Is the set of decreasing sequences on the unit ball of $l^1[0,\infty)$ a compact set under the sup norm? Formally, let
$$
\left\{\boldsymbol{x} = (x_1,x_2,x_3,\ldots)\in [0,1]^{\mathbb{N}^*} : 1 \geq x_1 \geq x_2 \geq \ldots \geq 0 ~; \sum_{i=1}^{\infty} x_i \leq 1\right\}.
$$
Is this a compact set with the sup norm $\|\boldsymbol{x}\|_{\infty} = \sup_{i\in \mathbb{N}} |x_i|$ ?
 A: Consider a sequence $y_n$,
$$y_n=(\frac1n,\frac1n,\frac1n,\cdots,\frac1n,0,0,\cdots).$$
A: The anwser is yes. We prove the conclusion by arbitrarily choosing a sequence $\{\boldsymbol{x}^{(n)}\}_n$ in $\{(x_1,x_2,\cdots)\in\mathbb{R}_+^\infty|\sum x_k\leq1\}$ and demonstrating it has a limit point in $\Vert\cdot\Vert_\infty$ sense.


*

*Due to Helly's selection theorem, there exists a subsequence, where for convenience we assumed to be $\{\boldsymbol{x}^{(n)}\}_n$, such that for every fixed positive integer $m$, $\{x^{(n)}_m\}_n$ converges to a number $x_m$ as $n\to\infty$. We denote $(x_1,x_2,\cdots)$ by $\boldsymbol{x}$, and apparently $$\boldsymbol{x}\in\{(x_1,x_2,\cdots)\in\mathbb{R}_+^\infty|\sum x_k\leq1\}.$$

*According to Fatou's lemma, we have
$$\sum_{m\geq1} x_m=\int \boldsymbol{x}d\mu\leq\liminf \int \boldsymbol{x}^{(n)}d\mu=\liminf \sum_{m\geq 1} x^{(n)}_m\leq 1,$$
where $\mu$ is the counting measure on $N_+$.

*Since the conponents is decreasing, 
$$x^{(n)}_m\leq1/m,\quad x_m\leq1/m,\quad n\in N_+,m\in N_+.$$
$\forall \epsilon>0$, choose $N\in N_+$ such that
$$|x^{(n)}_m-x_m|<\epsilon,\quad n>N,m\leq 2/\epsilon.$$
Now $\Vert\boldsymbol{x}^{(n)}-\boldsymbol{x}\Vert_\infty<\epsilon,\ n>N.$
