ForMyThunder said:
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.
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.
