How do you prove that $\mathsf{bd}(\mathsf{bd}(\mathsf{bd}(W)))= \mathsf{bd}(\mathsf{bd}(W))$

Question

$\newcommand{\bd}{\operatorname{bd}}$Prove that the $\bd(\bd(\bd(W)))=\bd(\bd(W))$ where $W$ is a subset of the topological space $(X,\mathscr{T})$.

@Rasmus: no. For example, consider $\mathbb{Q}$ in the ambient space $\mathbb{R}$. Then $\operatorname{bd}(\operatorname{bd}(\mathbb{Q}))=\operatorname{bd}(\mathbb{R})={\varnothing}$. — Chris Eagle, Oct 25 '11 at 12:19

score 5 · Answer 1 · answered Oct 25 '11 at 05:24

5

A set $A$ is equal to its own boundary if and only if it is closed and is contained in the closure of its complement. Now you only need to prove that this property is satisfied by the set in question.

answered Oct 25 '11 at 05:24

Shripad Garge

211

score 3 · Answer 2 · edited Dec 04 '11 at 04:57

3

$\DeclareMathOperator{\bd}{bd}$Prove that:

if $W$ is any subset of $X$, then $\bd(W)$ is closed;
if $Y$ is any closed subset of $X$, then $\bd(\bd(Y)) = \bd(Y)$.

edited Dec 04 '11 at 04:57

dfeuer

9,069

answered Oct 25 '11 at 05:30

Ted

33,788

score 1 · Answer 3 · edited Dec 04 '11 at 05:21

$\DeclareMathOperator{\cl}{cl} \DeclareMathOperator{\Int}{in} \DeclareMathOperator{\bd}{bd}$I will denote closure by $\cl$ and interior by $\Int$.

$\bd(W)=\cl(W)-\Int(W)$, so

$$\eqalign{\bd(\bd(W))&=\cl(\cl(W))-\Int(\Int(W))-\Int(\cl(W)-\Int(W))\cr&= \cl(W)-\Int(W)-\Int(\cl(W)-\Int(W)),}$$ but $$\eqalign{\bd(\bd(\bd(W)))&=\cl(\cl(W)-\Int(W)-\Int(\cl(W)-\Int(W)))\cr&-\Int(\cl(W)-\Int(W)-\Int(\cl(W)-\Int(W)))\cr &=\cl(W)-\Int(W)-\Int(\cl(W)-\Int(W))\cr&-(\Int(\cl(W)-\Int(W))+\Int(\cl(W)-\Int(W))\cr &=\cl(W)-\Int(W)-\Int(\cl(W)-\Int(W))\cr &=\bd(\bd(W)).}$$

How do you prove that $\mathsf{bd}(\mathsf{bd}(\mathsf{bd}(W)))= \mathsf{bd}(\mathsf{bd}(W))$

3 Answers3