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Suppose consider the space (R,d), where d is the absolute value metric, and f is a continuous function from R to R. Let A be a subset of R. Then is Cl(f(A)) a subset of f(Cl(A))? Prove or provide counterexample.

If yes, can you extend in for any arbitrary metric spaces I am very new to metric space and topology, so any help will be appreciated.

Martin Sleziak
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Doubts
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2 Answers2

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No.

Consider $f=\arctan$, and $A=\mathbb{R}$. You have $Cl(f(A))=\left[ -\frac{\pi}{2}, \frac{\pi}{2}\right]$, which is not a subset of $f(Cl(A))=\left( -\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Of course, if the property does not hold in $\mathbb{R}$, it has not any chance to hold in a general metric space.

TheSilverDoe
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Let $X$ be a topological space and let $f:X\to X$ be a function.

Under what condition do we have:$$\mathsf{Cl}(f(A))\subseteq f(\mathsf{Cl}(A))\text{ for every }A\subseteq X\tag1$$?

If $(1)$ is true then for a closed set $F$ we should have: $$\mathsf{Cl}(f(F))\subseteq f(\mathsf{Cl}(F))=f(F)$$showing that $f(F)$ is closed.

So $(1)$ implies that $f$ is a closed function.

Conversely if $f$ is a closed function then for every $A$ the set $f(\mathsf{Cl}(A))$ is a closed set with $f(A)\subseteq f(\mathsf{Cl}(A))$ so that consequently $\mathsf{Cl}(f(A))\subseteq f(\mathsf{Cl}(A))$.

Proved is now that $(1)$ is true if and only if $f$ is a closed function.

drhab
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