# Topology proof

1. Aug 5, 2009

### ForMyThunder

I can't seem to find out how to prove this theorem:

A collection {fa | a in A} of continuous functions on a topological space X (to Xa) separates points from closed sets in X if and only if the sets fa-1(V), for a in A and V open in Xa, form a base for the topology on X.

Could anyone help me out? Thanks.

2. Aug 6, 2009

### winter85

hi ForMyThunder,

first let's go over some basic definitions and facts,

1) A collection of continuous functions $$\{ f_{a} : a \in A \}$$ on a topological space X is said to separate points from closed sets in X iff for every closed set $$B \subset X$$, and every $$x \notin B$$ , $$\exists a \in A : f_{a}(x) \notin \overline{f_{a}(B)}$$

2) If for a collection $$\textbf{B}$$ of open sets of X, for every open set $$U \subset X$$ and every $$x \in U$$ there is an element $$B$$ of $$\textbf{B}$$ such that $$x \in B \subset U$$, then $$\textbf{B}$$ is a base for the topology of X.

Let $$\textbf{B} = \{f_{a}^{-1}(V) : a \in A, V \mbox{ open in } X_{a} \}$$. In light of 2, let $$U \subset X$$ be open, and let $$x \in U$$. since $$U$$ is open, $$U'$$ is closed, and $$x \notin U'$$, so by (1) $$\exists a \in A : f_{a}(x) \notin \overline{f_{a}(U')}$$.
Now $$\overline{f_{a}(U')}$$ is closed, being a closure, therefore $$\overline{f_{a}(U')}'$$ is open. Since $$f_{a}(x) \notin \overline{f_{a}(U')}$$, we have $$f_{a}(x) \in \overline{f_{a}(U')}'$$, so $$\exists V \subset \overline{f_{a}(U')}'$$ such that $$x \in V$$ and $$V$$ is open.
Since every set is a subset of its closure, now we have: $$V \subset (f_{a}(U'))'$$, we want to show that $$f_{a}^{-1}(V) \subset U$$. This is done by simple facts from set algebra, perhaps you can try to do it on your own?

This shows the first half of the theorem.

Last edited: Aug 6, 2009
3. Aug 6, 2009

### ForMyThunder

Awesome, thanks. I think I can figure out the second part. Thanks again.

4. Aug 6, 2009

### winter85

you're welcome. it helps to draw a diagram to get a feel for whats going on :)