# Convex hull of bounded set is bounded

I want to show that in a locally convex topological vector space $X$, the convex hull of a bounded set is bounded.

Apparantly this does not hold if $X$ is not locally convex. So the fact that that there exists a local base of absolutely convex absorbent sets is crucial.

Thanks for suggestions or tips.

• You have a bounded set $B$. You have a convex neighbourhood $V$ of $0$. That $B$ is bounded means that $B \subset t\cdot V$ for some $t > 0$. So what can you say about the convex hull of $B$? – Daniel Fischer Sep 23 '13 at 18:37
• That it is contained in the convex hull of tV, which is already convex, and bounded I guess :P – DinkyDoe Sep 23 '13 at 18:42
• $t\cdot V$ is a neighbourhood of $0$, so it is in general not bounded. But yes, it contains the convex hull of $B$. Now, what was the definition of a bounded set in a topological vector space? – Daniel Fischer Sep 23 '13 at 18:44
• Hmm, if for every open neighborhood $N$ of 0 there exista $\alpha$ such that B is contained in $\alpha N$. – DinkyDoe Sep 23 '13 at 18:48
• Right. You have: $B$ bounded. You want: $\operatorname{co}(B)$ bounded. All necessary parts are present, it remains to assemble them in the right way. – Daniel Fischer Sep 23 '13 at 18:50