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
- By definition, the $V$ can't be the closed convex hull of $X$, since it consists of infinite and not finite sums, right?
- How does theorem 1 follow from theorem 2?
- 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?
Best Answer
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.