This is a standard prelim question, to be done without appealing to any "big" theorems: prove that $B = \{x \in l^{2}: \Sigma_{n=1}^{\infty}n|x_{n}|^{2} \leq 1 \}$ is compact.
There is an answer here, which shows that $B$ is sequentially compact. So I wanted to do it by showing that it is totally bounded. But I ran into trouble, because I could not find a uniform bound for the series.
Now, if we had $B = \{x \in l^{2}: \Sigma_{n=1}^{\infty}n|x_{n}| \leq 1 \}$, the proof would go as follows:
If $x\in B$, then $\forall n\in \mathbb N,\ |x_n|\le 1/n$ we can choose an integer $N$ so that $n>N\Rightarrow \sum_{k=N+1}^{\infty}\frac{1}{n^{2}}<\epsilon/2.$ Define $B^N=\{x\in B:x_n=0,\ n>N\}.\ B^N$ is easily seen to be compact. Indeed, it is homeomorphic to a closed subset of the closed unit ball in $R^N.$ In particular, it is totally bounded, so we can find a finite $\epsilon/2-$net $E\subseteq B^N.$ But then by construction, $E$ is an $\epsilon-$ net for $B$:
let $x\in B$. Then, choose a $y\in E$ such that $\sum^N_{k=1} |x_n-y_n|^2<\epsilon/2.$ Then, $\sum |x_n-y_n|^2\le \epsilon/2+\sum^{\infty}_{k=N+1}|x_n|^2\le \epsilon/2+ \sum^{\infty}_{k=N+1} \frac{1}{n^2}<\epsilon.$
I'd like to see a proof showing total boundedness of (the original) $B$.
Best Answer
Note that $\sum_{n=N}^\infty n |x_n|^2 \ge N \sum_{n=N}^\infty |x_n|^2$. Given $\epsilon > 0$, we want to cover $B$ by balls of radius $\epsilon$.
Let $\pi_N$ be the orthogonal projection on coordinates $< N$, i.e. $\pi_N(x)_i = x_i$ if $i < N$, $0$ if $i \ge N$. Thus $\|\pi_N(x) - x\|^2 = \sum_{n=N}^\infty |x_n|^2 \le 1/N$ if $x \in B$. Take $N > 4/\epsilon^2$, so this says $\|\pi_N(x) - x\| \le \epsilon/2$.
Now, as in your proof, $\pi_N(B)$ is isometric to a compact subset of $\mathbb R^N$, so cover it by finitely many balls of radius $< \epsilon/2$. Then the corresponding balls of radius $\epsilon$ cover $B$.