# closed unit ball with radius 1

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

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

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.
• +1 @Davide,,,,,But im not getting how $d(e^{(m)} ,e^{(n)) } = 2$ ..i thinks It must be $d(e^{(m) },e^{(n)}) = 1$.. can u elaborate that – jasmine Sep 12 '18 at 16:36
• ,,@ Davide im not getting still confused according to ur edits ..im thinking like that $d(e^{(m},e^{(n)}) = d(1,1) = \sum_{n=1}^{\infty} |e^{m} -e^{n}| = \sum_{n=1}^{\infty} |1-1| = 0$ – jasmine Sep 12 '18 at 19:07
• What does zhe equalityy $d(e^{(m},e^{(n)}) = d(1,1)$ mean? Maybe the $m$ and $n$ canconfuse you. As an advice, you could try to compute $d\left(e^{(1)},e^{(2)}\right)$ to see what happens. – Davide Giraudo Sep 12 '18 at 19:14
• here $e^1= 1$ and $e^2 =1$...as im not getting still how $d(e^1 - e^2) = 2$ – jasmine Sep 12 '18 at 19:19