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 [tex]\{ f_{a} : a \in A \}[/tex] on a topological space X is said to separate points from closed sets in X iff for every closed set [tex]B \subset X[/tex], and every [tex]x \notin B[/tex] , [tex]\exists a \in A : f_{a}(x) \notin \overline{f_{a}(B)}[/tex]
2) If for a collection [tex]\textbf{B}[/tex] of open sets of X, for every open set [tex]U \subset X[/tex] and every [tex]x \in U[/tex] there is an element [tex]B[/tex] of [tex]\textbf{B}[/tex] such that [tex]x \in B \subset U[/tex], then [tex]\textbf{B}[/tex] is a base for the topology of X.
Let [tex]\textbf{B} = \{f_{a}^{-1}(V) : a \in A, V \mbox{ open in } X_{a} \}[/tex]. In light of 2, let [tex]U \subset X[/tex] be open, and let [tex]x \in U[/tex]. since [tex]U[/tex] is open, [tex]U'[/tex] is closed, and [tex]x \notin U'[/tex], so by (1) [tex]\exists a \in A : f_{a}(x) \notin \overline{f_{a}(U')}[/tex].
Now [tex]\overline{f_{a}(U')}[/tex] is closed, being a closure, therefore [tex]\overline{f_{a}(U')}'[/tex] is open. Since [tex]f_{a}(x) \notin \overline{f_{a}(U')}[/tex], we have [tex]f_{a}(x) \in \overline{f_{a}(U')}'[/tex], so [tex]\exists V \subset \overline{f_{a}(U')}'[/tex] such that [tex]x \in V[/tex] and [tex]V[/tex] is open.
Since every set is a subset of its closure, now we have: [tex]V \subset (f_{a}(U'))'[/tex], we want to show that [tex]f_{a}^{-1}(V) \subset U[/tex]. 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.