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So, this is something Rudin leaves to the reader, and I gave this my best shot. I'll have some questions at the end.

Proof.

Let $\epsilon \gt 0$ be given. Suppose $\{p_n\}$ is a sequence in metric space $X$ converging to some $p \in X$. Then there exists an integer $N$ such that $n \ge N$ implies $d(p,p_n) \lt \epsilon$. Let $n_1, n_2, n_3$, ... be a sequence of strictly increasing positive integers. Then $n_k \ge k$ for any integer $k$. Hence, given a subsequence $\{p_{n_k}\}$ of $\{p_n\}$, $k \ge N$ implies $d(p,p_{n_k}) \lt \epsilon$. Thus $p_{n_k} \rightarrow p $.

Now, suppose every subsequence $\{p_{n_{k_i}}\}$ of $\{p_n\}$ converges to some $p$, where $i$ ranges over the set $I$. Then to each $\{p_{n_{k_i}}\}$ corresponds a positive integer $N_i$ such that $n_{k_i}$ $\ge N_i$ implies $d(p_{n_{k_i}},p) \lt \epsilon$. Take $N = \max\limits_{i \in I}\{N_i\}$ so that for all $n_{k_i} \ge N, d(p_{n_{k_i}},p) \lt \epsilon.$ Then since to each $m \in \mathbb N$ corresponds an $n_{k_i}$, for all $m \ge N$ we have that $d(p_m, p) \lt \epsilon$. Hence $p_n \rightarrow p$, as desired. This completes the proof.

Questions:

  1. Is this correct?

  2. Can a sequence have an infinite number of subsequences?

2.5. Does my "$i \in I$" cover that? Do I need to bring in countability or define $I$ further?

  1. Is my assertion that to each $m \in \mathbb N$ corresponds an $n_{k_i}$ accurate? For instance, couldn't you just think of the two subsequences of positive and negative $n_k$ terms?

Thanks!

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  • $\begingroup$ The second part of your proof seems to me to be not entirely correct since you cannot prove that the maximum $N$ exists. This is essentially part of your 2nd question. Indeed, most sequences have infinite many subsequences (for example, consider $(x_{n + m})_n$ for natural numbers $m$). However, note that every sequence is a subsequence of itself, so if every subsequence converges, then the whole sequence converges by assumption and you do not have to prove anything. $\endgroup$ – Matthias Klupsch Sep 6 '16 at 6:32
  • $\begingroup$ The first part is correct. A sequence can have an infinite number of subsequences (to see this: note that $2^{\mathbb{N}}$ is uncountable by Cantor's Theorem), so taking the maximum over an infinite set is not necessarily well defined. $\endgroup$ – Jay Sep 6 '16 at 6:39
  • $\begingroup$ @MatthiasKlupsch Wow, I can't believe I missed that. It's that simple, huh? Thank you! Is there anyway, do you think, to modify my argument into something perfectly rigorous? Taking the max of a potentially infinite set was something I had qualms with too. $\endgroup$ – David Bowman Sep 6 '16 at 6:43
  • $\begingroup$ Isn't a sequence a subsequence of itself? $\endgroup$ – miracle173 Sep 6 '16 at 6:58
  • $\begingroup$ @miracle173 yeah, just as a set is a subset of itself. I didn't think of that. I'm trying now to think of some condition I can impose that would allow my idea to work $\endgroup$ – David Bowman Sep 6 '16 at 7:01
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Your attempt is excellent, however I think it is essential to point some things out:

1) Your proof of the first part is correct, but if you are a fan of clear writing, you could do with a little rephrasing and chopping as follows: Let $\epsilon > 0$. Then, consider a subsequence $\{p_{n_k}\}$. Note that for this $\epsilon$, there is a natural number $N$ such that $n > N \implies |p_n-p| < \epsilon$. Now, because $n_k \geq k$ for every $k$, it follows that for some $K$, $n_k > N$ for all $k > K$. Hence, considering this $K$ as your new $N$, conclude that $|p_{n_k}-p| < \epsilon$ for all $k > K$. Thus, $p_{n_k}$ converges to $p$.

2) A sequence certainly can have an infinite number of subsequences. For example, if you have a sequence $1,2,3,....$, then $2,3,...$ is a subsequence, so is $3,4,...$, so is $4,5,...$ etc. Hence it is possible. However, if your sequence is constant, for example, then you cannot have infinitely many subsequences, simply because they are all the same.

2.5) This is the contentious part. There is no finiteness condition on $I$, which creates the simple problem for the existence of the maximum. For example, suppose $N=1$ for $i_1$,$N=2$ for $i_2$, $N=3$for $i_3$ and so on, then $N$ doesn't have a maximum if $I$ is infinite, which it is most of the time. Hence, the argument breaks down here.

3) That's true,but then the argument has already broken down, so it is of no significance.

Now, how you do the second part may come as a bit of a farce. The reason is this:

Suppose I have a convergent sequence, $a_1,a_2,...$ that converges to $a$. I thought of the first number in my mind, called it $a_0$, and considered the sequence $a_0,a_1,a_2,...$,which I renumber and call $b_1,b_2,...$. Well, this sequence is also convergent!

Proof: Suppose you are given $\epsilon > 0$. Then, there exists $N \in \mathbb{N}$ such that $n > N \implies |a_n-a| <\epsilon$. But now, obviously, $b_{n+1}=a_{n}$, so therefore considering the number $N+1$, $n>N+1 \implies n-1>N \implies |a_{n-1}-a| <\epsilon \implies |b_n-a| <\epsilon$, so $N+1$ works for this $\epsilon$ and $b_n$. Hence, it follows that $b_n$ is convergent.

Now, all you need is this little trick: From your sequence $\{ a_n\}$, remove the first element $\{a_1\}$. Whatever is remaining is a subsequence of $\{a_n \}$, which is given to be convergent. But then, put back $a_1$, and then the resulting sequence, which is $\{ a_n\}$ , is convergent by the fact above! Hence, we are done.

This result is somewhat odd, because we did not seem to use the whole strength of the statement "all subsequences converge". But that's okay, we've got your result!

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For the second part, since every subsequence of $\{p_n\}$ converges and $\{p_n\}$ is a subsequence of itself, it follows that it converges.

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