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Topology - prove that X has a countable base

  1. Jan 11, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]X[/itex] - topological compact space

    [itex]\Delta = \{(x, y) \in X \times X: x=y \} \subset X \times X[/itex]

    [itex]\Delta = \bigcap_{n=1}^{\infty} G_{n}[/itex], where [itex]G_{1}, G_{2}, ... \subset X \times X[/itex] are open subsets.

    Show that the topology of [itex]X[/itex] has a countable base.

    2. Relevant equations

    3. The attempt at a solution

    i have no idea what to start with. i don't really want to get a solution, just some clues...
     
  2. jcsd
  3. Jan 11, 2012 #2

    micromass

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    Let's first find a candidate of a base, shall we??

    For each [itex](x,x)\in G_n[/itex], we can find [itex](x,x)\in U_x\times V_x\subseteq G_n[/itex]. Now apply compactness on the [itex]U_x\times V_x[/itex].
     
  4. Jan 11, 2012 #3
    meaning: choose a finite number of them?
    and compactness of what?
     
  5. Jan 11, 2012 #4

    micromass

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    Yes.

    Of X.

    You might also want to prove X to be Hausdorff...
     
  6. Jan 11, 2012 #5
    hm, ok, and i do this for each n, and get countable family of finite families of small open sets, and my base are all those small open sets?

    what do you mean?
     
  7. Jan 11, 2012 #6

    micromass

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    Not yet. You need to take all finite intersections as well.

    Now try to prove that it is indeed a base. (you will need to make one last modification to the base in the end)

    Certainly you know what Hausdorff means?
     
  8. Jan 11, 2012 #7
    :( i don't get it. why intersections? and... intersections of what?


    yes, but according to the definition of compactness i know, X is Hausdorff and i don't have to prove it.
     
  9. Jan 11, 2012 #8

    micromass

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    You found a collection of [itex]U\times V[/itex]'s. Now take all the finite intersections.

    We will eventually want a base such that

    [tex]\bigcap_{x\in G}{G}=\{x\}[/tex]

    Ah, ok. Never mind then.
     
  10. Jan 11, 2012 #9
    still don't get it... maybe it's too late. i'll think more about it in the morning.

    but: if i take only [itex]U[/itex]'s from those [itex]U\times V[/itex]'s... why isn't it already our base?
     
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