How do you prove that the convex hull of A is the smallest convex set containing A?
edit: definition of a convex hull: Given a set A ⊆ ℝn the set of all convex combinations of points from A is called the convex hull of A.
[Math] Prove that the convex hull of a set is the smallest convex set containing that set
convex-analysisconvex-hullsgeneral-topology
Related Solutions
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.
In my answer here: Proof that the Convex Hull of a finite set S is equal to all convex combinations of S I have given a proof that goes from the definition of the convex hull as the intersection of all convex sets containing the set of points to some of the other equivalent definitions, so you will work your way backwards if you are starting with a different equivalent definition. See the comment by julien.
Best Answer
Young, when you wrote
You meant that convex hull of A is the minimal convex set containing A, right?
To show this, which part is your definition? The linear-algebraic characterization?
You can see that any intersection of convex sets containing A is also a convex set containing A.