# Faulty argument on convergent subsequences

It is possible to show that given a bounded sequence in $\ell^\infty$ we can find a subsequence where each term in the sequences converges. However, this doesn't imply that the original sequence necessarily have a convergent subsequence. Why is that? Where does the argument break down?

Thanks.

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Coordinatewise convergence does not imply convergence in $\ell^\infty$.

Take the sequence

$x_1=(1,0,0,\ldots)$

$x_2=(1,1,0,0,\ldots)$

$x_3=(1,1,1,0,0\ldots)$

$\vdots$.

This sequence converges coordinatewise to the sequence $x=(1,1,1,\ldots)$ but $|| x-x_n ||_\infty = 1$ for each $n$. No subsequence of $\{ x_n\}$ can converge in $\ell^\infty$ because $x$ is the only candidate for a limit.

On the other hand, for each $j$, the $j^{\rm th}$ coordinate sequence $\{x_i(j)\}$ converges.

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Thanks, David!! –  David Nov 10 '11 at 17:44