Let $X,Y$ be metric spaces and define $f: X\to Y$. Show that $f$ is continuous iff $f(\overline{A})\subset\overline{f(A)}$ for each $A\subseteq X$.

My proof: $\Rightarrow$ Let $f:X\to Y$ be continuous and $A\subseteq X$.

Let $y\in f(\overline{A})$, that is, there exist $x\in\overline{A}$ such that $f(x)=y$ with $x\in\overline{A}$, $x\in A$ or $x$ is a limit point of $A$.

I) If $x\in A$, as $f(x)=y$, then $y\in f(A)\subseteq\overline{f(A)}\implies y\in \overline{f(A)}$.

II) If $x$ is a limit point of $A$, that is, there exist $(x_{N})\subseteq A$ such that $\lim_{N\to\infty}{x_{N}}=x$, so $f(x_{N})=y_{N}\in f(A)$

Take the limit $\lim_{N\to \infty}{f(x_{N})}=\lim_{N\to\infty}{y_{N}}$. Since $f$ is continouos, we can exchange the limit $f(\lim_{N\to\infty}{x_{N}})=\lim_{N\to\infty}{y_{N}}\implies f(x)=\lim_{N\to\infty}{y_{N}}$ but $f(x)=y$ by hypothesis, so $\lim_{N\to\infty}{y_{N}}=y$, then $y$ is a limit pointt of $f(A)$.

Therefore, $y\in\overline{f(A)}$. From I) and II), we can conlcuded that $f(\overline{A})\subseteq \overline{f(A)}$.

The other direction of the proof is clear to me, so I need verification of $\Rightarrow$ proof.

Question: Is this proof sufficient? Thanks!

  • $\begingroup$ What is exactly your question? $\endgroup$ – Severin Schraven Jan 20 at 21:44
  • $\begingroup$ @SeverinSchraven I think it is clear from the title the OP wants to know if this proof is correct or not. $\endgroup$ – Umberto P. Jan 20 at 21:45
  • $\begingroup$ The problem statement (statement of the claim), to prove it, requires that you prove: $\Big(f(\overline{A})\subset\overline{f(A)}\text{ for each } A\subseteq X\Big) \to (f: X\to Y \text{ is continuous })$ $\endgroup$ – jordan_glen Jan 20 at 21:54
  • $\begingroup$ Consider duplicates like (this)[math.stackexchange.com/questions/114462/…) $\endgroup$ – Henno Brandsma Jan 20 at 22:43

Your proof is fine, but here's another approach: $A\subseteq f^{-1}(\overline {f(A)})$ and since $f$ is continuous, $f^{-1}(\overline {f(A)})$ is closed. But then, $\overline A\subseteq f^{-1}(\overline {f(A)})\Rightarrow f(\overline A)\subseteq f(f^{-1}(\overline {f(A)}))=\overline {f(A)}.$

  • $\begingroup$ Matematieta.Nice. $\endgroup$ – Peter Szilas Jan 20 at 22:25
  • $\begingroup$ @PeterSzilas Thanks! $\endgroup$ – Matematleta Jan 20 at 22:32

Fine proof, although too verbose.

Suppose $y\in f(\overline{A})$. Then there exists $x\in\overline{A}$ such that $f(x)=y$. Let $(x_n)$ be a sequence in $A$ such that $\lim_{n\to\infty}x_n=x$. Then $$ y=f(x)=f\bigl(\lim_{n\to\infty}x_n\bigr)=\lim_{n\to\infty}f(x_n) $$ Since $f(x_n)\in f(A)$, we have proved that $y\in\overline{f(A)}$.

On the other hand, you can prove the result without sequences. Let $y\in f(\overline{A})$. Then $y=f(x)$, for some $x\in\overline{A}$. We want to prove that $y\in\overline{f(A)}$, so take a neighborhood $V$ of $y$; since $f$ is continuous, there exists a neighborhood $U$ of $x$ such that $f(U)\subset V$. As $x\in\overline{A}$, there exists $x'\in A\cap U$; therefore $f(x')\in f(A)\cap f(U)\subset f(A)\cap V$.

  • $\begingroup$ egreg.Very nice proof: On the other hand.... $\endgroup$ – Peter Szilas Jan 22 at 9:07

A more direct approach from the definition of closure:

Let $f$ be continuous and suppose $y \in f[\overline{A}]$. So $y=f(x)$ with $x \in \overline{A}$. Now let $O$ be an open neighbourhood of $y$, then $f^{-1}[O]$ is an open neighbourhood of $x$, so $f^{-1}[O] \cap A$ is non-empty (as $x \in \overline{A}$), say that $x' \in f^{-1}[O] \cap A$. But then $f(x') \in f[A]$ and $f(x') \in O$ so that $O$ intersects $f[A]$. As $O$ was an arbitrary open neighbourhood of $y$, $y \in \overline{f[A]}$ and the inclusion has been shown.


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