Total boundedness of $K \subset \ell^1$, iff $K$ is uniformly summable & bounded

Question

Let $K \subset \ell^1$ and define uniform summability of $K$ such that for any $\varepsilon >0$ exists a natural number $N \in \mathbb{N}$ such that for every $x \in K$ the $N$-tail of the series over $x$ is bounded, i.e. $\sum_{n=N}^\infty|x_n| \le \varepsilon$.

Claim: $K$ is totally bounded, iff $K$ is bounded and uniformly summable.

I was able to show that if $K$ is totally bounded, it is indeed bounded. By assumption $K$ can be covered by a finite number of $\varepsilon$-balls, thus it was easy to construct a ball centered around $0_{\ell^1}$, which covers all of $K$.

Now, under the same assumption, I want to show that $K$ is uniformly summable. By total boundedness of $K$, I know that for any $\delta >0$ and $x \in K$, I can find $y\in K$ such that $\|x-y\|_{\ell^1}< \delta$. I don't know how to translate this fact to an expression, which could be extended to bounding an $N$-tail of the series over $x$.

A hint for the backwards implication of the claim is also appreciated. Thanks

Answer 1

1. Suppose $K$ is totally bounded, you want to show that it is uniformly summable: Fix $\epsilon > 0$, then there are finitely many points $\{y_1,y_2,\ldots, y_m\} \subset K$ such that, for each $x\in K$, there is $1\leq i\leq m$ such that $$ \|x-y_i\| < \epsilon $$ For each $1\leq i\leq m$, there exists $N_i \in \mathbb{N}$ such that $$ \sum_{k=N_i}^{\infty} |y_i(k)| \leq \epsilon $$ Let $N := \max\{N_i : 1\leq i\leq m\}$, then show that, for any $x\in K$, $$ \sum_{k=N}^{\infty} |x(k)| \leq 2\epsilon $$ Hence, $K$ is uniformly summable.

2. Conversely, suppose $K$ is bounded and uniformly summable. You want to show that it is totally bounded. I will give an elaborate hint: Fix $\epsilon > 0$, and choose $N \in \mathbb{N}$ such that $$ \sum_{k=N}^{\infty} |x(k)| \leq \epsilon $$ for all $x\in K$. Furthermore, let $M > 0$ such that $\|x\| \leq M$ for all $x\in K$. Now consider the set $$ B := \{(x(1),x(2), \ldots, x(N), 0, 0, \ldots) : x\in K\} $$ This can be thought of as a bounded subset of $\mathbb{R}^N$, so $B$ is totally bounded. Now try to complete the proof from here.