I have a couple questions regarding the following proof:
2.30 Theorem: Suppse $Y \subset X$. 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 $X$.
Since $p \in V_p$ for all $p \in E$, it is clear that $E \subset G \cap Y$.
And the proof goes on...
My questions are the following:
Is the statement "$p \in V_p$ for all $p \in E$" true because if I have a point $p$, there is a point nearby $p'$ such that $p \in V_{p'}$?
I see why $E \subset G \cap Y$. But why is $E \neq G$?