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Is this subset compact in $l_1$ of all absolutely convergent real sequences, with the metric:$d_1(\{a_n\},\{b_n\})=\sum_{1}^{\infty}|a_n-b_n|$ closed unit ball centered at $0$ with radius $1?$ I guess not as it is not sequentially compact but I need one counter example.plz help

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    that it has a limit point outside2012-04-26

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Take $e^{(n)}\in\ell_1$ given by $e^{(n)}_k:=\begin{cases}1&\mbox{ if }n=k;\\\ 0&\mbox{otherwise.} \end{cases}$ Then $d\left(e^{(m)},e^{(n)}\right)=2$ if $m\neq n$ (because in the series giving the $\ell^1$ norm of $e^{(m)},e^{(n)}$,the terms with index $m$ and $n$ are equal to one while all the others are zero), which proves that the unit ball is not sequentially compact for $d$. A set which is not sequentially compact in a metric space cannot be compact.

In fact, the metric $d$ comes from a norm, namely $\left\lVert \left\{x_n\right\}\right\rVert=\sum_{n=1}^{+\infty}\left|x_n\right|$, and the unit ball of an infinite dimensional normed space is never compact for the topology of the norm.

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    $e^1$ is a sequence.2018-09-12