# Proof verification: Show that $f$ is continuous if $f(\overline{A})\subset\overline{f(A)}$.

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!

• What is exactly your question? – Severin Schraven Jan 20 at 21:44
• @SeverinSchraven I think it is clear from the title the OP wants to know if this proof is correct or not. – Umberto P. Jan 20 at 21:45
• The problem statement (statement of the claim), to prove it, requires that you prove: $\Big(f(\overline{A})\subset\overline{f(A)}\text{ for each } A\subseteq X\Big) \to (f: X\to Y \text{ is continuous })$ – jordan_glen Jan 20 at 21:54
• Consider duplicates like (this)[math.stackexchange.com/questions/114462/…) – 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, $$\overline A\subseteq f^{-1}(\overline {f(A)})\Rightarrow f(\overline A)\subseteq f(f^{-1}(\overline {f(A)}))=\overline {f(A)}.$$

• Matematieta.Nice. – Peter Szilas Jan 20 at 22:25
• @PeterSzilas Thanks! – 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 $$f(x')\in f(A)\cap f(U)\subset f(A)\cap V$$.

• egreg.Very nice proof: On the other hand.... – 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.