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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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Counter example of what? –  rotten Apr 26 '12 at 18:40
    
that it has a limit point outside –  El Angel Exterminador Apr 26 '12 at 18:42

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up vote 1 down vote accepted

Take $e^{(n)}\in\ell_1$ given by $e^{(n)}_k:=\begin{cases}1&\mbox{ if }n=k\\\ 0&\mbox{otherwise} \end{cases}$. Then $d(e^{(m)},e^{(n)})=2$ if $m\neq n$, 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 $\lVert \{x_n\}\rVert=\sum_{n=1}^{+\infty}|x_n|$, and the unit ball of an infinite dimensional normed space is never compact for the topology of the norm.

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thank you dear sir –  El Angel Exterminador Apr 26 '12 at 18:52

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