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I have reading about the closure interior and boundary operators in a topological space. I have been thinking about the following:

If $ b $ denotes boundary operator and $c$ , $i$ and $k$ denote closure and interior and complement respectively. Let $A \subseteq X$ where $X$ is a topological space. Does the following hold? If not when it will hold true and why?

$$bb(A)=b(A)$$

I think the above has a counter example $A= \mathbb{Q}$ and $X=\mathbb{R}$ because $bb(A)=\emptyset$ and $b(A)=\mathbb{R}$. But I would like if any one could suggest the case when equality holds and how? Also I would like to know when is $bbb(A)=bb(A)$ true for any subset $A$ of $X$?

Najib Idrissi
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User23
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1 Answers1

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This is true if and only if the boundary has empty interior.

It's easy to see the boundary of the boundary is always contained in the boundary. But if the boundary has an interior point, then that point will not be in the boundary of the boundary since it is not in the closure of the complement.

Edit: to answer your added question, it is always true that $bbbA=bbA$. This follows from the above and the fact that $ibcB$ is empty for any $B$ and the fact that the boundary of a set is closed, so letting $B=bA$, so $cB=B$, and $ibbA$ is empty. To see that $ibcB$ is empty, assume not and let $x$ be an interior point of the boundary of a closed set $C$. Then the boundary contains an open set $U$ containing $x$. Since $C$ is closed it contains its boundary so it contains $U$. So $U$ does not intersect $kC$, so $x$ is not in $ckC$, so it is not in $bC$, contradiction.

Callus - Reinstate Monica
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  • I got your answer.Thanks for Help. Basically you have proved $bA \subseteq bbA \iff ibA= \phi$ for space $X$ and $A \subseteq X$. Now by taking a point say $x \in ibA $ it generates a contradiction to the hypothesis that $bA \subseteq bbA$ we prove it one way which is $bA \subseteq bbA$ implies $ibA = \phi$.Am i getting it right?? – User23 Dec 13 '14 at 11:31
  • Yes. The other direction is not too tough. If you assume $x$ is a point in $bA$ not in $bbA$ then it is in $kckbA$ so it is in $ibA$. – Callus - Reinstate Monica Dec 13 '14 at 13:27
  • Got the converse part.Just edited the question a bit..if you could try that too :) – User23 Dec 15 '14 at 08:40
  • I would want to clarify two points in the additional answer (1)Wouldn't for showing $bbbA=bbA$ we will have to show $ibbA=\phi$ which basically refers to the fact that the interior of boundary of the set $bA$ is empty in agreement to the fact proved by you in the previous question instead of showing $ibkA=\phi$ (2) Next to get the contradiction in the end step we will have to show $x \notin ckB$ because $bB=cB \cap ckB$ instead of showing $x \notin kcB$? and are you refering to the closed set $B$ as $bA$? I may be getting it wrong.Please elaborate. – User23 Dec 22 '14 at 09:30
  • Oops, I mixed up $c$ and $k$ a few times. I'll edit again. – Callus - Reinstate Monica Dec 22 '14 at 11:25
  • just one question In the last step if $U \cap kC=\phi$ will not imply $U \cap ckC=\phi$ as $ckC \supseteq kC$??So how come you conclude $x \notin ckC$ – User23 Dec 22 '14 at 12:17
  • I think by definition, if we find a neighborhood $U$ of $x$ such that $U$ has empty intersection with $V$, then $x$ is not in the closure of $V$. – Callus - Reinstate Monica Dec 23 '14 at 02:10
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    got it as $U$ was arbitrary open set so it holds.I suggest this way of proof,As $i=kck$, $i $ is interior, $c$ is closure and $k$ is complement ,For any closed set $C$,$ibC=i(cC \cap ckC)=icC \cap ickC=iC \cap ickC=kckC \cap kckckC=\phi$ because $kckC \subseteq ckckC$.Then let $C=bA$ as $bA $ is closed for any $A$ subset of $X$,Does this method seem topologically correct?? – User23 Dec 26 '14 at 11:28
  • It looks correct to me. However, I find it hard to read. It's very formal. That is a matter of taste, of course. – Callus - Reinstate Monica Dec 26 '14 at 11:31
  • Wait, how does a boundary have an interior point? If it did, then there would be a point in the boundary that has a neighborhood containing either no points in $A$ or only points in $A$. My logic might be wrong though. – Arturo don Juan Jul 09 '15 at 03:32
  • @ArturoDonJuan The boundary is the intersection of the closure and the complement of the interior. I think you're describing the intersection of the closure and the interior of the complement? Consider $\mathbb{Q}$ inside $\mathbb{R}$. The boundary of $\mathbb{Q}$ is $\mathbb{R}$ because $\mathbb{Q}$ is dense but has no interior. – Callus - Reinstate Monica Jul 15 '15 at 23:02
  • I was actually just thinking in terms of neighborhoods about a point on the boundary (i.e. the boundary is defined as the set of all points that contain both points in the set and not in the set for every neighborhood about that point, I thought). However, I understand where I was going wrong. I totally forgot that a boundary of a set doesn't have to also be the boundary of itself. – Arturo don Juan Jul 16 '15 at 13:01