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UPDATED: In preparation for an exam I am struggling with the following problem. We have $A:=\{x=(x_{n})_{n}\in \ell^{2}| \phantom{x} \|x\|\leq 1\}$. Consider the metric $d:A\times A \rightarrow \mathbb{R}_{+}$ defined by $$d(x,y)=\sum_{n=1}^{\infty}(1/3)^{n}|x_{n}-y_{n}|.$$ Prove that $(A,d)$ is sequentially compact.

What I have done so far: I have updated what I have done so far using the hints below. But I still have some questions as I want to make sure I understand it thorougly.

Let $(x_{n})_{n}$ be an arbitrarily chosen sequence in $A$. Now first take a subsequence $(x_{n1})_{n}$ such that the first component converges. Now take a subsequence of this subsequence $(x_{n2})_{n}$ such that the second component converges. Going on in the same way we get a sequence of nested subsequences $(x_{n(k+1)})_{n}\subset (x_{nk})_{n}$. Now take the sequence $(x_{kk})_{k}$, this construction guarantees that every component of this sequence converges, it is equivalent to the convergence in the metric. Also note that for each $k,n$ that $x_{nk}$ is a sequence of real numbers so we have in fact three layers of sequences. Since the elements are sequences and we take sequences of these elements and than a sequence of subsequences of the sequence of elements in $A$. Hence, $(A,d)$ is sequentially compact.

(Question: Have I now formally proved that indeed $(A,d)$ is sequentially compact)

I don't have much experience with these kind of proofs so any help is much appreciated.

Lech121
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  • Made updates using the hints below. – Lech121 Sep 20 '13 at 19:27
  • You need to use an indexed subscript to keep track of the various subsequences to get it totally right. Something like this: The first subsequence is given by indices $n_{j1}$ so that $x_{n_{j1}}1$ converges as $j\to\infty$. The next one will be given by $n_{j2}$ being a subsequence of $(n_{j1})$ so that $x_{n_{j2}2}$ converges, and so forth. You may prefer to write $y_{jk}=x_{n_{jk}k}$ instead, so that $(y_{j(k+1)})j$ is a subsequence of $(y{jk})j$, then take the diagonal $y{jj}$. This awful mess is the reason I wrote the note I refer to in my answer. – Harald Hanche-Olsen Sep 20 '13 at 20:09

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(Note: The question was substantially altered since I wrote this answer. If you wish to see what I really answered, use the history mechanism to see the original question.)

You got it wrong from the start. Let's drop the comma and just write $x_n=(x_{nj})_j$, shall we? Now you want convergence in the first component, so you'd like to have $x_{n1}$ convergent as $n\to\infty$. It won't be convergent necessarily, but you can pick a subsequence that converges. Note that this really means picking a subsequence of $(x_n)$ so that $(x_{n1})$ converges. Next, pick a subsequence of the subsequence so that $(x_{n2})$ converges, and so on. Finally, “take the diagonal” to finish the proof.

Incidentally, I think the conventional notation for subsequences is a horrible mess, and makes this kind of proof much harder than necessary, both to write and to read. So I have written up an alternative view of the diagonal method. The basic idea: All the difficulties stem from the insistence of relabeling indices when taking a subsequence, so that the sequence is always indexed on the set of natural numbers. By allowing subsets of the natural numbers as index sets, we simplify this tremendeously.

Harald Hanche-Olsen
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  • Thank you very much for your help. I have updated my question using your hints and I want to make sure that I understand it thoroughly so if you could give me feedback if I have now really finished the proof I would really appreciate it. – Lech121 Sep 20 '13 at 19:23
  • Much better. Unfortunately, I don't have time for a detailed reading right now. I think you have the right idea, but your notation at the end still seems somewhat deficient. (It is notoriously hard to get right, so don't feel bad about it.) I'll add a comment to the question with a bit more detail. – Harald Hanche-Olsen Sep 20 '13 at 20:02
  • Thank you. I look forward to your comment. – Lech121 Sep 20 '13 at 20:07
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You need to show something that's bounded with respect to the $d$ metric has a convergent subsequence, not that something that's bounded with respect to the normal $l^2$ norm has a convergent subsequence.

Matt Rigby
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