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I know that: $x \in \partial E \Longleftrightarrow $ every neighborhood of x contains at least one point in the set and at least one point not in the set.

And I also know that $$\partial (\partial E) \subseteq \partial E $$ assuming that $E \subset X$, and $(X, d)$ metric space.

The question is: When is it that $$\partial (\partial E) \subset \partial E$$ ?

It depends on the interior points of $\partial E$ , but how can I prove that? And, above all, how can I imagine a set with this kind of boundary?

Hints are accepted, thanks in advance!

moonknight
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1 Answers1

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By the definition

$$\partial A=\overline{A}\cap\overline{X\backslash A}$$

Take $X=\mathbb{R}$ and $E=[0,1]\cap\mathbb{Q}$. Then $\overline{E}=[0,1]$ since $\mathbb{Q}$ is dense. Also $\mathbb{R}\backslash\mathbb{Q}$ is dense so $\overline{\mathbb{R}\backslash E}=\mathbb{R}$. Therefore $\partial E=[0,1]$. Moreover $\partial(\partial E)=\{0, 1\}$.

Lemma. $\partial(\partial E)=\partial E$ if and only if $\partial E$ has empty interior.

Proof.

"$\Leftarrow$" Since $\partial E$ is closed, then $\overline{\partial E}=\partial E$. Since $\partial E$ has empty interior then the complement is dense $\overline{X\backslash \partial E}=X$. Thus

$$\partial(\partial E)=\overline{\partial E}\cap\overline{X\backslash \partial E}=\partial E\cap X=\partial E$$

"$\Rightarrow$" Assume that $\partial E$ has a nonempty interior and take $x$ from the interior. Note that $x\not\in\overline{X\backslash\partial E}$ because if $U\subseteq\partial E$ is such that $x\in U$ and open, then $X\backslash\partial E\subseteq X\backslash U$ and thus $\overline{X\backslash\partial E}\subseteq X\backslash U$ because the set on the right side is closed.

Since $x\not\in \overline{X\backslash\partial E}$ then obviously $x\not\in\partial(\partial E)$ which contradicts the assumption. $\Box$

So for example this always holds in case when $E$ is closed or open.

freakish
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