Real Analysis: Seemingly Obvious Question about Proving that points of S are an open subset of Metric Space E

Question

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?

Answer 1

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.

Answer 2

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, $z \in B_r(x)$.