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?


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 $X$?

  • $\begingroup$ How about singletons in $ℝ$? $\endgroup$
    – k.stm
    Dec 13, 2014 at 10:33
  • $\begingroup$ @k.stm its correct as $bb\{x\}=b\{x\}=\{x\}$ for any singleton $\{x\}$ in $\mathbb{R}$. $\endgroup$
    – User23
    Dec 13, 2014 at 11:37
  • $\begingroup$ See here for a criterion when $bb(A) = b(A)$. $\endgroup$
    – ComFreek
    Apr 24, 2019 at 13:18

1 Answer 1


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.

  • $\begingroup$ 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?? $\endgroup$
    – User23
    Dec 13, 2014 at 11:31
  • $\begingroup$ 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 $ibA$. $\endgroup$ Dec 13, 2014 at 13:27
  • $\begingroup$ Got the converse part.Just edited the question a bit..if you could try that too :) $\endgroup$
    – User23
    Dec 15, 2014 at 8:40
  • $\begingroup$ 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. $\endgroup$
    – User23
    Dec 22, 2014 at 9:30
  • 1
    $\begingroup$ 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?? $\endgroup$
    – User23
    Dec 26, 2014 at 11:28

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