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Math Help - Analysis

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Analysis

    Hello everyone. I am posting this not because the normal proof is difficult, but because I think I have come up with an alternative way of doing it and am wondering if I made any fatal errors. Also, half the reason I am putting this up here is to improve my often remarked upon lack of coherency. So any constructive criticism in either category is much appreciated.

    Definitions: If \chi is a metric space and \left\{p_n\right\} is a sequence under \chi, then we define an infinite subsequence of \left\{p_n\right\} as follows: If we have a infinite sequence of natural numbers n_1,n_2,\cdots,n_k where n_1<n_2<\cdots<n_k then the sequence \left\{p_{n_k}\right\} is a subsequence of \left\{p_n\right\}.

    So now that we have defined an infinite subsequence, let's get to the question.

    Question: If \left\{p_n\right\} is a convergent sequence under \chi which converges to p then prove that every infinite subsequence \left\{p_{n_k}\right\} of \left\{p_n\right\} converges to p.

    Answer: I think I know how they intended it to be done, but I came up with a different method and would appreciate any crticism.

    1. In the definition of the subsequence we defined the n_k in \left\{p_{n_k}\right\} to be sequence of natural numbers such that n_1<n_2<\cdots<n_k. So instead of n_k let us define it as follows: let n_k be represented by \varphi: \mathbb{N}\mapsto\mathbb{N} which is a function of n and has the quality that it is monotonically increasing, so then we may represent a subsequence of \left\{p_n\right\} as p_{\varphi}.


    2. We know because \left\{p_n\right\} converges that for every \varepsilon>0 there exists and integer N such that \forall{N\leqslant{n}}~d_{\chi}\left(p,p_n\right)<  \varepsilon.

    3. So from this we can make the observation that d_{\chi}\left(p,p_{f}\right)<\varepsilon whenever N\leqslant{f^{}\left(n\right)}\implies{\lceil{\til  de{\varphi}\left(N\right)\rceil}}=N_0\leqslant{n}. Where \tilde{\varphi} is the generalized inverse function.

    4. Stating the end part of 4. more explicitly: for every \varepsilon>0 there exists an integer N_0 such that d_{\chi}\left(p,p_f\right)<\varepsilon whenever N_0\leqslant{n}. This the definition of a sequence which converges to the value p which is what was asked to be proven.


    Any comments greatly appreciated,
    Mathstud.
    Last edited by Mathstud28; December 10th 2008 at 05:39 AM.
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  2. #2
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    Quote Originally Posted by Mathstud28 View Post
    2. So now here is where I might have a problem if there is one.

    Based on how we defined f above it is clear that f^{-1} exists.
    You're right, that's where you have a problem. f is increasing, hence f is one-to-one, but it not onto, so you can't define f^{-1}(N) for any N.

    You can patch this by saying: let us define the "generalized inverse function" \widetilde{f} by \widetilde{f}(N)=\min\{n\geq 0\,|\,f(n)\geq N\}. By doing so, you still can say:

    3. We know because \left\{p_n\right\} converges that for every \varepsilon>0 there exists and integer N such that \forall{N\leqslant{n}}~d_{\chi}\left(p,p_n\right)<  \varepsilon.

    4. So from this we can make the observation that d_{\chi}\left(p,p_{f}\right)<\varepsilon whenever N\leqslant{f^{}\left(n\right)}\implies\widetilde{f  }(N)=N_0\leqslant{n}.
    and it works.

    But honestly, it seems to me that this is exactly the usual proof, written a bit differently... (with an explicit N_0)
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Laurent View Post
    You're right, that's where you have a problem. f is increasing, hence f is one-to-one, but it not onto, so you can't define f^{-1}(N) for any N.

    You can patch this by saying: let us define the "generalized inverse function" \widetilde{f} by \widetilde{f}(N)=\min\{n\geq 0\,|\,f(n)\geq N\}. By doing so, you still can say:


    and it works.

    But honestly, it seems to me that this is exactly the usual proof, written a bit differently...
    Hmm, maybe it is the standard proof, but not the one my book supplies. Thank you for your time Laurent!

    EDIT: After you said something I looked back and realized I was thinking of the wrong proof, my book "Principles of Mathematical Analysis" omits this proof. So I'm kind of happy that it is the common proof...all be it with the wrong notation
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  4. #4
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    Quote Originally Posted by Laurent View Post
    You're right, that's where you have a problem. f is increasing, hence f is one-to-one, but it not onto, so you can't define f^{-1}(N) for any N.
    It's true that f does not have an inverse, but since it is injective it does have a left inverse, which is all that is needed for this argument. Personally, I found Mathstud28's suggestion to be a succinct version of a proof that is sometimes presented in a much less clear way.
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