# Question regarding Rudin's proof that for $Y \subset X$, $E \subset Y$ is open relative to $Y$ iff $E = Y \cap G$ for some open $G \subset X$

I am reading Rudin's Principles of Mathematical Analysis and I have some questions regarding Theorem 2.30 (p.36) about open relative sets.

2.30 Theorem. Suppose $$Y \subset X$$. A subset $$E$$ of $$Y$$ is open relative to $$Y$$ if and only if $$E = Y \cap G$$ for some open subset $$G$$ of $$X$$.

Proof. Suppose $$E$$ is open relative to $$Y$$. To each $$p \in E$$ there is a positive number $$r_p$$ such that the conditions $$d ( p,q ) < r_p$$, $$q \in Y$$ imply that $$q \in E$$. Let $$V_p$$ be the set of all $$q \in X$$ such that $$d ( p,q ) < r_p$$, and define $$G = \bigcup_{p \in E} V_p.$$ Then $$G$$ is an open subset of $$X$$, by Theorems 2.19 and 2.24.

Since $$p \in V_p$$ for all $$p \in E$$, it is clear that $$E \subset G \cap Y$$.

By our choice of $$V_p$$, we have $$V_p \cap Y \subset E$$ for every $$p \in E$$, so that $$G \cap Y \subset E$$. Thus $$E = G \cap Y$$, and one half of the theorem is proved.

Conversely, if $$G$$ is open in $$X$$ and $$E = G \cap Y$$, every $$p \in E$$ has a neighborhood $$V_p \subset G$$. Then $$V_p \cap Y \subset E$$, so that $$E$$ is open relative to $$Y$$.

The question is: after Rudin defines $$G$$, he immediately states that $$G$$ is open in $$X$$, by Theorems 2.19 (every neighborhood is an open set) and 2.24 (every union of open sets is an open set). By Theorem 2.24 I know that the union of open sets is still an open set. However, I don't know how Theorem 2.19 applies in this situation and thus implies $$V_p$$ is open. Because $$V_p$$ is a set of all $$q$$, not $$p$$, if it is the set of all $$p$$, I know it must be open.

Can somebody please help me explain why $$G$$ is open? And please be very specific.

$V_p$ is a ball-neighbourhood (he states the definition for it: all $q \in X$ such that $d(p,q) < r_p$, so it's just the open ball with centre $p$ and radius $r_p$), so it's open by 2.19.
And $G$ is just a union of these open sets, hence open too.