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How can one prove this proposition:

proposition 5.15: $f$ is continuous iff $f(\overline{A}) \subset \overline{f(A)}$, where $\overline{A}$ denotes the closure of a set $A.$

It seems to be a small result, but on the chapter on continuity, it just states this proposition, but I have little clue on how to show this is true. I would appreciate the help.

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marked as duplicate by drhab, user 170039, Community Mar 2 '18 at 15:12

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    $\begingroup$ see here $\endgroup$ – rbird Mar 2 '18 at 14:39
  • $\begingroup$ @rbird Thank you very much for the source, I will try to study the proof. $\endgroup$ – Aurora Borealis Mar 2 '18 at 14:41
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Let, $f:(X,\tau)\rightarrow(Y,\tau')$ be continuous, then for any closed set $F$ in $(Y,\tau')$ $f^{-1}(F)$ is closed in $X$.

Now let $A\subset X$ be any set. Then $\overline{f(A)}$ is a closed set in$(Y,\tau')$ also $f^{-1}(\overline{f(A)})$ is closed in $(X,\tau)$ with, $A\subset f^{-1}(\overline{f(A)})$.

But, $\overline{A}$ is the smallest subset containing $A$. So, $\overline{A}\subset f^{-1}(\overline{f(A)})$ i.e. $f(\overline{A})\subset \overline{f(A)}$.

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  • $\begingroup$ That is only one side of it. $\endgroup$ – drhab Mar 2 '18 at 14:44
  • $\begingroup$ @drhab - did I make any mistake? please edit if necessary $\endgroup$ – Sujit Bhattacharyya Mar 2 '18 at 14:46
  • $\begingroup$ You did not make a mistake, but what lacks in your answer is a proof of the other side: if $f(\overline A)\subseteq\overline{f(A)}$ for every $A\subseteq X$ then $f$ is continuous. $\endgroup$ – drhab Mar 2 '18 at 14:49
  • $\begingroup$ oh! yes. sorry for that $\endgroup$ – Sujit Bhattacharyya Mar 2 '18 at 16:17

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