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Let $E\subset\mathbb{R}^n$ be an arbitrary subset. Is it true that $$ \partial\partial E=\partial E $$ The inclusion $\partial\partial E\subset \partial E$ is pretty clear to me since $\partial E$ is closed, however I fail to see the other inclusion. How to prove it?

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  • $\begingroup$ What about the rationals in the reals? $\endgroup$ – Michael Burr Apr 7 '16 at 11:08
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It's False:

Consider $\mathbb Q\subset \mathbb R$ then $\partial \mathbb Q=\mathbb R$ but $\partial\partial \mathbb Q=\emptyset$ .

However you cold prove that $\partial \partial \partial A= \partial\partial A$ for any set A.

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Wikipedia says:

$\partial\partial E=\partial E$ iff $\partial E$ has no interior points

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