Alternative proof of a theorem in Rudin

[identity1]

In Rudin, theorem 3.7 is that:

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

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

 

[jgens]

You have the right idea here, but the way things are written currently, your 'proof' is not correct. The way you have chosen {p_nk} forces it to be the constant sequence {q}, which does not work since we do not know q in E.

 

[identity1]

Could you elaborate? I only see that my construction shows that d(p_{n_1}, q)<2, d(p_{n_2}, q)<1, \ldots, d(p_{n_k}, q)<\frac{2}{k}.

 

[jgens]

Fix p_nk. Then d(p_nk,q_m) < m^-1. But since m is arbitrary, this forces p_nk = q.

 

[identity1]

But that's not now I defined p_{n_k}. Each index, n_k, corresponds to a specific n; you can't fix p_{n_k} but then alter the index of the q in the expression like that.

 

[jgens]

But that's not now I defined p_{n_k}. Each index, n_k, corresponds to a specific n; you can't fix p_{n_k} but then alter the index of the q in the expression like that.

Nope. In fact, n_k is indexed only on k. Be more careful.

 

[identity1]

Yea, but p_{n_k} is indexed by n and k. And in my construction, I write for each q_n there is a p_{n_k} for which something holds. So each p_{n_k} explicitly corresponds to each q_n..

 

[identity1]

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 n_k&gt;n_{k-1} so that the subsequence keeps moving forward (we are guaranteed to be able to do this also by the definition of convergence).

 

[jgens]

Yea, but p_{n_k} is indexed by n and k. And in my construction, I write for each q_n there is a p_{n_k} for which something holds. So each p_{n_k} explicitly corresponds to each q_n..

Wrong again! As I said before, p_nk 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?

 

[identity1]

OH. Wow. I should've written p_{k_n}. My bad xP. Barring that notational slip-up, how's the proof look to you.

 

[jgens]

With that change of notation, your proof looks good to me :) Out of curiosity, how does Rudin do this proof differently?

 

[identity1]

Well, he starts by constructing p_{n_1} similarly, and defines \delta=d(q, p_{n_1}). Then he proceeds by induction: have having constructed p_{n_1}, \ldots, p_{n_{i-1}}, he constructs p_{n_i} so that n_i&gt;n_{i-1}, d(q, p_{n_i})&lt;2^{-i}\delta. My main point of confusion I guess, was that I wasn't sure why induction was needed. I'm still thinking about it, actually.

 

[identity1]

Actually, talking about it made me realize the answer :P

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