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Math Help - Hard Proof

  1. #1
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    Hard Proof

    Hi!

    I have been struggling with an proof for days now and would like to see if there is someone in here who can help me out.

    the excersise is as follows;
    We have a Banach Space A. Let {e_alpha}_(alpha belongs to A) be a base when A is considered being a linear space and show with Baires category theorem that the index set A is either finite or not countable?

    Someone who is indulged to try?

    Peter
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  2. #2
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    Quote Originally Posted by Peter_L View Post
    We have a Banach Space A. Let {e_alpha}_(alpha belongs to A) be a base when A is considered being a linear space and show with Baires category theorem that the index set A is either finite or not countable?
    I don't think you meant to use the same letter for the name of the space and the name of the index set, did you? I'll use E for the space and A for the index set.

    If A is finite then E is finite-dimensional. Suppose, for a contradiction, that E is infinite-dimensional but has a countable base {e_n} (indexed by the natural numbers). For n=1, 2, ..., let E_n be the subspace of E spanned by the first n basis vectors e_1, ..., e_n. Let A_n be the complement of E_n in E. Then A_n is open and dense (easily proved). So by Baire's theorem the intersection of the A_n is dense. But every element of E is in E_n for n large enough, so the intersection of the A_n is empty. Contradiction.
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  3. #3
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    thanks for the reply! I have two questions though,

    1. is it ok to just take a space A_n which is the complement of the subspace E_n in E?

    2. so, we have the complement A_n to E_n in the Banach Space E, and since E_n is a sub space spanned by the n first base vectors, A_n must be open and dense? is it because we know that E_n is closed?
    Last edited by Peter_L; March 8th 2008 at 09:41 AM.
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  4. #4
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    Quote Originally Posted by Peter_L View Post
    thanks for the reply! I have two questions though,

    1. is it ok to just take a space A_n which is the complement of the subspace X_n in X?

    2. so, we have the complement A_n to X_n in the Banach Space X, and since X_n is a sub space spanned by the n first base vectors, A_n must be open and dense? is it because we know that X_n is closed?
    We seem to have another change of notation. The Banach space has become X, and presumably X_n is what I called E_n, the finite-dimensional subspace of X spanned by the first n basis vectors.

    Just to be clear, the set A_n is not a subspace of X. It is the set-theoretic complement of X_n in X. So I prefer not to call it a "space": it's just a set. It is open because it's the complement of X_n, which is closed because it is finite-dimensional.

    The reason that A_n is dense is that if a point x is not in A_n then it must be in X_n. Then we can approximate x by elements of the form x+\epsilon e_{n+1} (where ε is small but nonzero). These points are not in X_n and are therefore in A_n.
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  5. #5
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    oops, mixed the notation up with another exercise..hehe,

    but thank you very much!! it was to a lot of help for me!! find it pretty hard with this kind of abstract math...
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