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 $X$?
Best Answer
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.