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Conway says if $K$ is a nonempty compact convex subset of a locally convex Hausdorff space $X$, then ext$K \neq \emptyset$ and $K = \bar{\text{co}}(\text{ext}K)$.

Here the compactness of $K$ and closed convex hull depends on the topology of $X$. But it seems strange that the statement on nonempty ext$K$ does not depend on the topology of $K$, since the definition of extreme point has nothing to do with the topology.

More formally, if $K$ is compact for any topology on $X$ (making it into a locally convex space) then it has a nonempty extreme point?

Jiya
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    The answer is yes, by the result you quote. What more are you after? – J. De Ro Jan 19 '22 at 23:14
  • I don’t understand the question. A subset $K$ of a vector space may or may not have extreme points. If there exists a topology on $X$ turning it into a locally convex space and such that $K$ is a compact subset, then $K$ has extreme points. – Plop Jan 19 '22 at 23:15
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    The locally convex topology should be Hausdorff, of course. It is indeed a fascinating feature of the theorem that a purely topological assumption implies a geometric conclusion. – Jochen Jan 20 '22 at 06:40
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    You can turn the result around: if the set $K$ has no extreme points or the set of extreme points is too small then there is no topology under which $K$ is compact. – daw Jan 20 '22 at 07:50
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    In particular, this is a very basic criterion in testing if a given Banach space is a dual space: If the unit ball doesn't have any extreme points, then it cannot be a dual space (because the unit ball of the dual space is weak$^\ast$ compact by the Banach-Alaoglu theorem). – MaoWao Jan 20 '22 at 12:39

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