The closure of convex hull compact 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 $A_1$?
 A: 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 
$\overline{\mathrm{co}}\,(K\cup\{x_0\})$.
A: It is well-know that the closure of the convex-hull $\overline{\mathrm{Co}(K)}$ is compact, if $K$ is compact and $E$ is a Banach space. Note that $\overline{A_n} \subset \overline{\mathrm{Co}(A_n)}$. Thus, if $\overline{\mathrm{Co}(A_n)}$ is compact, then $\overline{A_n}$ is compact too. Since $$A_1 \subset \{x_1,\ldots, x_{n-1}\} \cup \overline{A_n} =:K$$ and the latter set $K$ is compact, we get that $\overline{\mathrm{Co}(A_1)}$ is compact. In fact, note that
$$\overline{\mathrm{Co}(A_1)} \subset \overline{\mathrm{Co}(K)}$$
and the last set is compact (because $K$ is compact).
