Relative compactness criterion in $\mathfrak{c}_{0}$ I'm having trouble proving the following relative compactness criterion in $\mathfrak{c}_{0}$, the space of all sequences of complex numbers converging to $0$, with the metric $d((x_{n})_{n=1}^{\infty}, (y_{n})_{n=1}^{\infty}) = \sup_{n \in \mathbb{N}} |x_{n} - y_{n}|$:
$T \subseteq \mathfrak{c}_{0}$ is relatively compact iff there exists a sequence of positive real numbers $(x_{n})_{n=1}^{\infty}$ such that $(x_{n})_{n=1}^{\infty} \in \mathfrak{c}_{0}$ for all $y \in T$ and $n \in \mathbb{N}$, $|y_{n}|<x_{n}$.
I've only had an idea for one implication of the proof, and I'll outline it here (even though it's wrong, it's all I've got):
Let $\varepsilon >0$. For each $n \in \mathbb{N}$, we construct a finite $\frac{\varepsilon}{2^{n-1}}$-net $N_{n} = \{a^{(n1)},...,a^{(nm_{n})}\}$, and take $$x_{n} = \max_{1 \leq i \leq m_{n}}\left(|a_{n}^{(ni)}|+\frac{\varepsilon}{2^{n-2}}\right).$$
Such a sequence is indeed greater by modulo than any $y \in T$ for every $n \in \mathbb{N}$, however, there is no guarantee that $x_{n} \to 0$, because I take out a new net for every $n$.
What would be a good approach to make sure that $x_{n} \to 0$? Also, what would the proof of the other implication look like?
 A: Suppose first that $T$ is compact. For each $y\in\mathfrak{c}_0$ and each natural $n$, let $f_n(y)=|y_n|$. The function $f_n$ is continuous and therefore $f_n(T)$ is compact. Let $x_n=\sup_{y\in T}f_n(y)$. It is then trivially true that$$(\forall n\in\mathbb N)(\forall y\in T):|y_n|\leqslant x_n.\tag{1}$$I will prove that $\lim_{n\to\infty}x_n=0$.
Suppose otherwise. Then there is a $\varepsilon>0$ such that the inequality $x_n\geqslant\varepsilon$ holds infinitely many times. On the other hand, since $T$ is compact, it is totally bounded. Therefore, there is a finite subset $\bigl\{y(1),y(2),\ldots,y(k)\bigr\}$ of $T$ such that$$T\subset\bigcup_{j=1}^kB\left(y(j),\frac\varepsilon3\right).\tag{2}$$Since each sequence $y(j)$ belongs to $\mathfrak{c}_0$, there is a natural number $N$ such that$$(\forall j\in\{1,2,\ldots,k\})(\forall n\in\mathbb{N}):n\geqslant N\implies\bigl|y(j)_n\bigr|<\frac\varepsilon3.\tag{3}$$But it follows from $(2)$ and $(3)$ that $(\forall y\in T)(\forall n\in\mathbb{N}):n\geqslant N\implies|y_n|<\frac23\varepsilon$, thereby contradicting the assertion that $x_n\geqslant\varepsilon$ infinitely often.
Let us see now the general case. If $T$ is relatively compact, then $\overline T$ is compact and therefore, as we have seen, there is a sequence $(x_n)_{n\in\mathbb N}\in\mathfrak{c}_0$ of non-negative real numbers such that$$(\forall n\in\mathbb N)\left(\forall y\in\overline T\right):|y_n|\leqslant x_n.$$Since $T\subset\overline T$, $(1)$ holds.
In order to prove the reverse implication, suppose that there is a sequence $(x_n)_{n\in\mathbb N}\in\mathfrak{c}_0$ of numbers greater than or equal to $0$ and let $T$ be the set of all sequences $y\in\mathfrak{c}_0$ such that $(\forall n\in\mathbb{N}):|y_n|\leqslant x_n$. Then $T$ is closed and therefore, since $\mathfrak{c}_0$ is complete, $T$ is complete. So, in order to prove that $T$ is compact, all that remains to be proved is that $T$ is totally bounded, which is easy from the definition of $T$ and the fact that every bounded subset of $\mathbb{C}^n$ is totally bounded. And, since $T$ is compact, every subset of $T$ is relatively compact.
