In our lecture, we used the

Theorem 1 (Mazur) Let $X$ be a Banach space and $M \subset X$. If $M$ is relatively compact, $\overline{\text{co}(M)}$, the closed convex hull of $M$, is, too.

We defined the convex hull of some subset $M \subset X$ to be $$ \text{co}(M) := \left\{ \sum_{i = 1}^{N} \lambda_i x_i: N \in \mathbb{N}, \lambda_i \in [0,1], \sum_{i = 1}^{N} \lambda_i = 1, x_i \in M \ \forall i \in \{1, \ldots, N\} \right\} $$ and the closed convex hull $\overline{\text{co}}(M)$ to be the closure of $\text{co}(M)$ with respect to the norm on $X$.

When attempting to find a proof of theorem 1 on the internet, I found this pdf arcticle by Mazur himself in German. The precise statement proven is the following:

Theorem 2 (Mazur) Let $Z$ be a compact subset of a Banach space. Then the smallest convex superset of $Z$, $W$ compact as well.

His proof starts like this: Since the statement is trivial for finite $Z$, we assume that $Z$ is infinite. Since $Z$ is compact it is separable. Let $A = (x_k)_{k\in \mathbb{N}} \subset Z$ be an enumerate of the countable dense subset. We define $$ V := \left\{ \sum_{n \in \mathbb{N}} a_n x_n: a_k \ge 0 \ \forall k \in \mathbb{N}, \sum_{n \in \mathbb{N}} a_n = 1 \right\}. $$ He now goes on to show $V$ is compact. Since $V$ is compact and convex, so is $\overline{V}$. Since we have $Z = \overline{A}$ and therefore $Z \subset \overline{A}$ and also $A \subset V$ we have $Z \subset \overline{V}$.

Since $W$ is the smallest convex superset of $Z$, we have $W \subset \overline{V}$ which is compact.

My Questions

  1. By definition, the $V$ can't be the closed convex hull of $X$, since it consists of infinite and not finite sums, right?
  2. How does theorem 1 follow from theorem 2?
  3. Why is $W$ compact? If it where closed, it would be clear, since closed subsets of compact sets are compact. But since convex sets aren't closed in general (right?), this is not the case, right?
  • $\begingroup$ There are many errors in your statements. Perhaps these are caused by translation. Theorem 2 is not true. The definition of $V$ does not make sense since the series in the definition may not converge. (In Theorem 2 you have to take the smallest closed convex superset of $Z$). $\endgroup$ – Kavi Rama Murthy Jul 22 '19 at 23:34
  • $\begingroup$ @KaviRamaMurthy From the answer below we know that in the above "compact" has to be replaced by "totally bounded" or rather "precompact". Does this solve the issue of the sum convergence? $\endgroup$ – Viktor Glombik Jul 23 '19 at 17:43

Your objections are all correct; Theorem 2 is wrong as stated. It appears that Mazur uses the word "compact" to mean what is now called "totally bounded" (indeed, if you look at the details, that's what he proves about $V$). A subset of a complete metric space is totally bounded iff its closure is compact, so this statement is equivalent to Theorem 1. In general, you should not be surprised to find terminological discrepancies like this when reading 90-year-old papers.

  • $\begingroup$ Thanks for for this input. Do you maybe have a source detailing how terminology changed? Furthermore, is the following reasoning correct? The statement is trivial for finite Z since the convex hull of Z can be chosen as a closed ball around one of the points with a finite radius large enough to capture all points which is closed and bounded. $\endgroup$ – Viktor Glombik Jul 23 '19 at 10:30
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    $\begingroup$ I don't know the history; I just inferred the meaning from what Mazur wrote. For finite $Z$, your reasoning is incorrect as stated sinice a closed and bounded set is not necessarily compact. Rather, you must first restrict your attention to the finite-dimensional subspace spanned by $Z$ and then apply your argument, since in a finite-dimensional normed vector space a closed and bounded set is compact. $\endgroup$ – Eric Wofsey Jul 23 '19 at 14:58
  • $\begingroup$ It is trivial that a totally bounded set is separable; in fact, this is how you prove a compact metric space is separable. Choose a finiite $\epsilon$-net for each $\epsilon$ in a sequence going to $0$, and they form a countable dense subset. $\endgroup$ – Eric Wofsey Jul 23 '19 at 14:59
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    $\begingroup$ I don't know what problem you are referring to. The span of $Z$ is not bounded, since it's an entire vector subspace. My point was that if $Z$ is finite, you first prove that the convex hull of $Z$ is closed and bounded as a subset of its span, which implies it is compact only because the span is finite-dimensional. $\endgroup$ – Eric Wofsey Jul 24 '19 at 15:05
  • $\begingroup$ You might be interested in this question of mine. $\endgroup$ – Viktor Glombik Jul 24 '19 at 23:52

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