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Suppose $C_1,\cdots,C_n$ are closed, compact, convex subsets of a locally convex topological vector space, then is $\text{Conv}(C_1 \cup \cdots \cup C_n) = \overline{\text{Conv}(C_1 \cup \cdots \cup C_n)}$, i.e. is it closed? My intuition tells me that it is since $C_i$'s are required to contain its limit points, and I vaguely remember limit points are extreme points (Is it? Please correct me if I am wrong). So the convex hull of the finite union contains all the extreme points of the union and hence it is closed.

Anyway, I was told by a friend that I might be able to prove it by using nets, which I am not very sure how to construct one if that is even possible. Any help is appreciated.

Edit: I forgot to mention, $C_i$ are subsets of a locally convex topological vector space, so the norm is not really defined here.

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  • $\begingroup$ Since the $C_k$ are convex, you have $$\operatorname{Conv}(C_1\cup \dotsc \cup C_n) = s\left(\Delta \times \prod_{k = 1}^n C_k\right)$$ where $\Delta$ is the simplex $\left\{(t_1,\dotsc,t_n) : t_k \geqslant 0,\; \sum_{k = 1}^n t_k = 1\right\}$ and $s(t_1,\dotsc,t_n,x_1,\dotsc,x_n) = \sum_{k = 1}^n t_k\cdot x_k$. $\endgroup$ – Daniel Fischer Nov 14 '15 at 22:16
  • $\begingroup$ @DanielFischer I am not sure what you are trying to achieve here. In fact, that is an equality I proved before, and I don't really see how that helps. ): $\endgroup$ – Guo Xian Yau Nov 14 '15 at 22:35
  • $\begingroup$ What properties of $\Delta \times \prod\limits_{k = 1}^n C_k$ would you expect to be relevant here? $\endgroup$ – Daniel Fischer Nov 14 '15 at 22:37
  • $\begingroup$ Oh shit, that's compact. Then you can apply Krein-Milman! :O $\endgroup$ – Guo Xian Yau Nov 14 '15 at 22:37
  • $\begingroup$ What has Krein-Milman to do with it? It's compact, $s$ is continuous, hence $\operatorname{Conv}(C_1 \cup \dotsc \cup C_n)$ is compact [hence closed, if the space is Hausdorff]. No extreme points mentioned. $\endgroup$ – Daniel Fischer Nov 14 '15 at 22:39
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Hint: Apply the theorem of Krein-Milman

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  • $\begingroup$ How do you know the convex hull of the union itself is compact in that case? $\endgroup$ – Guo Xian Yau Nov 14 '15 at 20:40
  • $\begingroup$ I the spaces are separated, you can easily show this by considering sequences. $\endgroup$ – Tsemo Aristide Nov 14 '15 at 20:41
  • $\begingroup$ Okay, I will try both methods and keep you updated. $\endgroup$ – Guo Xian Yau Nov 14 '15 at 20:44
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Credits to both @Tsemo and @Daniel.

Basically using the fact that Daniel mentioned, $\text{Conv}(C_1 \cup \cdots\cup C_n)$ is compact. Now LCTVS are Hausdorff, so it is a compact subset of a Hausdorff space which is closed.

Otherwise since the convex hull is itself compact, proceed to prove that $\text{Conv}(\text{Ext}(\text{Conv}(C))) = \text{Conv}(C)$, and apply Krein-Milman.

Please upvote Tsemo's answer instead, or Daniel's if he decided to post the solution.

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