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Suppose that $X$ is a nonempty compact metric space. It is well-known that the hyperspace of all nonempty, closed subsets of $X$, endowed with the Hausdorff metric topology, is a compact metric space. But can one state a similar result for the hyperspace of nonempty, closed, convex subsets of $X$. That is to say, if $X$ is a nonempty, compact, convex subset of a metric linear space, is the hyperspace of nonempty, closed, convex subsets of $X$, endowed with the Hausdorff metric topology, a compact metric space?

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    $\begingroup$ Can you prove that in this context a convergent sequence of non-empty closed convex sets has a convex limit? Then the collection of these sets is a closed set in a compact Hausdorff space (where closed and compact coincide). $\endgroup$ – drhab Jun 23 '17 at 8:18
  • $\begingroup$ @drhab. Thanks. Yes, I think I can prove that. Let me think about it. $\endgroup$ – mo15 Jun 23 '17 at 8:24
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Here is an attempted answer. As per the comment, it suffices to show that a convergent sequence of nonempty closed convex sets has a convex limit. So let $\{A_n\}$ be a sequence of nonempty closed convex subsets of $X$ and suppose that it converges to a (nonempty, closed) subset $A$ of $X$. Pick two points $x$ and $y$ in $A$ and a scalar $\lambda$ in $[0,1]$. We need to show that $\lambda x+(1-\lambda)y\in A$. It is well-known that because $A_n\to A$ and $x$ and $y$ belong to $A$, there are sequences $\{x_n\}$ and $\{y_n\}$ with $x_n\in A_n$ and $y_n\in A_n$ for each $n$ such that $x_n\to x$ and $y_n\to y$. Since each $A_n$ is convex it follows that $\lambda x_n+(1-\lambda)y_n\in A_n$ for each $n$, and since vector addition and scalar multiplication are continuous we have $\lambda x_n+(1-\lambda)y_n\to\lambda x+(1-\lambda)y$. But then $\lambda x+(1-\lambda)y$ is in the topological limit inferior of $\{A_n\}$, and since $A$ equals the topological limit inferior of $\{A_n\}$ (because $A_n\to A$) it follows that $\lambda x+(1-\lambda)y\in A$.

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