Prove that for any given set $ A \subset X$ where $X$ is topology space this is true $ bd(bd(bdA)) = bd(bdA)$

Question

I will appreciate any hint in proofing that.

I tried to use this two definitons: $bdA = clA \setminus int(A)$ and $bdA = clA \cap cl(X\setminus A)$. But I get very long equation and I do not know what to do with this. I tried to prove $\subseteq$ and $\supseteq$ but I do not know how to start with that.

And I know that in some cases $bd(bdA) \neq bdA$ for example $\mathbb{Q}$ in Euclidean topology.

Answer 1

First note several things:

• For any $A \subseteq X$ we have that $\partial A$ is closed.
• If $U$ is an open set, then $\operatorname{Int}(\partial U) = \emptyset.$ Indeed, \begin{align} \operatorname{Int}(\partial U) &= \operatorname{Int}(\overline{U} \cap (\operatorname{Int} U)^c) =\operatorname{Int}(\overline{U}) \cap \operatorname{Int}((\operatorname{Int} U)^c) \\ &= \operatorname{Int}(\overline{U}) \cap \operatorname{Int}(U^c) = \operatorname{Int}(\overline{U}) \cap (\overline{U})^c = \emptyset \end{align} where we used $\operatorname{Int}(A \cap B) = \operatorname{Int}(A) \cap \operatorname{Int}(B)$ and $\operatorname{Int}(A^c) = (\overline{A})^c$ which holds for all $A,B \subseteq X$.
• If $F$ is a closed set, then since $\partial F = \partial(F^c)$ and $F^c$ is open, we conclude that $\operatorname{Int}(\partial F) = \emptyset$ as well.

Now for any $A \subseteq X$ we have that $\partial A$ is closed and hence $\operatorname{Int}(\partial \partial A)$ is empty. Therefore $$\partial \partial \partial A = \overline{\partial \partial A } \cap \operatorname{Int}(\partial \partial A)^c = \partial \partial A \cap X = \partial \partial A$$ since $\partial \partial A$ is closed as well.