Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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.

share|cite|improve this question
To get you started on 1.: could you show that $\mathrm{int}(S)$ is open? So, let $x$ in $\mathrm{int}(S)$, what do you know and what should you prove? – Did Nov 8 '12 at 17:33
OP: How do you define interior? – copper.hat Nov 8 '12 at 17:33
@copper.hat: The definition is given in the problem statement – Hagen von Eitzen Nov 8 '12 at 17:47
@HagenvonEitzen: Thanks, I missed that completely... I prefer the definition of the interior as the largest open set contained in the set and the closure as the smallest closed set containing the set. It is 'visually' more appealing to me. – copper.hat Nov 8 '12 at 17:49
@copper.hat: I totally agree. – Hagen von Eitzen Nov 8 '12 at 17:57
up vote 1 down vote accepted


(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.

share|cite|improve this answer
Are you sure about (2)? After all $(0,1)\cup (1,2)$ is already open. I suggest $\{0\}$ instead. – Hagen von Eitzen Nov 8 '12 at 17:49
@Hagen: Sometimes I really want a bigger screen. I misremembered the question, essentially reversing closures and interiors. Thanks. – Brian M. Scott Nov 8 '12 at 17:57
I just need a bigger brain. – copper.hat Nov 8 '12 at 17:58
I frequently misinterpret questions. I could blame age, but that would be unfair to age :-). – copper.hat Nov 8 '12 at 18:01
@copper.hat: Wait till next birthday, when you get the joker. :-) – Brian M. Scott Nov 8 '12 at 18:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.