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Math Help - More Metric Spaces

  1. #1
    is up to his old tricks again! Jhevon's Avatar
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    More Metric Spaces

    Here are two more questions that have been bothering me. for the first, i don't think my solution is good enough, there's somethign too "obvious" about it. for the second, i haven't a clue

    1) Let X and Y be metric spaces. Assume that X is complete and that there exists f: X \to Y which is bijective and continuous. Suppose that f^{-1} is uniformly continuous. Prove that Y is complete

    2) Prove that any open subset of a complete metric space is homeomorphic to a complete metric space. (Hint: If A is the complement of the open set, use the metric
    \left. D_1(x,y) = D(x,y) + \left| \frac 1{D(x,A)} - \frac 1{D(y,A)} \right| \right)


    Here are definitions/theorems i figured would come in handy:

    Defintion: A metric space M is complete if every Cauchy sequence in M is convergent to a point of M.

    Theorem: Any convergent sequence in a metric space is a Cauchy sequence.

    Definition: A function f from a metric space X to a metric space Y is uniformly continuous if for every \epsilon > 0 there exists a \delta > 0 such that, for all x,x' \in X, D(x,x') < \delta implies D[f(x),f'(x)] < \epsilon.

    Theorem: A uniformly continuous function carries Cauchy sequences into Cauchy sequences.

    Definition: Let f_i,f be functions from a metric space X to a metric space Y. We say that the sequence f_i converges uniformly to f if the following is true: For any \epsilon > 0 there exists N (depending on \epsilon but independent of x) such that D[f_i(x),f(x)]< \epsilon for all i \ge N and all x \in X.

    Definition: Two metric spaces are said to be homeomorphic if there exists between them a bijection which is continuous in both directions.


    Whew! ok, here's what i did.

    ...oh, wait, i have to go to class. I'll post my "proof" to question 1) later, there's not enough time.

    for question 2)

    Proof:

    I have no clue, but I have a gut feeling it is true (they wouldn't ask me to prove something that was false, after all)

    QED

    Ok, so of course that proof was a joke, and a bad one, i know. but i'm not sure what to do
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Jhevon View Post
    1) Let X and Y be metric spaces. Assume that X is complete and that there exists f: X \to Y which is bijective and continuous. Suppose that f^{-1} is uniformly continuous. Prove that Y is complete
    If (y_n) is a Cauchy sequence in Y, then the uniform continuity of f^{-1} implies that (f^{-1}(y_n)) is Cauchy in X, and therefore converges to some point x_0\in X. The continuity of f then tells you that y_n\to f(x_0). So Y is complete.
    Quote Originally Posted by Jhevon View Post
    2) Prove that any open subset of a complete metric space is homeomorphic to a complete metric space. (Hint: If A is the complement of the open set, use the metric  D_1(x,y) = D(x,y) + \left| \frac 1{D(x,A)} - \frac 1{D(y,A)} \right| .)
    Call the space X, and the open subset Y. There are two things to prove here. First, that the identity map from Y with the metric D to Y with the metric D_1 is a homeomorphism. Second, that D_1 is a complete metric on Y.

    Since  D(x,y)\leqslant D_1(x,y), it is clear that the identity map is continuous from (Y,D_1) to (Y,D). For the other direction, suppose that x_n\to y in (Y,D). Then D(x_n,A)\to D(y,A), and it follows that D_1(x_n,y)\to0. That is (an outline of) the proof that the identity is a homeomorphism.

    To show that (Y,D_1) is complete, let (x_n) be a Cauchy sequence in (Y,D_1). Then (x_n) is Cauchy in (Y,D) and so has a limit x_0\in X. You have to show that x_0\in Y. For this, notice that \left| \frac 1{D(x_m,A)} - \frac 1{D(x_n,A)} \right|\leqslant D_1(x_m,x_n), and so \left(1/{D(x_n,A)}\right) is a Cauchy sequence of real numbers. So it converges to a limit c. Therefore D(x_n,A)\to1/c\neq0. Hence D(x_0,A) = 1/c>0, which shows that x_0\in Y.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Opalg View Post
    If (y_n) is a Cauchy sequence in Y, then the uniform continuity of f^{-1} implies that (f^{-1}(y_n)) is Cauchy in X, and therefore converges to some point x_0\in X. The continuity of f then tells you that y_n\to f(x_0). So Y is complete.
    my solution was very close to that. somehow i didn't see why uniform convergence was needed, but now i do. thanks

    Call the space X, and the open subset Y. There are two things to prove here. First, that the identity map from Y with the metric D to Y with the metric D_1 is a homeomorphism. Second, that D_1 is a complete metric on Y.

    Since  D(x,y)\leqslant D_1(x,y), it is clear that the identity map is continuous from (Y,D_1) to (Y,D). For the other direction, suppose that x_n\to y in (Y,D). Then D(x_n,A)\to D(y,A), and it follows that D_1(x_n,y)\to0. That is (an outline of) the proof that the identity is a homeomorphism.

    To show that (Y,D_1) is complete, let (x_n) be a Cauchy sequence in (Y,D_1). Then (x_n) is Cauchy in (Y,D) and so has a limit x_0\in X. You have to show that x_0\in Y. For this, notice that \left| \frac 1{D(x_m,A)} - \frac 1{D(x_n,A)} \right|\leqslant D_1(x_m,x_n), and so \left(1/{D(x_n,A)}\right) is a Cauchy sequence of real numbers. So it converges to a limit c. Therefore D(x_n,A)\to1/c\neq0. Hence D(x_0,A) = 1/c>0, which shows that x_0\in Y.
    thanks, i had no clue about this one. well, i didn't get a chance to think about it much
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