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.
[Math] The convex hull of every open set is open
convex-analysisfunctional-analysistopological-vector-spaces
Related Solutions
This is not true.
Let $K$ be, say, the unit ball in a Hilbert space, equipped with its weak topology. Let $\xi$ be a norm-one linear functional, and consider $U := K \cap \{\xi \in [-1, -c) \cup (c, 1]\}$ for $0 < c < 1$. It is an open set, however, its convex hull is $\{x \in K \, | \, \Vert \mathrm{pr}_{\xi^\perp} x \Vert < \sqrt{1 - c^2}\}$, where $\mathrm{pr}_{\xi^\perp}$ is the orthogonal projection on the hyperplane $\xi^\perp$. This set is clearly not open in the weak topology.
There is a theorem in van de Vel's Theory of Convex Structures that may be of interest. First define a closure system. Suppose that $X$ is a set and $\mathcal{C}$ is a collection of subsets of $X$. Then $\langle X, \mathcal{C} \rangle$ is a closure system if and only if for all $\mathcal{A} \subseteq \mathcal{C}$ we have $\cap \mathcal{A} \in \mathcal{C}$. Sometimes people require $\varnothing \in \mathcal{C}$. I can't remember if van de Vel does. For $A \subseteq X$ define $$\mathsf{cl}(A) = \cap \{ C \in \mathcal{C} \colon A \subseteq C \} .$$ In any event the convex subsets of a real vector space satisfy these properties.
Theorem Suppose that $\langle X, \mathcal{C} \rangle$ is a closure system. Then the following statements are equivalent:
- For all $A \subseteq X$ we have $\mathsf{cl}(A) = \cup \{ \mathsf{cl}(F) \colon F \text{ is a finite subset of } A \} $.
- For all $\mathcal{D}$ which are collections of subsets of $X$ that are directed by inclusion (see below for a definition of directed by inclusion) we have $\mathsf{cl}(\cup \mathcal{D}) = \cup \{ \mathsf{cl}(D) \colon D \in \mathcal{D} \} $.
- For all $\mathcal{T}$ which are collections of subsets of $X$ that are totally ordered by inclusion (see below for a definition of totally ordered by inclusion) we have $\mathsf{cl}(\cup \mathcal{T}) = \cup \{ \mathsf{cl}(T) \colon T \in \mathcal{T} \} $.
A collection $\mathcal{D}$ of subsets is directed by inclusion if and only if for all $D_{0}, D_{1} \in \mathcal{D}$ there is a $D \in \mathcal{D}$ with $D_{0}, D_{1} \subseteq D$.
A collection of subsets is totally ordered by inclusion if and only if for all $T_{0}, T_{1} \in \mathcal{T}$ we have $T_{0} \subseteq T_{1}$ or $T_{1} \subseteq T_{0}$.
Edit
This is in response to Tim's comment.
Suppose that $X$ is a set and $\mathcal{F}$ is the collection of finite subsets of $X$. Suppose also that $f \colon \mathcal{F} \rightarrow \mathcal{P}(X)$ where $\mathcal{P}(X)$ is the collection of all subsets of $X$. For each $n \in \mathbb{N}$ define \begin{align} g_{n} \colon \mathcal{P}(X) &\rightarrow \mathcal{P}(X) \\ \text{for all $A \subseteq X$ by the assignment} \\ g_{0} \colon A &\mapsto A \\ g_{n} \colon A & \mapsto g_{n-1}(A) \cup (\cup \{ f(F) \colon F \subseteq g_{n-1}(A) \text{ is finite} \} ) \end{align}
Then $A \mapsto \cup \{ g_{n}(A) \colon n \in \mathbb{N} \} $ is a convex hull operator. The function $f$ generalizes the notion of the points between $x, y \in \mathbb{R}^{k}$. If you think of this construction in this manner then the $g_{n}$ functions just accumulate line segments.
Best Answer
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.