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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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    Ignore last comment I edited in my original post2012-10-21

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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 understand you are doing this by induction, but doesn't it matter how you construct your set at each inductive step?2012-10-21