# Are convex hulls of closed sets also closed?

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?

• Does Carathéodory's theorem (convex hull) has any hints here? – Anusha Manila May 17 '15 at 20:13
• 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. – Matt Rosenzweig May 17 '15 at 21:29
• Thank you, Hagen von Eitzen and Matt Rosenzweig – 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.
• 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. – Rob Arthan May 17 '15 at 21:29
• garret: my pleasure! are you sure you need an $(n+1)$-simplex rather than just an $n$-simplex? – Rob Arthan May 17 '15 at 21:41
The convex hull of $\{\,(x,y)\mid x^2y=1\,\}$ is not closed in $\mathbb R^2$ (it is the open upper half plane).