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Continuity and Closure

$f\colon M\to N$ is continuous iff for all $X\subset M$ we have that $f\left(\overline{X}\right)\subset\overline{f(X)}$.

I only proved $\implies$.

If $f$ is continuous then for any $X\subset M$,

$$X\subset f^{-1}[f(X)]\subset f^{-1}\left[\overline{f(X)}\right]=\overline{f^{-1}\left[\overline{f(X)}\right]}$$


$$\overline{X}\subset f^{-1}\left[\overline{f(X)}\right]\implies f\left(\overline{X}\right)\subset \overline{f(X)}.$$

The other side must be the same idea but I don't know why I can't prove it.

Added: With exactly same idea when I proved $\implies$ I did proved $\Longleftarrow$,

Let $F\subset N$ any closed set then:

$$f\left[f^{-1}(F)\right]\subset f\left[ \overline{f^{-1}(F)}\right]\subset \overline{f\left[ f^{-1}(F)\right]}\subset \overline{F}=F$$

in particular

$$f\left[ \overline{f^{-1}(F)}\right]\subset F\implies f^{-1}(F)\supset\overline{f^{-1}(F)}$$

then $f^{-1}(F)=\overline{f^{-1}(F)}$ and $f$ is continuous.

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marked as duplicate by Asaf Karagila, Zhen Lin, t.b., Chris Eagle, Martin Sleziak Jun 15 '12 at 7:51

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.

Is this for all subsets $X$ of $M$? – ncmathsadist Jun 13 '12 at 23:13
Also, what are $M$ and $N$? Topological spaces? Metric spaces? Bound this discussion with some context. – ncmathsadist Jun 13 '12 at 23:15
@ncmathsadist Yes is it. And both are topological spaces. (general topology) tag. – Gastón Burrull Jun 13 '12 at 23:16
It looks like an exact duplicate, David. – ncmathsadist Jun 14 '12 at 0:26
up vote 7 down vote accepted

Suppose that $f$ is not continuous. Then there is a closed set $C\subseteq N$ such that $f^{-1}[C]$ is not closed in $M$. Let $X=f^{-1}[C]$. Since $X$ is not closed, there is some $p\in\operatorname{cl}X\setminus X$. Then $f(p)\in f[\operatorname{cl}X]$, but $f(p)\notin C\supseteq\operatorname{cl}f[X]$, so $f[\operatorname{cl}X]\nsubseteq\operatorname{cl}f[X]$.

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Thanks Brian you always make it all look easy. Could you tell me if my (edited) solution is right? – Gastón Burrull Jun 14 '12 at 3:17

First, a general fact about maps of sets: $$X \subseteq f^{-1} Y \iff f X \subseteq Y$$ Now, suppose for all $X$ in $M$ we have $f \overline{X} \subseteq \overline{f X}$. Let $Y$ be a closed subset of $N$ and let $X = f^{-1} Y$. A map $f : M \to N$ is continuous if and only if the preimage of every closed set is closed, so we need to show $X$ is closed. Clearly, $f X \subseteq Y$ and $X = f^{-1} f X$. Consider $\overline{X}$: we have $X \subseteq \overline{X}$, so $f X \subseteq f \overline{X} \subseteq \overline{f X} \subseteq Y$, hence $\overline{X} \subseteq f^{-1} Y = X$, i.e. $\overline{X} = X$.

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Thanks, your answer is good and simple. I've edited some mistakes. Can you see my edited answer that you was corrected before? – Gastón Burrull Jun 14 '12 at 3:53

Let $V\subset N$ be an open set around $f(x)$. Then its complement $V^c$ is closed. Let $U=cl({f^{-1}(V^c)})^c$. Then it is an open set. Because of the property of the function: $$ f(cl({f^{-1}(V^c)}))\subset cl(f({f^{-1}(V^c)}) \subset V^c $$ we see that $x\in U$, and that $f(U)\subset V$. Then f is continuous.

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Why do you say $\overline{f f^{-1} V^c} = V^c$? It is possible that $V^c \ne \emptyset$ but $f^{-1} V^c = \emptyset$, if $f$ is not surjective. – Zhen Lin Jun 13 '12 at 23:33
You are right, I changed it. – guaraqe Jun 13 '12 at 23:36
Nice answer! Thanks for your more basic approach. – Gastón Burrull Jun 14 '12 at 3:40

I think that I have the answer. Let $F\subset N$ any closed set then:

$$f\left[f^{-1}(F)\right]\subset f\left[ \overline{f^{-1}(F)}\right]\subset \overline{f\left[ f^{-1}(F)\right]}\subset \overline{F}=F$$

in particular

$$f\left[ \overline{f^{-1}(F)}\right]\subset F\implies f^{-1}(F)\supset\overline{f^{-1}(F)}$$

then $f^{-1}(F)=\overline{f^{-1}(F)}$ and $f$ is continuous.

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Your first containment is not true if $F$ does not lie in the image of $f$. – Zhen Lin Jun 13 '12 at 23:34
@ZhenLin You're right I've fixed. – Gastón Burrull Jun 13 '12 at 23:44
@ZhenLin am I right now? – Gastón Burrull Jun 14 '12 at 2:37
Not quite: $f\left[f^{-1}[F]\right]$ may not be $F$. But it’s a subset of $F$, so $\operatorname{cl}f\left[f^{-1}[F]\right]\subseteq\operatorname{cl}F=F$, and that’s good enough to make the rest of your argument work. – Brian M. Scott Jun 14 '12 at 4:09
Looks good now. – Brian M. Scott Jun 14 '12 at 4:24

Suppose $f$ is not continuous. Then for some $U$ open in $N,$ $f^{-1} (U)$ contains no neighborhood $N_x$ about some point $x\in f^{-1} (U).$ In other words, $N_x \cap M / f^{-1} (U) $ is nonempty for each $N_x,$ so that $x \in \overline{M / f^{-1} (U)}$ - thus $f(x) \in f(\overline{M / f^{-1} (U)} ).$ However, $\overline{ f( M / f^{-1} (U) )} \subset \overline{N / U} = N/U,$ and $x\notin N/U.$ Thus $f(\overline{M / f^{-1} (U)} )$ is not contained in $\overline{ f( M / f^{-1} (U) )}.$

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Good answer thanks, but I edited a mistake. – Gastón Burrull Jun 14 '12 at 4:16

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