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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 $A_1$?

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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).

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  • $\begingroup$ Great, thank you. Now we know that $A_1$ is relatively compact. in the paper that i'm reading, they say that $\{x_n\}$ has a convergent subsequence, do you have any idea why? $\endgroup$ – Motaka Nov 26 '18 at 11:15
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    $\begingroup$ $\overline{A_1}$ is compact. Since the notation of compactness and sequential compactness is equivalent in metric spaces, your statement is a direct consequence of compactness. $\endgroup$ – p4sch Nov 26 '18 at 11:20
  • $\begingroup$ I will rephrase my statement. $\overline{A_1}=\overline{\{x1,x_2,...\}}$ is compact so its sequential compact, but why $A_1$ is sequential compact $\endgroup$ – Motaka Nov 26 '18 at 12:15
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    $\begingroup$ Any sequence in $A_1$ also also a sequence in $\overline{A_1}$. Thus it has a convergent subsequence. However, the limes lies in $\overline{A_1}$ and can be not in $A_1$. (The paper doesn't say that the limes is in $A_1$!) $\endgroup$ – p4sch Nov 26 '18 at 12:17
  • $\begingroup$ Exactly, I forget this one , Thank you so much $\endgroup$ – Motaka Nov 26 '18 at 12:18
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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 $\overline{\mathrm{co}}\,(K\cup\{x_0\})$.

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  • $\begingroup$ Thank you @Yiorgos S. Smyrlis $\endgroup$ – Motaka Nov 26 '18 at 13:58

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