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Topology proof

  1. Aug 5, 2009 #1
    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. jcsd
  3. Aug 6, 2009 #2
    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.
     
    Last edited: Aug 6, 2009
  4. Aug 6, 2009 #3
    Awesome, thanks. I think I can figure out the second part. Thanks again.
     
  5. Aug 6, 2009 #4
    you're welcome. it helps to draw a diagram to get a feel for whats going on :)
     
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