Rudin PMA p.45 problem 23
A collection $\{V_\alpha\}$ of a subsets of $X$ is said to be base for $X$ if the following is true: For every $x\in X$ and every open set $G\subset X$ such that $x\in G$, $x\in V_\alpha \subset G$ for some $\alpha$. " In other words, every open set in X is the union of a subcollection of $\{V_\alpha\}$. "
I don't understand why those two statements are equivalent.
Let $G$ be an open set. Let $I=\{\alpha | (\exists x\in G) x\in V_\alpha \subset G\}$. Then by the first definition, $G\subset \bigcup_{\alpha \in I} V_\alpha$.
I don't understand why $V_\alpha \cap G \in \{V_\alpha\}$. (I think this is critical to show the equivalence)