$\newcommand{\bd}{\operatorname{bd}}$Prove that the $\bd(\bd(\bd(W)))=\bd(\bd(W))$ where $W$ is a subset of the topological space $(X,\mathscr{T})$.
How do you prove that $\mathsf{bd}(\mathsf{bd}(\mathsf{bd}(W)))= \mathsf{bd}(\mathsf{bd}(W))$
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0@Rasmus: no. For example, consider $\mathbb{Q}$ in the ambient space $\mathbb{R}$. Then $\operatorname{bd}(\operatorname{bd}(\mathbb{Q}))=\operatorname{bd}(\mathbb{R})={\varnothing}$. – 2011-10-25
3 Answers
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.
$\DeclareMathOperator{\bd}{bd}$Prove that:
- if $W$ is any subset of $X$, then $\bd(W)$ is closed;
- if $Y$ is any closed subset of $X$, then $\bd(\bd(Y)) = \bd(Y)$.
$\DeclareMathOperator{\cl}{cl} \DeclareMathOperator{\Int}{in} \DeclareMathOperator{\bd}{bd}$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 $\eqalign{\bd(\bd(\bd(W)))&=\cl(\cl(W)-\Int(W)-\Int(\cl(W)-\Int(W)))\cr&-\Int(\cl(W)-\Int(W)-\Int(\cl(W)-\Int(W)))\cr &=\cl(W)-\Int(W)-\Int(\cl(W)-\Int(W))\cr&-(\Int(\cl(W)-\Int(W))+\Int(\cl(W)-\Int(W))\cr &=\cl(W)-\Int(W)-\Int(\cl(W)-\Int(W))\cr &=\bd(\bd(W)).}$
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2Please try to break lines. – 2011-10-25