If $X$ is a compact set in $\mathbb R^n$, how can we show that $\operatorname{conv} X$ is compact as well? Can we say something similar without assuming boundedness, i.e., are convex hulls of closed sets also closed?

  • $\begingroup$ Does Carathéodory's theorem (convex hull) has any hints here? $\endgroup$ – Anusha Manila May 17 '15 at 20:13
  • $\begingroup$ As demonstrated in the answers below, it is not true that the convex hull of a closed set is closed. Regarding necessary and sufficient conditions for the convex hull of an unbounded closed set to be closed, this note may be of use. $\endgroup$ – Matt Rosenzweig May 17 '15 at 21:29
  • $\begingroup$ Thank you, Hagen von Eitzen and Matt Rosenzweig $\endgroup$ – Anusha Manila May 18 '15 at 10:25

One approach is to prove that a point in the convex hull of a set in $\mathbb R^n$ is actually a convex combination of at most $n+1$ points, by a natural induction.

Then the convex hull is the continuous image of the cartesian product of $n+1$ copies of your set and the $n$-simplex, which is compact. A continuous image of compact is compact.

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    $\begingroup$ The fact you mention in your first paragraph is Caratheodory's theorem. In the second paragraph I think you need the cartesian product of $n+1$ copies of the set and an $n$-simplex. $\endgroup$ – Rob Arthan May 17 '15 at 21:29
  • $\begingroup$ @RobArthan, thanks for the reference to Cartheodory: I hadn't realized this fact was due to him! (And thx for noting the typo!!!) $\endgroup$ – paul garrett May 17 '15 at 21:34
  • $\begingroup$ garret: my pleasure! are you sure you need an $(n+1)$-simplex rather than just an $n$-simplex? $\endgroup$ – Rob Arthan May 17 '15 at 21:41
  • $\begingroup$ @RobArthan :) :) $\endgroup$ – paul garrett May 17 '15 at 21:44
  • $\begingroup$ Thank you, Paul Garrett and RobArthan $\endgroup$ – Anusha Manila May 18 '15 at 10:25

The convex hull of $\{\,(x,y)\mid x^2y=1\,\}$ is not closed in $\mathbb R^2$ (it is the open upper half plane).

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