Let $S$ be a subset of the metric space $E$. A point $p \in S$ is called interior point of $S$ if there is an open ball in $E$ of center $p$ which is contained in $S$. Prove that the set of interior points of $S$ is an open subset of $E$ (called the interior of $S$).

It seems as if the statement has already proven what it wants me to show. In order to prove that something is an open set I need to find an open ball centered at some $p$ with radius $r$. If the set of open balls is contained in subset $S$ then $S$ is open. Yet I feel that I'm oversimplifying the actual process of proving this.

Anyone have any idea on this?


2 Answers 2


Denote by $S^\circ$ the interior of $S$.

A set $E$ is open if for every $x \in S$ there is a ball $B$ with $x \in B \subset E$.

Now let $S$ be an arbitrary set and let $x \in S^\circ$. By definition there is a ball $B$ satisfying $x \in B \subset S$. You are being asked to prove that there is a ball $B$ satisfying $x \in B \subset S^\circ$. It is a subtle distinction.

Here it goes. Let $x \in S^\circ$ (assuming without loss of generality that $S^\circ$ is nonempty) and let $B$ be a ball satisfying $x \in B \subset S$. Now let $y \in B$. Then also $y \in B \subset S$ so by definition $y \in S^\circ$. Thus $$ y \in B \implies y \in S^\circ$$ so that $B \subset S^\circ$. We conclude $x \in B \subset S^\circ$ and thus $S^\circ$ is open.

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    $\begingroup$ I'm thinking perhaps the definition of "open" being used is subtly different again: a subset $S$ of $E$ is open if for every $x \in S$, there is a ball centred at $x$, contained in $S$! I think the exercise should involve some use of the triangle inequality. $\endgroup$ Feb 13, 2018 at 0:19
  • $\begingroup$ @TheoBendit: yes you are right. The answer uses the fact that the ball $B$ should contain $x$ and not necessarily be centered at $x$. This works but one should be able to use the definition based on ball centered at $x$. Thus if $B=B(x, r), y\in B$ then ball $B'=B(y, r'), 0<r'\leq r-d(x, y) $ is contained in $B$ and we are done. $\endgroup$
    – Paramanand Singh
    Feb 13, 2018 at 3:35

I will prove that an open ball is an open set. Can you use this fact, along with Umberto's answer to prove that the interior of a set is open?

Let $B_r(x)$ be the open ball centred around $x$ with radius $r>0$. We wish to show that $B_r(x)$ is open.

Let $y\in B_r(x)$, and let $s =r - d(x,y)$. $s>0$ since $d(x,y)<r$. We will now argue that $B_s(y) \subset B_r(x)$.

Let $z \in B_s(y)$. We have that $d(x,z) \le d(x,y)+d(y,z) < d(x,y) + (r- d(x,y))=r$. Hence, $z \in B_r(x)$.

  • $\begingroup$ I understand this now. The use of Triangle Inequality was quite cunning. I see that my confusion also stemmed from the fact that I looked at the problem as $S$ being the set of interior points which was obviously false. Thank you again. Excellent answer. $\endgroup$
    – Nicklovn
    Feb 13, 2018 at 22:51

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