TITLE: The convex hull of every open set is open
QUESTION [7 upvotes]: Let $X$ be a topological vector space. Prove that the convex hull of every open subset of $X$ is open.
I tried using definition of Convex Hull and Open Set, but I couldn't prove the statement.

REPLY [13 votes]: Let $A$ be an open subset of $X$.
If $\,x\,$ is an element of the convex hull of $A$, then there exist $\,x_1,\dots,x_n$ in $A$ and $\lambda_k \ge 0$ with $\sum_{k=1}^n \lambda_k=1\,$ and $\,x=\sum_{k=1}^n \lambda_k x_k$.
At least one $\lambda_k$ does not vanish: say $\lambda_1$.
The function $f: X \rightarrow X$, defined by $f(z)=\lambda_1^{-1}(z-\sum_{k=2}^n \lambda_k x_k)$, is continuous (it is a homothety), so $f^{-1}(A)=\lambda_1 A+\sum_{k=2}^n \lambda_k x_k\,$ is open.
Now $x \in f^{-1}(A)$ and $f^{-1}(A)$ is a subset of the convex hull of $A$, so the hull is open.