The convex hull of every open set is open

Question

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.

Answer 1

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.

Answer 2

1. In the case we have a metric:

Let $z \in [A]$ ($[A]$ denotes the convex hull of $A$).

Therefore for some $n \in \mathbb{N}$ there are $x_1,x_2 \ldots, x_n \in A$ and $\lambda_1,\lambda_2 \ldots,\lambda_n \in (0,1)$ such that $\lambda_1 + \lambda_2 + \ldots + \lambda_n =1$ and $$z = \lambda_1 x_1 + \lambda_2 x_2 + \ldots +\lambda_n x_n.$$

Now $\exists \epsilon: \, \forall i \in \{1 \ldots n\}$: $B(x_i,\epsilon) \subset A$ and for $w \in B(0,\epsilon)$ $x + w \in B(x_i,\epsilon)$ and $x_i + w \in A$.

Note that $$z + w = \lambda_1 (x_1 + w) + \lambda_2(x_2 + w) + \ldots + \lambda_n(x_n + w)\Rightarrow z+ w \in A.$$ Therefore $B(z, \epsilon) \subset [A]$ and $[A]$ is open.

2. In the case we just have open sets and X is a topological space:

Good references are Proof that the convex hull of S is open if S is open and Topological vector space.

In this case by definition the scalar multiplication is open. This means that in the case $A$ is open consider $\lambda \in \mathbb{R}$ (or any other field) and $f_{\lambda}:X \to X$ $f_{\lambda}(x) = \lambda^{-1} x$ $${\lambda x: x \in A} = f_{\lambda}^{-1}(A) \quad \text{ is open}$$ consider also $x \in X$ and $g_{x}: X \to X$ $g_{x}(a) = a - x$.

$$A + x = {a + x: a \in A} = g_{x}^{-1}(A) \quad \text{ is also open.} $$

Therefore $$[A] = \bigcup_{n \in \mathbb{N}} \underset{\lambda_1 + \ldots + \lambda_n = 1}{\bigcup_{\lambda_{i} \in \mathbb{R}}}f_{\lambda_1}^{-1}(A) + \ldots + f_{\lambda_n}^{-1}(A).$$

Finally note that if $A$ is open $$A + B := \{x+ y: x \in A, y \in B\} = \bigcup_{y \in B} (A + y)$$ is also open. Therefore $$f_{\lambda_1}^{-1}(A) + \ldots + f_{\lambda_n}^{-1}(A)$$ is also open.

So we conclude that $[A]$ is open.