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Proving that for a continuous map $f: \overline A \rightarrow F $, where $F$ is a metric space, it holds that $f[ \overline A ] \subset \overline{ f[A] }$.

Proof:

$\overline{ f[A] }$ is closed, since $f$ is continuous then the inverse of the set must be closed. We have:

$f^{-1}[\overline{ f[A] }]$ is closed and it holds that $A \subset f^{-1}[\overline{ f[A] }]$.

So it must hold that $\overline A \subset f^{-1}[\overline{ f[A] }]$ because $\overline A$ is the smallest closed set that contains $A$.

So, therefore it actually holds that $\overline A = f^{-1}[\overline{ f[A] }]$ because $\overline A$ is the largest set in the domain of $f$.

Now we have $f[\overline A] = f \circ f^{-1}[\overline{ f[A] }] \subset \overline{ f[A] } $.

So it holds that $f[\overline A] \subset \overline{ f[A] }$.

$\blacksquare$

Is the proof correct? If so, is there a way to somehow solve it in a different way using sequences? Also, my proof seems not to utilise the fact that $F$ is a metric space. If it is right, then the condition must be more general then. The proof also must imply that if $f$ is surjective then an equality will hold.

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    $\begingroup$ A similar question was asked recently: math.stackexchange.com/q/3953731/42969 (but with a different proof). $\endgroup$
    – Martin R
    Dec 18 '20 at 12:08
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    $\begingroup$ Have I not witnessed a similar problem moments ago? Is this part of an active assignment you have with your classmates? At least you provided your own version of proof which is something good. Note that if you want to change from $\subseteq$ to $\subset$ provide an example where the inclusion is proper. $\endgroup$ Dec 18 '20 at 12:08
  • $\begingroup$ Wow, that must be the Universe syncing our minds?! :-) If my proof is right, then I'll reference this in the other question. $\endgroup$
    – СССР
    Dec 18 '20 at 12:11
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    $\begingroup$ Your proof is good, I meant you should provide an example showing that $f(\overline {A})\neq\overline{f(A)}$ $\endgroup$ Dec 18 '20 at 12:24
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    $\begingroup$ That look good. If you want a simple one consider this : $f: \mathbb{R} \longrightarrow \mathbb{R}$ and define $f(x)=\frac{1}{1+x^{2}} .$ Therefore, $A=\bar{A}=[0,+\infty) .$ We have that $f(\bar{A})=f(A)=(0,1]$ However, $\overline{f(A)}=[0,1] .$ Thus, $f(\bar{A}) \neq \overline{f(A)}$ $\endgroup$ Dec 18 '20 at 13:19
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The proof is fine and you are right, you don't need $F$ metric space. It holds in general that if $X, Y$ are topological spaces, then $$f: X \rightarrow Y \text{ is continuous }$$ if and only if for every subset $A \in X$ we have $$f(\overline{A}) \subset \overline{f(A)}$$

For the proof with sequences, then you may want the domain is a metric space (first countability would be enough) so that you can use the characterization of the closure of sets by means of sequences. The codomain needs not to be assumed metrizable for your purposes.

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