# Exercise on relatively open subsets

This question is from Christopher Tapp's Matrix groups for Undergraduates:

Let $S \subset G \subset \mathbb R^n$ be subsets. Prove the following:
$S$ is open in $G$ if and only if for all $p \in S$, there exists $r>0$ such that $\left\{ q \in G | \text{ dist}(p,q) < r \right\} \subset S$, where dist$(p,q)$ is the usual Euclidean distance in $\mathbb R^n$.

I have attempted the $\impliedby$ direction as follows:

$\exists r > 0$ such that $\left\{ q \in G | \text{ dist}(p,q) < r \right\} \subset S$. Further, we will assume that $S$ is closed, and aim for a contradiction.

Then there exists a closed subset $E$ of $\mathbb R^n$ such that $E \cap G = S.$

Since $E$ is closed it contains all of its boundary points, so there is at least one boundary point of $E$ that is in $S$.

And clearly at this boundary point $\nexists r >0$ such that $\left\{ q \in G | \text{ dist}(p,q) < r \right\} \subset S$. Contradiction.

Would anyone be able to tell me if this attempt has a vague relation to correctness? Also, any thoughts on the reverse implication. Thanks very much.

What you do is proving that $S$ is not closed in $G$.

That is not the same as proving that $S$ is open in $G$.

For a legitimate proof of $\impliedby$ note at first that set $\{q\in G\mid\text{dist}(p,q)<r\}=G\cap\{q\in\mathbb R^n\mid\text{dist}(p,q)<r\}$ is open in $G$ since $\{q\in\mathbb R^n\mid\text{dist}(p,q)<r\}$ is an open subset of $\mathbb R^n$.

Secondly note that: $$S=\bigcup_{p\in S}\{q\in G\mid\text{dist}(p,q)<r_p\}\tag1$$ where $r_p>0$ such that $\{q\in G\mid\text{dist}(p,q)<r_p\}\subseteq S$ for every $p\in S$.

This tells us that $S$ is a union of open sets in $G$ hence is open in $G$.

The reverse.

Let $S\subseteq G$ be open in $G$ and let $p\in S$.

We have $S=G\cap U$ where $U$ is an open subset of $\mathbb R^n$.

Then $p\in U$ and some $r>0$ exists with $\{q\in\mathbb R^n\mid\text{dist}(p,q)<r\}\subseteq U$.

Then $p\in G\cap\{q\in\mathbb R^n\mid\text{dist}(p,q)<r\}=\{q\in G\mid\text{dist}(p,q)<r\}$.

No, this is not correct at all. You had to prove that $S$ is open. What you tried to do was to prove that $S$ is not closed. Why? Not being closed is not the same thing as being open; think about $[0,1)$ in $\mathbb R$, for instance.

For each $p\in S$, take $r(p)>0$ such that $\left\{q\in G\,\middle|\,\operatorname{dist}(p,q)<r(p)\right\}\subset S$. This is the same thing as asserting that $D_{r(p)}(p)\cap G\subset G$. Consider the set$$O=\bigcup_{p\in S}B_{r(p)}(p),$$which is an open subset of $\mathbb{R}^n$. By the definition of $r(p)$, $O\cap G=S$. This expresses $S$ as the intersection between $G$ and an open subset of $\mathbb{R}^n$. Therefore, $S$ is an open subset of $G$.

If $S$ is open in $G$, the $S=O\cap G$, for some open subset $O$ of $\mathbb{R}^n$. If $p\in S$, there is some $r>0$ such that $D_r(p)\subset O$. Therefore $D_r(p)\cap G\subset O\cap G=S$. But$$D_r(p)\cap G=\left\{q\in G\,\middle|\,\operatorname{dist}(p,q)<r(p)\right\}.$$