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I hope this question is not duplicate here: Let $X$ be a topological space and $A$ be "any" subset of $X$, then $$\partial \partial \partial A = \partial \partial A.$$ (Here, $\partial B$ denotes the boundary of the subset $B$).

My work: I started with the definition and had $\partial \partial \partial A=\partial \partial A -(\partial \partial A)^{\circ}$. Then I am trying to show that the interior $(\partial \partial A)^{\circ}$ is empty.

Any ideas? Can it be a simple argument? Mine is not. Thanks in advance.

A simpler version: Boundary of a boundary of an open set

Bob Dobbs
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    Somewhat related: https://math.stackexchange.com/questions/105745/boundary-operators-idempotence – also, if you can see the deleted answer at https://math.stackexchange.com/questions/443404/boundary-of-the-boundary-of-a-set-is-empty – Gerry Myerson Nov 02 '22 at 11:54
  • Can you not simply use that $\partial \partial S \subseteq \partial S$, where equality holds iff S has no interior, then since after applying the boundary operation twice this set really has no interior the claim follows by the above. – a.s. graduate student Nov 02 '22 at 11:58
  • I think Henno Brandsma's answer in the first link from Gerry's comment settles this. After all $S=\partial A$ is closed, and Henno shows that $\partial S=\partial\partial S$, which is just what's needed here. – Jyrki Lahtonen Nov 02 '22 at 11:59
  • See also https://math.stackexchange.com/questions/313028/iterated-boundary-in-a-metric-space and one of the answers of https://math.stackexchange.com/questions/43627/boundary-in-the-topological-space – Gerry Myerson Nov 02 '22 at 12:01
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    Some define boundary as $\partial X=\bar X\setminus X^\circ$. You seem to define it differently. – Ivan Neretin Nov 02 '22 at 12:02
  • Boundary is closed. – Bob Dobbs Nov 02 '22 at 12:03

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Reamark that $\partial A$ is closed for arbitrary subset $A$, since $\partial A=C_{X}A^{\mathrm{o}} \cap C_{X}(C_{X}A)^{\mathrm{o}} $,in particular $\partial \partial A \subset \partial A $. Now ,suppose to the contrary that there is $x$ an interior point of $\partial \partial A $,then there is some open subset $x\in V\subset \partial \partial A $,but this is impossible since $\partial \partial A \subset \partial A $, so $V\subset \partial A $ and $x$ is boundary point of $\partial A$,so every neighboorhood of it must meet $\partial A$ and its complement.

belkacem abderrahmane
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  • $\partial A$ can be the whole space and what did you contradict then? – Bob Dobbs Nov 02 '22 at 12:33
  • If $\partial A$ is the whole space, then $\partial \partial A=\emptyset $ ! – belkacem abderrahmane Nov 02 '22 at 12:51
  • So you deduced the boundary of $A$ has no complement? – Bob Dobbs Nov 02 '22 at 13:31
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    It's a proof by Contradiction, I assumed that $\partial \partial A $ has non-empty interior, picked a point in this interior, So this point has two properties 1) it's an interior point of $\partial \partial A $, in particular it is an interior point of $\partial A$, 2)it is a boundary point of $\partial A$, so every nbd of it must meet both $\partial A$ and its complement, and this is not satisfied for the nbd $V$, So I arrived at a Contradiction, I concluded that the interior of $\partial \partial A $ must be empty. – belkacem abderrahmane Nov 02 '22 at 13:59