A Question concerning $\ell^1$ Let $a_n >0$ and $\sum_{n=1}^\infty a_n < \infty$. 
Define $$
M=\{ \{x_n\}_{n=1}^\infty \in \ell^1 : \forall n, |x_n|\le a_n\}
$$
Then I want to prove that $M$ is compact and $M$ cannot be included any finite-dimensional subspace of $\ell^1$.
 A: First of all $M$ contains the sequences
$$
(\overline x^n)_m:=\delta_{m,n} a_n,\quad a_n>0
$$
Therefore, any subspace which contains $M$, has to include all the linearly independent sequences $(e^n)_m:=\delta_{m,n},$ therefore cannot be finite dimensional.
As for the compactness, the direct way is to show that any sequence $\{x^n\}$ in $M$ (which is necesessarly bounded) has a convergent subsequence $\{\bar x^m\}$. Here you can use a standard diagonal argument. Let $x_1^{k_1}$ be a convergent subsequence of $ (x^n)_1$, $x_2^{k_2}$ be a convergent subsequence of $ (x^{k_1})_2$, $x_3^{k_3}$ be a convergent subsequence of $ (x^{k_2})_3$,... This subsequences exist,  since for any $k$, $|(x^n)_k|\leq a_k $. Then define
$$
\overline x^\ell=((x^{k_\ell})_1, (x^{k_\ell})_2,(x^{k_\ell})_3,\ldots).
$$
A little of work is needed to show the convergence of  $\{\bar x^\ell\}$. Here you have to use that $ |(\bar x^\ell)_k|\leq a_k$ and so
$$
\sum_{k=N}^\infty|(\bar x^\ell)_k|\leq \sum_{k=N}^\infty a_k=R_N,\qquad \lim_{N\to \infty} R_N=0.
$$
