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(Converse of my last question)

If $A \subseteq \ell_\infty$, and $A=\{l\in \ell_\infty: |l_n| \le b_n \}$, where $b_n$ is a sequence of real, non-negative numbers, then if $\lim (b_n) = 0$ it must mean that $A$ is compact subset of $X$.

Take any sequence of sequences $(x_n)$, our goal is to construct a subsequence, $(x_{n_k})$ which will converge. Suppose for $n \geq N$ we have $|b_n|<\epsilon$ (by convergence of $(b_n)$). For the first $N-1$ points, however, I thought we could use Bolzanno-Weirstrass on each of the $N-1$ first terms (since $|x_n| \le b_n \forall n$) in the following way: apply Bolzano Weirstrass on all the first terms, then for the second terms apply it on the subsequence we got from the first terms, and so on... and claiming this subsequence will converge to $\{x_1,x_2,...,x_{N},...\}$ - edited, where $x_i$ is the limit of the $i_{th}$ subsequence.

However, the subsequence created could have a few cases where we end up with no terms at the end of this inductive argument. My teacher told me to use Cantor Diagonalization to avoid this, but I don't see how this would work.

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Sorry this might make no sense, the subsequence maybe should converge to $\{x_1,x_2,x_3,...,x_{N-1},x_N,x_{N+1},...\}$ where $x_i$ is the limit of the $i^{th}$ subsequence in the Bolzano Weirstrass argument. – Simon Sehayek Oct 20 '12 at 15:52
Ignore last comment I edited in my original post – Simon Sehayek Oct 21 '12 at 2:37
up vote 2 down vote accepted

With diagonal argument: the sequence $\{x_n^{(1)}\}$ is bounded, hence we can find $A_1$, an infinite subset of $\Bbb N$, such that $\{x_n^{(1)}\}_{n\in A_1}$ is convergent. Then construct by induction a decreasing sequence $\{A_k\}$ of infinite subsets of $\Bbb N$ such that $\{x_n^{(k)}\}_{n\in A_k}$ converges to some $x^{(k)}$. Then denote $j_k$ the $k$-th element of $A_k$, and consider the sequence $x_{j_k}$. It's a subsequence of $\{x_n\}$, and for each $j$, $x_{n_k}^{(j)}\to x^{(j)}$. The fact that $b_n\to 0$ gives convergence in $\ell_{\infty}$.

Using pre-compactness: fix $\varepsilon>0$, and $N$ such that $|b_n|<\varepsilon$ if $n\geq N$. Then we use the fact that $\prod_{j=1}^N[-b_j,b_j]\subset \Bbb R^n$ is precompact to get $v_1,\dots,v_l$ such that each element of $\prod_{j=1}^N[-b_j,b_j]$ is in some $B(v_j,\varepsilon)$. Finally, define $w_j:=(v_j,0,\dots,0)$. As $A$ is closed and precompact in a Banach space, it's a compact set.

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I don't see how this is different from my argument and how you are using a diagonalization argument? How are you constructing $\{A_k\}$ and doesn't it need to be an infinite subset of $\{A_{k-1}\}$? – Simon Sehayek Oct 20 '12 at 16:50
In what you did, you extract a subsequence which depends on $\varepsilon$ (as the $N$ depends on it). – Davide Giraudo Oct 20 '12 at 16:54
Yeah I didn't mean that, I stated that it should converge to $\{x_1,x_2,...,x_{N-1},x_{N},x_{N+1},...\}$ in the comment under my question. But the problem with this argument is what if the subset that occurs at every step will yield the empty set at infinity (e.g. after the first inductive step we get all terms divisible by 2, the next step we get all terms divisible by 4, the next step we get all terms divisible by 8, etc. then at $\infty$ we are left with the empty set) – Simon Sehayek Oct 20 '12 at 16:59
You are right, I forget to say that the sequence $\{A_k\}$ is decreasing. Check that the sequence $n_k$ I define is increasing. (I don't need the intersection of the $A_k$) – Davide Giraudo Oct 20 '12 at 17:03
What does it mean for a sequence of sets to be decreasing, is it that $A_{k} \subseteq A_{k-1}$? Also why don't you need the intersection of the $A_k$ (we can't suddenly add new terms in the kth step)? – Simon Sehayek Oct 20 '12 at 17:27

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