Let $A_n=\{x_n,x_{n+1},…\}\subset E$ for each $n\in \mathbb N$, such that $E$ is a Banach space.
If the closure of the convex hull of $A_n$ is compact i.e $\overline{co}(A_n)$ compact, is $\overline{co}(A_1)$ compact?
what we can say about $A_1$?
The closure of convex hull compact
compactnessfunctional-analysisreal-analysis
Best Answer
It suffices to show the following (and then use induction)
Claim. If $K$ is a convex and compact subset of a Banach space $E$ and $x_0\in E$, then $\,\overline{\mathrm{co}}\,(K\cup\{x_0\})$ is also compact.
Proof of the Claim. Let $\{z_n\}\subset \overline{\mathrm{co}}\,(K\cup\{x_0\})$, we need to show that there exists a converging subsequence of $\{z_n\}$ with limit in $\overline{\mathrm{co}}\,(K\cup\{x_0\})$.
First of all, there exists another sequence, $\{w_n\}\subset {\mathrm{co}}\,(K\cup\{x_0\})$, such that $$ \|z_n-w_n\|<\frac{1}{n}\qquad\text{and}\qquad w_n=t_nk_n+(1-t_n)x_0, $$ where $k_n\in K$ and $t_n\in[0,1]$. Clearly, $\{t_n\}$ possesses and converging subsequence $\{t_{n_j}\}$ and $k_{n_j}$ possesses also a converging subsequence. For simplicity, and economy of indices, say that $$ t_\ell\to t\in[0,1]\qquad\text{and}\qquad k_\ell\to k\in K. $$ Then $$ w_\ell\to tk+(1-k)x_0\in \overline{\mathrm{co}}\,(K\cup\{x_0\}) $$ and as $z_\ell-w_\ell\to 0$, we also have that $$ z_\ell\to tk+(1-k)x_0\in \overline{\mathrm{co}}\,(K\cup\{x_0\}). $$ Hence, every sequence in $\overline{\mathrm{co}}\,(K\cup\{x_0\})$ possesses a converging subsequenc in $\overline{\mathrm{co}}\,(K\cup\{x_0\})$.