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Math Help - [SOLVED] Metric spaces and Baire's theorem.

  1. #1
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    [SOLVED] Metric spaces and Baire's theorem.

    Hello! Here I go again with Baire's theorem.
    I'm solving this infinitely long question with millions of sub questions, and I'm stuck on the last one.
    I won't (for now) go into detail with all of the question because it's kinda complex (lots of definitions etc...). I'll try however, to describe the last question:
    I need to prove that a function f:R--->R which is continous exactly in every rational point can't exist. In other words, there isn't a function f that satisfies: {x in R | f is continous in x} = Q.
    There's a hint: assume there is one, and contradict using Baire's theorem.

    Now, this is the last question after many more, which were obviously supposed to build the way to the solution... It tingles, but I can't reach a contradiction.
    What I have proved in the past questions, and I think is relevant, is:
    1) The set of all points in which f is continous is a "G-delta" set in R. Meaning, it can be presented as a countable intersection of open sets in R.
    2) Q is a "F-sigma" set in R. Meaning, it can be presented as a countable union of closed sets in R.

    The rest of the questions seem to only point out the first fact I described, and they contain some bothersom definitions, so I'll hope my intuition is right and leave you with just that, unless someone tells me there's just not enough info or something.

    Baire's theorem, as I've studied it, means:
    Let (X,d) be a complete metric space, and Let {Di} be no-where-dense sets (=> int{cl(Di)} = empty set).
    Then, int(union(Di)) = empty set (that's a contable union, sorry for the wording of math).

    That's it really. The claim is simple, but I'm not sure the proof is... although I feel close. if I assume by contraction, I get that Q is a "G-delta" set and a "F-sigma" set. But I just cannot progress any further, nor can I see how to use baire's theorem.
    Is that fact that int (intersection{Ui}) = empty set a contradiction, where Ui are open, and the intersection is countable? If I were to know that, I would solve the problem, but I can't seem to prove that.

    I'm done - I'm waiting for you genious people to save me!
    Sorry for the length.

    Bless you!
    Tomer.
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  2. #2
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    The set of rational numbers is not a G_\delta-set. Proof here.
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  3. #3
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    Bless you, good ol' Opalg!
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