How $d\left(e^{(m)},e^{(n)}\right)=2$ ??

Question

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 have found this question on here closed unit ball with radius 1

My confusion is

My attempt : take $e^{(m)} = (0,0,0,.....,1,0,......) $ and $e^{(n)} = (0,1,0,0..............,)$ now $d\left(e^{(m)},e^{(n)}\right)= \sum_{n=1}^{\infty} |e^{(m)} - e^{(n)}| = \sum_{n=1}^{\infty} |1-1|=0$ Im getting $0$ , How $d\left(e^{(m)},e^{(n)}\right)=2 ?$

Answer 1

The metric is defined by:

$d_1((a_n),(b_n))=\sum_{n=1}^\infty |a_n-b_n|$

Now we want to evaluate $d_1((e_k^{(n)}), (e_k^{(m)}))$ for $n\neq m$.

Example:

$n=1$ and $m=2$ then $(e^{(n)})=(1,0,0,\dotso )$ and $(e^{(m)})=(0,1,0,\dotso )$

So $\sum_{k=1}^\infty |e_k^{(n)}-e_k^{(m)}|=|1-0|+|0-1|+|0-0|+|0-0|+\dotso =2$

[Disclaimer: The notation above is just for the intuition and mathematically not sound.]

The general prove is basically the same. If you understand the given example, you understand why the result has to be $2$.

Answer 2

The distance is not $\sum_n |e^{m}-e^{n}|$. It is $\sum_i |e^{m}_i-e^{n}_i|$ where $x_i$ denoted the $i-$th coordinate of $x \in \ell^{1}$. Note that two of the terms in this sum are $1$ and the rest are $0$.