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Math Help - [SOLVED] How do I prove R is a baire space?

  1. #1
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    [SOLVED] How do I prove R is a baire space?

    Hello! Thanks for reading.
    I need to use the fact that R is a baire space somewhere, but I just don't seem to be able to prove it. I'm not even sure if it's trivial or not (meaning, if it's even required to prove it), but I guess I can't take the risk. Moreover, it's disturbing I don't know how too.
    I know that R is a complete metric space.
    I've read that complete spaces are baire spaces, but I haven't got the proof for that.
    My definition for baire space is one of the following (which are equivalent, I've proved that):
    1) every countable intersection of dense open sets is a dense set.
    2) every countable unification of non-dense (not sure of the term) sets has an empty interior.

    Can't see why R satisfies that, nor how to prove the more general claim, that complete space => Baire space.

    Bless you!
    Tomer.
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  2. #2
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    Have you looked at a site such as this: Baire Space -- from Wolfram MathWorld?
    It seems that you have a problem with definitions.
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  3. #3
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    I don't know what's the deal with my lecturer, but all the definitions I get are for some reason theorems, and the theorems I need to prove are definitions
    I've got an excercise "teaching" me what a baire space is, using the definition I wrote down. I need to prove some things.
    Later on, I need to prove a more "practical" question, regarding R, but I need to use the fact that it's a baire space. I cannot just say "R is complete (or homeomorphic to itself = a complete space) and therefore it's a baire space", because we have not studied that. All we studied in class is Baire's theorem, but it mentions nothing of Baire spaces (though I can see the connection).

    In other words, I'm still stuck .

    Thank you very much for responding!
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  4. #4
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    Quote Originally Posted by aurora View Post
    I need to use the fact that R is a baire space somewhere, but I just don't seem to be able to prove it. I'm not even sure if it's trivial or not (meaning, if it's even required to prove it), but I guess I can't take the risk. Moreover, it's disturbing I don't know how too.
    I know that R is a complete metric space.
    I've read that complete spaces are baire spaces, but I haven't got the proof for that.
    My definition for baire space is one of the following (which are equivalent, I've proved that):
    1) every countable intersection of dense open sets is a dense set.
    2) every countable unification of non-dense (not sure of the term) sets has an empty interior.
    Here is the proof of Baire's theorem, as given in Gert Pedersen's excellent book Analysis Now.

    Theorem If {A_n} is a sequence of open, dense subsets of a complete metric space (X,d), then the intersection \bigcap A_n is dense in X.

    Proof. Let B_0 be an closed ball in X with radius r>0, and let B_0^o denote its interior. Since A_1 is dense in X and open, the set A_1\cap B_0^o contains a closed ball B_1 with radius < r/2. Since A_2 is also dense and open, A_2\cap B_1^o contains a closed ball B_2 with radius < r/4. By induction we find a sequence (B_n) of closed balls in X, such that B_n\subseteq A_n\cap B_{n-1}^o and radius (B_n)<2^{-n}r for every n. Since X is complete, there is evidently a point x in X such that \{x\} = \bigcap B_n\subseteq B_0\cap\bigl(\bigcap A_n\bigr). This shows that \bigcap A_n intersects every nontrivial ball in X, ensuring its density.
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  5. #5
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    Thank you very much! That's exactly what I needed!
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