The convex hull of a precompact set in a Banach space is precompact

At the begining of the Chapter 21 of Peter Lax's functional analysis, he says that

If C is a precompact set in a Banach space, so is its convex hull.

can be deduced from the following properties:

(a) S is precompact if and only if every sequence of points of S contains a Cauchy subsequence.

(b) S is precompact iff for every $\epsilon >0$ it can be covered by a finite number of balls of radius $\epsilon$.

I can not see how we can deduce it from these properties, can someone help me? Thanks!

Let $n_k$ be a finite $\epsilon$-net for $C$ and let $A= \operatorname{co} C$.

Then $N=\operatorname{co} \{n_k \}_k$ and note that $N$ is contained in a finite dimensional subspace and hence is compact. In particular, $N$ has a finite $\epsilon$-net $\nu_k$.

I claim that this is an $2\epsilon$-net for $A$.

If $x \in A$, then $x = \sum_i \lambda_i c_i$ with $\sum_i \lambda_i = 1$, $\lambda_i \ge 0$, and $c_i \in C$.

Choose $n_{k_i}$ to be an element of $C$'s $\epsilon$-net nearest $c_i$, then $n= \sum_k \lambda_i n_{k_i}$ satisfies $\|x-n\| < \epsilon$. Now choose $\nu_k$ such that $\|n-\nu_k\| < \epsilon$, then $\|x-\nu_k\| < 2 \epsilon$.

• Didn't you mean to say that $N$ is only precompact? We can conclude that $\overline{N} \subset \text{Span}((n_k)_{k = 1}^{\ell})$ is a closed and bounded subset of a finite dimensional subspace (then span) and therefore compact, implying the precompactness of $N$, right? This would also make sense in context because this would imply that $N$ is totally bounded and therefore contains a finite $\varepsilon$-net. – Viktor Glombik Jul 24 '19 at 22:52
• @ViktorGlombik: Well, $N$ is compact. View it as the range of $\mu \mapsto \sum_k \mu_k n_k$ over the set $\{ \mu | \sum_k \mu_k = 1 , \mu_k \ge 0 \}$. Since the latter is compact, so is the range. – copper.hat Jul 24 '19 at 22:57
• Also in the second last paragraph the summation over $i$ is a finite sum, right? – Viktor Glombik Jul 24 '19 at 22:59
• @ViktorGlombik: Yes, all the sums are finite. – copper.hat Jul 24 '19 at 23:04
• When saying $n := \sum_{k} \lambda_i n_{k_i}$, do you mean to actually sum over $i$? – Viktor Glombik Jul 24 '19 at 23:07