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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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    @Katlus: You can find a brief explanation about topological spaces at [Wikipedia](http://en.wikipedia.org/wiki/Topological_space). The relationship between topological spaces and metric spaces, in a nutshell, is that a metric induces a topology, namely the one generated by the open balls with respect to the metric. So a metric space is a topological space in a natural way, but a topological space may or may not admit a metric (be metrizable), and if it does, the metric is far from unique. All concepts in your question are topological and not metric.2012-08-06

1 Answers 1

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  • (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.)