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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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    $\begingroup$ Not necessarily. Proof by google, counterexample here. $\endgroup$ – Francois Ziegler Jan 31 '14 at 19:45
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    $\begingroup$ Via Theorem 5.35 on that page, the answer is "yes" if one replaces "convex hull" with "closed convex hull". $\endgroup$ – Tom LaGatta Jan 31 '14 at 22:42
  • $\begingroup$ (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 LaGatta Feb 1 '14 at 0:52
  • $\begingroup$ mathoverflow.net/questions/6627/convex-hull-in-cat0 this related question is purportedly open $\endgroup$ – Paul Fabel Feb 9 '14 at 17:47
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    $\begingroup$ 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 Furber Apr 18 '19 at 2:48

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