Proving every open set in metric space $X$ is the union of a subcollection of base 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)
 A: *

*(1) 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 $\alpha$. 

*(2) Every open set in $X$ is union of a subcollection of $\{V_\alpha\}$. 
You are asking about (2) $\Rightarrow$ (1), right?

If $G\subseteq X$  is open, then by (2) there is a set $I$ such that $G=\bigcup_{\alpha\in I} V_\alpha$. 
Clearly, for every $\alpha\in I$ we have $V_\alpha\subseteq G$. (Union of some system contains all sets in this systems.)
From the equality $G=\bigcup_{\alpha\in I} V_\alpha$ we see that every $x\in G$ is contained in $V_\alpha$ for some $\alpha\in I$ (simply by definition of union).
Thus for every $x$ we have an $\alpha\in I$ such that:


*

*$x\in V_\alpha$

*$V_\alpha\subseteq G$

You also asked:

I don't understand why $V_\alpha \cap G \in \{V_\alpha\}$.

This is not true in general. I.e. an intersection of a basic set and open set need not be a basic set.
E.g. the intervals with rational endpoints form a basis for the real line with the usual topology.
The set $G=(0,\sqrt2)$ is open. But intersection of $G$ with a basic set is not necessarily a basic set, take $G\cap(1,2)=(1,\sqrt2)$ for example.
Of course, $V_\alpha\cap G=V_\alpha$ whenever $V_\alpha\subseteq G$. (Which is probably what you wanted to use.)
