Let $A$ be a compact set in a separable Hilbert space $H$, and let $\bar A$ denote its convex hull. Is $\bar A$ compact?
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asked Jan 31, 2014 at 19:38
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11$\begingroup$ Not necessarily. Proof by google, counterexample here. $\endgroup$– Francois ZieglerJan 31, 2014 at 19:45
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4$\begingroup$ Via Theorem 5.35 on that page, the answer is "yes" if one replaces "convex hull" with "closed convex hull". $\endgroup$– Tom LaGattaJan 31, 2014 at 22:42
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$\begingroup$ (note: that theorem requires that $H$ be locally convex and completely metrizable, which is satisfied for a Hilbert space $H$. The assumption of separability is not necessary) $\endgroup$– Tom LaGattaFeb 1, 2014 at 0:52
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$\begingroup$ mathoverflow.net/questions/6627/convex-hull-in-cat0 this related question is purportedly open $\endgroup$– Paul FabelFeb 9, 2014 at 17:47
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4$\begingroup$ In any locally convex space $E$, the closed convex hull of a precompact set $X$ is precompact (see Schaefer's Top. Vect. Sp., Chapter II, Section 4.3). It follows that if $E$ is quasi-complete (every closed bounded set is complete - automatically true if $E$ is complete), then the closed convex hull of $X$ is compact, being precompact and complete. $\endgroup$– Robert FurberApr 18, 2019 at 2:48
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