Alternative proof of a theorem in Rudin

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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!
 

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  • #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.
 
  • #3
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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].
 
  • #4
jgens
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Fix pnk. Then d(pnk,qm) < m-1. But since m is arbitrary, this forces pnk = q.
 
  • #5
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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.
 
  • #6
jgens
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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.

Nope. In fact, nk is indexed only on k. Be more careful.
 
  • #7
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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]..
 
  • #8
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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).
 
  • #9
jgens
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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]..

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.

If it helps you to move over that point, Rudin does the same thing in his proof as well.

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?
 
  • #10
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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.
 
  • #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?
 
  • #12
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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.
 
  • #13
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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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