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Special Case of the Baire Category Theorem

  1. Dec 7, 2012 #1
    1. So the solutions claim a different and better proof of this, but I just wanted to see if mine made sense.

    Question - Let X be a compact Hausdorff space; let {An} be a countable collection of closed sets of X. Show that if each set An has empty interior in X, then the union [itex]\bigcup[/itex]An has empty interior in X.

    2. Relevant equations

    If X is compact Hausdorff with no isolated points, for any nonempty open set U of X and any point x of X, there exists a nonempty open set V contained in U such that x[itex]\notin[/itex][itex]\overline{V}[/itex].

    3. The attempt at a solution

    Now that I've seen the solution it seems obvious, but here is what I did.

    If we let Int([itex]\bigcup[/itex]An)=U and take a point xi of Ai, then there exists an open set Vi [itex]\subset[/itex] Int([itex]\bigcup[/itex]An) s.t. x[itex]\notin[/itex][itex]\overline{V}[/itex]i. We can form this construction because we can choose a point in Int([itex]\bigcup[/itex]An) different from x, because if not, then we would either have an empty interior (in the case where x[itex]\notin[/itex] Int([itex]\bigcup[/itex]An) ) or we would have a one-point open set (in the case where x[itex]\in[/itex] Int([itex]\bigcup[/itex]An) , which would mean that Ai for some i has non-empty interior. Then by the Hausdorff property, there is a W1 and W2 about x and y respectively, that are disjoint. Then W2[itex]\bigcap[/itex]Int([itex]\bigcup[/itex]An) is an open set contained in Int([itex]\bigcup[/itex]An), not containing x. It follow that the closure of this set, which I'll call V, does not contain x.

    Then we can choose [itex]\overline{V}[/itex]1 [itex]\supset[/itex] [itex]\overline{V}[/itex]2 [itex]\supset[/itex] .....

    And since X is compact, there exists an x in this infinite intersection. Then, since U is the union of {An}, x must be contained in Ai for some i. But then [itex]\overline{V}[/itex]i [itex]\cap[/itex] Ai is a closed set containing x but not xi. Then the complement of [itex]\overline{V}[/itex]i [itex]\cap[/itex] Ai in Ai is open and non empty in Ai, implying that Ai has non-empty interior, a contradiction.
     
    Last edited: Dec 7, 2012
  2. jcsd
  3. Dec 7, 2012 #2

    micromass

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    It doesn't really seem obvious to me that the x that you constructed actually lies in U. Sure, every [itex]V_i[/itex] lies in U. But you're taking the intersections of the [itex]\overline{V_i}[/itex]. Do those necessarily lie in U?

    also, you seem to make LaTeX more difficult than it needs to be. If you want to type [itex]\overline{V_i}[/itex], then you can do this by

    Code (Text):

    [itеx]\overline{V_i}[/itex]
     
    This seems a lot easier than using the [NOPARSE]...[/NOPARSE] tags.

    Here is a LaTeX FAQ: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3
     
    Last edited: Dec 7, 2012
  4. Dec 7, 2012 #3
    O, I did not see that. Its not obvious to me that x lies in U, and I don't know whether or not its true. Thanks for pointing that out.

    Maybe I don't actually need x to be in U, only for x to be in [itex]\bigcup[/itex]{An}. Still though, its not clear to me that is true.

    Anyway, maybe this proof just doesn't work.
     
  5. Dec 7, 2012 #4

    micromass

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    I'm thinking that you might be able to fix it by taking an open set [itex]U^\prime[/itex] such that [itex]\overline{U^\prime}\subseteq U[/itex] that is nonempty. Then you might take care that all the [itex]V_i\subseteq U^\prime[/itex]. That would for the intersection of the [itex]\overline{V_i}[/itex] to be in U.

    I didn't really work it out, but you can try some trickery like this.
     
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