# 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)

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Each $g\in G$ is contained in $V_{g \alpha } \subset G$ for some $\alpha .$ Hence $\displaystyle\bigcup_{g\in G} V_{g \alpha } \subset G \subset \displaystyle\bigcup_{g\in G} V_{g \alpha }.$ The reverse implication is easier. –  user17794 Aug 6 '12 at 7:13
You talk about a metric space in the title, but the question never mentions metric spaces and seems to be about topological spaces in general. –  joriki Aug 6 '12 at 7:13
Did you want to write $I=\{\alpha (\exists x\in G) x\in V_\alpha\subset G\}$ instead of $I=\{\alpha | x\in G ⇒ x\in V_\alpha \subset G\}$? –  Martin Sleziak Aug 6 '12 at 7:14
@Martin That looks much better. Would you edit my post? –  Katlus Aug 6 '12 at 7:17
@joriki honestly i don't know which one is right since i haven't learned what is topological space yet. I guess it is a metric space. Would you give me a brief explanation about topological space? –  Katlus Aug 6 '12 at 7:21

• (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:

• $x\in V_\alpha$

• $V_\alpha\subseteq G$

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