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Alternative proof of a theorem in Rudin

  1. Dec 23, 2011 #1
    In Rudin, theorem 3.7 is that:

    The set of subsequential limits of a sequence [itex]\{p_n\}[/itex] in a metric space X is a closed subset of X.

    I tried proving this myself, and my proof came out to be similar to the one in Rudin, yet to me, simpler. I wanted to post it and ask if it seems correct.

    Like in Rudin, call the set of subsequential limits of [itex]\{p_n\}[/itex] [itex]E^*[/itex]. Then if q is a limit point of [itex]E^*[/itex], we need to show it is in [itex]E^*[/itex], meaning that there is some subsequence of [itex]\{p_n\}[/itex] which converges to q. Now, for every [itex]n\in\mathbb{N}[/itex], choose some [itex]q_n\in E^*[/itex] such that [itex] d(q_n, q)<\frac{1}{n}[/itex]. This is possible since we assumed that q was a limit point of [itex]E^*[/itex].

    Now, since each [itex]q_n\in E^*[/itex], this means that there is some [itex] p_{n_k}[/itex] such that [itex]d(p_{n_k}, q_n)<\frac{1}{n}[/itex] (this is from the definition of convergence). Thus, [itex]d(q, p_{n_k})\le d(q, q_n)+d(q_n, p_{n_k})<\frac{2}{n}[/itex], showing that this subsequence [itex]\{p_{n_k}\}[/itex] converges to q, proving the theorem.

    I'm just curious to see if this proof is valid, since it seems to me a bit simpler than the one given in the textbook. Thanks!
     
  2. jcsd
  3. Dec 23, 2011 #2

    jgens

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    You have the right idea here, but the way things are written currently, your 'proof' is not correct. The way you have chosen {pnk} forces it to be the constant sequence {q}, which does not work since we do not know q in E.
     
  4. Dec 23, 2011 #3
    Could you elaborate? I only see that my construction shows that [itex] d(p_{n_1}, q)<2, d(p_{n_2}, q)<1, \ldots, d(p_{n_k}, q)<\frac{2}{k}[/itex].
     
  5. Dec 23, 2011 #4

    jgens

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    Fix pnk. Then d(pnk,qm) < m-1. But since m is arbitrary, this forces pnk = q.
     
  6. Dec 24, 2011 #5
    But that's not now I defined [itex]p_{n_k}[/itex]. Each index, [itex]n_k[/itex], corresponds to a specific n; you can't fix [itex]p_{n_k}[/itex] but then alter the index of the q in the expression like that.
     
  7. Dec 24, 2011 #6

    jgens

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    Nope. In fact, nk is indexed only on k. Be more careful.
     
  8. Dec 24, 2011 #7
    Yea, but [itex]p_{n_k}[/itex] is indexed by n and k. And in my construction, I write for each [itex]q_n[/itex] there is a [itex]p_{n_k}[/itex] for which something holds. So each [itex]p_{n_k}[/itex] explicitly corresponds to each [itex]q_n[/itex]..
     
  9. Dec 24, 2011 #8
    If it helps you to move over that point, Rudin does the same thing in his proof as well.

    One thing I did forget to write is that we choose [itex]n_k>n_{k-1}[/itex] so that the subsequence keeps moving forward (we are guaranteed to be able to do this also by the definition of convergence).
     
  10. Dec 24, 2011 #9

    jgens

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    Wrong again! As I said before, pnk is indexed only on k. It has no correspondence with n. If you do not know this, then you need to review the definition of a subsequence.

    I do not have my copy of Rudin with me, but I would guess Rudin is more careful with his construction. If he does the exact same thing, then why do you think that your proof is an improvement?
     
  11. Dec 24, 2011 #10
    OH. Wow. I should've written [itex]p_{k_n}[/itex]. My bad xP. Barring that notational slip-up, how's the proof look to you.
     
  12. Dec 24, 2011 #11

    jgens

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    With that change of notation, your proof looks good to me :) Out of curiosity, how does Rudin do this proof differently?
     
  13. Dec 24, 2011 #12
    Well, he starts by constructing [itex]p_{n_1}[/itex] similarly, and defines [itex]\delta=d(q, p_{n_1})[/itex]. Then he proceeds by induction: have having constructed [itex]p_{n_1}, \ldots, p_{n_{i-1}}[/itex], he constructs [itex]p_{n_i}[/itex] so that [itex]n_i>n_{i-1}, d(q, p_{n_i})<2^{-i}\delta[/itex]. My main point of confusion I guess, was that I wasn't sure why induction was needed. I'm still thinking about it, actually.
     
  14. Dec 24, 2011 #13
    Actually, talking about it made me realize the answer :P

    In my correction, where I realized I forgot to asset that [itex] k_n>k_{n-1}[/itex], I'm implicitly using a type of weak induction. Induction is necessary, and he does it more completely in his proof.
     
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