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


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 Dec 27 '18 at 22:47

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.

  • $\begingroup$ 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. $\endgroup$ – Hagen von Eitzen Feb 7 '17 at 12:36
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    $\begingroup$ 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. $\endgroup$ – TonyK Feb 7 '17 at 12:37
  • $\begingroup$ True!thanks, I think it is ok now. $\endgroup$ – Gerard Gracia Feb 7 '17 at 16:10

The proof for the second part is fine as it stands, but you could give a proof in the same spirit as for the other direction:

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.

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    $\begingroup$ This is a misuse of MSE: posting an answer to "slowly" "explain" the proof in the OP's question. $\endgroup$ – Alex M. Jun 7 '18 at 17:24
  • $\begingroup$ @AlexM. Then vote it also down. Note, with newbies we should be so polite as we can, on obvious reasons. $\endgroup$ – peterh Jun 7 '18 at 17:29
  • $\begingroup$ @peterh: It seems to me that, maybe involuntarily, the poster is not very polite to the readers when he chooses to "explain everything slowly" as if for idiots. $\endgroup$ – Alex M. Jun 7 '18 at 18:44

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