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The definition of the closure of a set $S$ is the intersection of all the closed subsets of the topological space $X$ which contain $S$ as a subset.

The definition of boundary of $S$ is $\text{Comp}\{\text{Int}(S) \cup \text{Ext}(S)\}$.

Since the boundary of $S$ does not contain $\text{Int}(S)$ as a subset, it is not a closed set that contains $S$, so the boundary of $S$ is not in the closure of $S$, but apparently it is. Why is this?

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  • $\begingroup$ What is $\operatorname{Comp}$? $\endgroup$ – Asaf Karagila Apr 1 '15 at 12:24
  • $\begingroup$ $\text{Comp}(S)$ is the complement of the set $S$ $\endgroup$ – mr eyeglasses Apr 1 '15 at 12:27
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The argument in the last paragraph of the post is wrong. If it were true, you would be able to apply it and prove that no point lies in the closure.

Now let $p\in\partial S$. Then $p\not\in\mathrm{Ext} S$, and so any open neighborhood of $p$ intersects $S$. It follows that $p$ lies in any closed set which contains $S$, i.e. $p$ is in the closure.

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  • $\begingroup$ How do you know that any open neighborhood of $p$ intersects $S$? $\endgroup$ – mr eyeglasses Apr 1 '15 at 12:03
  • $\begingroup$ @ᴇʏᴇs otherwise, by definition we have $p\in\mathrm{Ext}S$. $\endgroup$ – Amitai Yuval Apr 1 '15 at 12:10
  • $\begingroup$ Sorry, which definition are you using? Our definition of $\text{Ext}S$ is $\text{Int}(\text{Comp}S)$, and our definition of neighborhood of $p$ is a subset of the topological space that contains an open set such that $p$ is contained in the open set $\endgroup$ – mr eyeglasses Apr 1 '15 at 12:12
  • $\begingroup$ @ᴇʏᴇs ... and the definition of open neighbourhood of $p$ should be: an open set containing $p$. $\endgroup$ – Hagen von Eitzen Apr 1 '15 at 12:36

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