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Math Help - Proof that a function space is complete

  1. #1
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    Proof that a function space is complete

    Another one that I'm kind of uncomfortable with my solution:

    B(X) vector space of all bounded functions in X. Show that it's going to be defined under the sup norm:

    d(f,g) = sup (mod(f(x)-g(x))


    What I did was to show that every Cauchy sequence in B(X) converges to an f, (so mod(f-fn) gets arbitrarily small with n --> infinity);

    and the terms fn and fm get arbitrarily close to each other with a high n. So for a high enough n, we have

    mod(f) <= mod(fn) + mod(f - fn) < E, for an arbitrary E, so f , the limit of the cauchy sequence, is a bounded function, belonging to B(X). Hence, the space is complete.

    I feel that I'm missing something here, since i didn't use the sup norm directly.
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  2. #2
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    What are the domain and codomain of the functions in your space?
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  3. #3
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    Domain is X, codomain is not defined in the question.
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  4. #4
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    Quote Originally Posted by galahad View Post
    Domain is X, codomain is not defined in the question.
    It has to be any metric space X and the codomain needs to be \mathbb{R}.

    Now, you've taken analysis before, right?
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    It has to be any metric space X and the codomain needs to be \mathbb{R}.

    Now, you've taken analysis before, right?

    I'm taking my first analysis course right now. Sorry if my questions are too basic.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by galahad View Post
    I'm taking my first analysis course right now. Sorry if my questions are too basic.
    I guess that's my fault for not asking the question clearly. Your question is actually a rephrasing (with slight modification) of a theorem you should have covered when you did uniformly convergent sequences of functions.
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  7. #7
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    The codomain could be any Banach space couldn't it?

    edit: does X have to be a metric space?
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by maddas View Post
    The codomain could be any Banach space couldn't it?
    Yes.

    edit: does X have to be a metric space?
    It could be any topological space.


    My reasoning for my response was the OPs previous posts. But, noticing that he said "mod" maybe not a Banach space and \mathbb{C}?
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  9. #9
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    Does it have to be a topological space?
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by maddas View Post
    Does it have to be a topological space?
    For any fruitful, non-smartass discussion haha, yes.
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  11. #11
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    Why? You don't actually use the topology on X in the proof I don't think.
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  12. #12
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by maddas View Post
    Why? You don't actually use the topology on X in the proof I don't think.
    It's possible I'm being old-fashioned, but I think it's in the spirit of the question. Usually one talks at most generally about \mathcal{C}\left[X,\mathcal{B}\right] with X\in\text{TOP} (:P) since all of the fun things you can do with it are in that context. If you want to try to do it without, go ahead.
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