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 $n_k>n_{k-1}$ 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 $n_i>n_{i-1}, d(q, p_{n_i})<2^{-i}\delta$. 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 $k_n>k_{n-1}$, I'm implicitly using a type of weak induction. Induction is necessary, and he does it more completely in his proof.