Understanding the definition of a base of a metric space

Question

Definition: Let $(X,d)$ be a metric space. A collection $\{v_n\}$ of subsets is said to be a base for $X$ if for every $x \in X$ and every open set $G \subset X$, such that $x \in G$ we have $x \in \{V_n\} \subset G$ for some $N$.

The notation is what is giving me a hard time here. I just don't understand how it translates into "Every open set in $X$ is the union of a subcollection of $\{V_n\}$".

I don't understand how this is the case if that collection is always a subset of $G$, not equal to $G$?

Answer 1

"Every open set in X is the union of a subcollection of $\{V_n\}$" means that any open set $U$ can be written as $$U=\bigcup V_i,$$ where $V_i$ belong to the base for any $i$. E.g. in $\mathbb{R}$ with Euclidean topology, a base could be $$\{B(x,r):x,r\in \mathbb{Q}\}.$$

Answer 2

Take an open set $O$, and $x\in O$. By definition, there exists $V=V_x$ in your base such that $x\in V\subseteq O$. This means that $O=\bigcup \{x:x\in O\}\subseteq \bigcup V_x\subseteq O$; so $O=\bigcup V_x$ is a union of open basic sets.