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Let $S\subseteq \mathbb{R}$. A point $x\in S$ is an interior point of $S$ if there is an open interval $(a,b)$ for which $x\in (a,b) \subseteq S$. Denote the set of all interior points of $S$ by $\mathrm{int}(S)$.

  1. Prove that $S$ is an open set if and only if $S=\mathrm{int}(S)$.
  2. Prove that $\overline{S}$ and $\overline{\mathrm{int}(S)}$ need not be equal.

The definition of an open set:

A subset $U$ of a metric space $(M, d)$ is called open if, given any point $x \in U$, there exists a real number $\epsilon > 0$ such that, given any point $y \in M$ with $d(x, y) < \epsilon$, $y$ also belongs to $U$.

I have been thinking about 1. and 2. for 30 minutes and I am completely blank. Sorry.

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    @copper.hat: I totally agree.2012-11-08

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$\newcommand{\int}{\operatorname{int}}$HINTS:

(1) Clearly it’s always true that $\int S\subseteq S$. Suppose that $S$ is open; you want to show that for each $x\in S$, $x\in\int S$, so let $x\in S$. Since $S$ is open there is an $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subseteq S$, because $d(x,y)<\epsilon$ if and only if $y\in(x-\epsilon,x+\epsilon)$. Why not just take $a=x-\epsilon$ and $b=x+\epsilon$; will that work?

Now suppose that $\int S=S$; to prove that $S$ is open, you must show that for each $x\in S$ there is an $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subseteq S$, so let $x\in S$. Then $x\in\int S$, so there are $a$ and $b$ such that $x\in(a,b)\subseteq S$. What happens if you let $\epsilon=\min\{x-a,b-x\}$?

(2) Consider the set $\Bbb Q$, or any non-empty finite set.

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    @Joakim: If you set $a=x-\epsilon$ and $b=x+\epsilon$, you have $x\in(a,b)\subseteq S$. But $(a,b)$ is open, so $(a,b)\subseteq\int S$, and therefore $x\in\int S$.2012-11-11