Proving a Theorem on Point-Set Topology

[ForMyThunder]

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

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

Could anyone help me out? Thanks.

 

[winter85]

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

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

 

[ForMyThunder]

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

 

[winter85]

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