# Boundary of Boundary of a set?

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 $X$?

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

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.
• 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?? Dec 13, 2014 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 $ibA$. Dec 13, 2014 at 13:27
• 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. Dec 22, 2014 at 9:30
• 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?? Dec 26, 2014 at 11:28