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

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    Is $\mathsf{bd}$ the boundary?2011-10-25
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    @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

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