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 $$A_1$$?

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).

• 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? – Motaka Nov 26 '18 at 11:15
• $\overline{A_1}$ is compact. Since the notation of compactness and sequential compactness is equivalent in metric spaces, your statement is a direct consequence of compactness. – p4sch Nov 26 '18 at 11:20
• 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 – Motaka Nov 26 '18 at 12:15
• 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 $A_1$!) – p4sch Nov 26 '18 at 12:17
• Exactly, I forget this one , Thank you so much – Motaka Nov 26 '18 at 12:18

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 $$\overline{\mathrm{co}}\,(K\cup\{x_0\})$$.

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