm going to explain everything slowly.  "$\Rightarrow$" let $f:X->Y$ be continuous $\Rightarrow\forall F\subset Y$ such that $y$ is closed in $Y$ $f^{-1} (F)$ is closed in X. Pick any $A\in X$. It is "obvious" $f(A)= f(A) \Rightarrow f(A)\subset\overline {f(A)}$ hence $A\subset f^{-1}\left( \overline {f(A)}\right)$ $\Rightarrow \bar A\subset \overline{ f^{-1}\left( \overline {f(A)}\right)}$. Here it is important to note that $\overline{f(A)}$ is closed in Y hence it's inverse image is closed in X. Thus$\overline{ f^{-1}\left( \overline {f(A)}\right)}= f^{-1}\left( \overline {f(A)}\right)$ hence subbing this back into the original formula we see that $\bar A \subset f^{-1}\left( \overline {f(A)}\right)$ and the result follows.  "$\Leftarrow$" Now suppose $f(\bar A)\subset \overline {f(A)}$ pick any $C_Y\subset Y$ such that $C_Y$ is closed in Y. Let $A=f^{-1}(C_Y)$, $f(\bar A)\subset \overline{f(A)}\subset \overline{C_Y}\subset C_Y$. Taking the first and last parts of that working gives: $f(\bar A) \subset C_Y$. Finally, taking inverse images gives $\bar A \subset f^{-1}(C_Y)$ using our definition at the start we see $\bar A\subset A$. This show us $A$ is closed and hence the pre-image of a closed set is closed and thus our function is continuous.