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Let $Y$ be a subset of a topological space $X$. How can I prove that $$\operatorname{cl}(\operatorname{int}(Y))=\operatorname{cl}(Y)\text{?}$$

I know that since $\operatorname{int}(Y)\subset Y$, then $\operatorname{cl}(\operatorname{int}(Y))\subset \operatorname{cl}(Y)$. But I don't know how to prove the other inclusion.

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    I don't think that's true in general. – 2017-01-07
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    Counterexample: if $Y$ is a single point in $\mathbb{R}$, then $cl(int(Y))=\varnothing$ while $cl(Y)=Y$. – 2017-01-07
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    An interesting true one is Cl(Int(Cl(Int (A))))=Cl(Int(A)). In other words, if f(A)=Cl(Int (A)) for all A then f(f(A))=f(A). – 2017-01-07
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    If Cl(y)=Cl(Int (Cl(y))) then Cl(y) is called a regular closed set – 2017-01-07

2 Answers 2

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The statement is false. Let $Y=\Bbb Q$, the set of all rationals. Interior of $Y$ is empty.

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The statement is false in general. E.g. $Y = \mathbb{Q} \subseteq \mathbb{R}$ in the uusal topology, then $\operatorname{int}(Y) = \emptyset$, and so $\operatorname{cl}(\operatorname{int(Y)}) = \emptyset$ while $\operatorname{cl}(Y) = \mathbb{R}$.

A set $Y$ that does satisfy $\operatorname{cl}(\operatorname{int}(Y)) = Y$ is called regular closed.

And for all subsets $Y$ we do have: $$\operatorname{cl}(\operatorname{int}(\operatorname{cl}(\operatorname{int}(Y)))) = \operatorname{cl}(\operatorname{int}(Y))$$

(see my notes here for a proof).

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    That's quite an interesting little sets of notes, @Henno. I've only skimmed for now, but the $14$ jumped out at me. I'm probably falling victim to the law of small numbers, but it reminded me of the $14 = 2^{3 + 1} - 1$ faces of the three dimensional permutohedron. I'll definitely be looking more closely at some point! – 2017-01-07
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    @pjs36 it's a classical fact proved by Kuratowski. The 14 is just what follows from the basic facts in the beginning of the notes.. – 2017-01-07
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    Fair enough, thank you. – 2017-01-07