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

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


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

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

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

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

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

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

    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
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    Opalg's Avatar
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    Quote Originally Posted by Jhevon View Post
    1) Let $\displaystyle X$ and $\displaystyle Y$ be metric spaces. Assume that $\displaystyle X$ is complete and that there exists $\displaystyle f: X \to Y$ which is bijective and continuous. Suppose that $\displaystyle f^{-1}$ is uniformly continuous. Prove that $\displaystyle Y$ is complete
    If $\displaystyle (y_n)$ is a Cauchy sequence in Y, then the uniform continuity of $\displaystyle f^{-1}$ implies that $\displaystyle (f^{-1}(y_n))$ is Cauchy in X, and therefore converges to some point $\displaystyle x_0\in X$. The continuity of f then tells you that $\displaystyle y_n\to f(x_0)$. 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 $\displaystyle A$ is the complement of the open set, use the metric $\displaystyle D_1(x,y) = D(x,y) + \left| \frac 1{D(x,A)} - \frac 1{D(y,A)} \right|$ .)
    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 $\displaystyle D(x,y)\leqslant D_1(x,y)$, it is clear that the identity map is continuous from (Y,D_1) to (Y,D). For the other direction, suppose that $\displaystyle x_n\to y$ in (Y,D). Then $\displaystyle D(x_n,A)\to D(y,A)$, and it follows that $\displaystyle D_1(x_n,y)\to0$. That is (an outline of) the proof that the identity is a homeomorphism.

    To show that (Y,D_1) is complete, let $\displaystyle (x_n)$ be a Cauchy sequence in (Y,D_1). Then $\displaystyle (x_n)$ is Cauchy in (Y,D) and so has a limit $\displaystyle x_0\in X$. You have to show that $\displaystyle x_0\in Y$. For this, notice that $\displaystyle \left| \frac 1{D(x_m,A)} - \frac 1{D(x_n,A)} \right|\leqslant D_1(x_m,x_n)$, and so $\displaystyle \left(1/{D(x_n,A)}\right)$ is a Cauchy sequence of real numbers. So it converges to a limit c. Therefore $\displaystyle D(x_n,A)\to1/c\neq0$. Hence $\displaystyle D(x_0,A) = 1/c>0$, which shows that $\displaystyle x_0\in Y$.
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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 $\displaystyle (y_n)$ is a Cauchy sequence in Y, then the uniform continuity of $\displaystyle f^{-1}$ implies that $\displaystyle (f^{-1}(y_n))$ is Cauchy in X, and therefore converges to some point $\displaystyle x_0\in X$. The continuity of f then tells you that $\displaystyle y_n\to f(x_0)$. 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 $\displaystyle D(x,y)\leqslant D_1(x,y)$, it is clear that the identity map is continuous from (Y,D_1) to (Y,D). For the other direction, suppose that $\displaystyle x_n\to y$ in (Y,D). Then $\displaystyle D(x_n,A)\to D(y,A)$, and it follows that $\displaystyle D_1(x_n,y)\to0$. That is (an outline of) the proof that the identity is a homeomorphism.

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