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I'm trying to prove this result. Could you verify if my attempt is fine? Is there a shorter proof?

Theorem: Let $(E, |\cdot|)$ be a normed space. If $K \subset E$ is totally bounded, then so is its convex hull $\operatorname{conv} K$.

I post my proof separately as below answer. If other people post an answer, of course I will happily accept theirs. Otherwise, this allows me to subsequently remove this question from unanswered list.

Analyst
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    It dosn't help anyone to post proofs of dozens of standard theorems like this. This is Theorem 3.24 in Rudin's RCA an your proof is not simpler. – Kavi Rama Murthy Jun 03 '22 at 09:18

1 Answers1

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We need the following lemma.

Lemma: Let $(E, |\cdot|)$ be a normed space. Let $x_1, \ldots, x_n \in E$. Then the convex hull of $\{x_1, \ldots, x_n\}$ is totally bounded.

Proof: We proof by induction. The case $n=2$ is trivial. Assume that $X:=\operatorname{conv} \{x_1, \ldots, x_n\}$ is totally bounded. We want to prove $Y:=\operatorname{conv} \{x_1, \ldots, x_n, x_{n+1}\}$ is totally bounded. Fix $r>0$. There are $y_1, \ldots, y_m \in X$ such that $\{B(y_i, r/2)\}_{i=1}^m$ covers $X$.

Let $[y_i, x_{n+1}]$ be the line segment connecting $y_i$ and $x_{n+1}$. Then $[y_i, x_{n+1}]$ is totally bounded. There are $z_{i1}, \ldots, z_{iM} \in [y_i, x_{n+1}]$ such that $\{B(z_{ij}, r/2)\}_{j=1}^M$ covers $[y_i, x_{n+1}]$.

If $y \in Y$, then $y = \lambda x + (1-\lambda)x_{n+1}$ for some $\lambda \in [0, 1]$ and $x \in X$. There is $i_0$ such that $x \in B(y_{i_0}, r/2)$. There is $j_0$ such that $\lambda y_{i_0} + (1-\lambda)x_{n+1} \in B(z_{i_0j_0}, r/2)$. We have \begin{align} |y-z_{i_0j_0}| &= |(\lambda x + (1-\lambda)x_{n+1})-z_{i_0j_0}| \\ &\le |(\lambda y_{i_0} + (1-\lambda)x_{n+1})-z_{i_0j_0}| + |\lambda (x- y_{i_0})|\\ &\le \frac{r}{2} + \frac{r}{2} = r. \end{align}

It follows that $\{B(z_{ij}, r/2) \mid i=1, \ldots, m \text{ and } j=1, \ldots, M\}$ covers $Y$. This completes the proof.

Fix $r>0$. There are $x_1, \ldots, x_n \in K$ such that $\{B(x_i, r/2)\}_{i=1}^n$ covers $K$. Let $X := \operatorname{conv} \{x_1, \ldots, x_n\} \subset \operatorname{conv} K$. By Lemma, $X$ is totally bounded. There are $z_1, \ldots, z_N \in X$ such that $\{B(z_i, r/2)\}_{i=1}^N$ covers $X$. Fix $y \in \operatorname{conv} K$. There are $y_1, \ldots, y_m \in K$ and $\lambda_1, \ldots, \lambda_m \in \mathbb R_{>0}$ such that $\sum_{i=1}^m \lambda_i=1$ and $y = \sum_{i=1}^m \lambda_i y_i$.

Assume $y_i \in B(x_{\varphi(i)}, r/2)$ with $\varphi(i) \in \{1, \ldots, n\}$ for all $i = 1, \ldots, m$. Clearly, $z:=\sum_{i=1}^m \lambda_i x_{\varphi(i)} \in X$. There is $i_0$ such that $z \in B(z_{i_0}, r/2)$. Then \begin{align*} |y-z_{i_0}| &\le |y-z| + |z-z_{i_0}| \\ &= \left | \sum_{i=1}^m \lambda_i y_i - \sum_{i=1}^m \lambda_i x_{\varphi(i)} \right | + |z-z_{i_0}| \\ &\le \sum_{i=1}^m \lambda_i |y_i - x_{\varphi(i)}| + |z-z_{i_0}| \\ &\le r. \end{align*}

It follows that $\{B(z_i, r/2)\}_{i=1}^N$ covers $\operatorname{conv} K$.

Analyst
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    To prove that the simplex $\overline{\mathrm{conv}}({x_1,\ldots,x_n})$ is totally bounded you can observe that it's compact as a continuous image of a compact set. – Mateo Jun 03 '22 at 11:49
  • @Mateo Could you confirm if my understanding of your idea is correct? Let $S := \operatorname{span} {x_1,\ldots,x_n}$. Then $S$ is a finite-dimensional normed vector space. Then $S$ is topologically isomorphic to some $\mathbb R^d$. A subset of $\mathbb R^d$ is compact if and only if it is closed and bounded. Clearly, $\overline{\mathrm{conv}}({x_1,\ldots,x_n})$ is closed and bounded, so it's compact. – Analyst Jun 03 '22 at 12:31
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    it's correct. Since compact sets are totally bounded, you are done. Of course your inductive proof seems also correct and has some advantages, so it's up to you, which one you use in your book. :-) – Mateo Jun 03 '22 at 12:52