TITLE: Are convex hulls of closed sets also closed?
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: 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.