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For a point $p$ in a simplex we define $ \text{supp (p)}= \{ i \in I_n \ | \ p_i \neq 0\}$
For a convex set $C$ we define $F$ to be its face if $F$ is convex and $\forall x,y \in C, \lambda \in (0,1) : \lambda x + (1- \lambda ) y \in F \Rightarrow x,y \in F$
For $A \subset \{1,2,...,n \}$ we define $F(A) : = \{ p \in \Delta_n \ | \ \text{supp(p)} \subset A \}$
Could you tell me how to prove that $$\text{conv} (F(A_1) \cup F(A_2)) = F(A_1 \cup A_2)$$
Here this inclusion is quite evident "$\subset$" and I've proved it using $2)$ below and the fact that for $A_1 \subset A_2$ we have $F(A_1) \subset F(A_2)$ (precisely the fact that both sets are contained in their union)
I know that:
$1) F \text{ is a face } \iff \exists A \subset I_n : F = F(A)$
$2) \text{conv} (\{ e_i, i \in A \} ) = F(A)$ (canonical basis of $\mathbb{R}^n$)
$3) F(A_1 \cup A_2) = F(A_1) \cup F(A_2)$ if $A_1 \subset A_2$ or $A_2 \subset A_1$
But I don't know how and if to put those facts together.
Could you help me a bit?
For a convex set $C$ we define $F$ to be its face if $F$ is convex and $\forall x,y \in C, \lambda \in (0,1) : \lambda x + (1- \lambda ) y \in F \Rightarrow x,y \in F$
For $A \subset \{1,2,...,n \}$ we define $F(A) : = \{ p \in \Delta_n \ | \ \text{supp(p)} \subset A \}$
Could you tell me how to prove that $$\text{conv} (F(A_1) \cup F(A_2)) = F(A_1 \cup A_2)$$
Here this inclusion is quite evident "$\subset$" and I've proved it using $2)$ below and the fact that for $A_1 \subset A_2$ we have $F(A_1) \subset F(A_2)$ (precisely the fact that both sets are contained in their union)
I know that:
$1) F \text{ is a face } \iff \exists A \subset I_n : F = F(A)$
$2) \text{conv} (\{ e_i, i \in A \} ) = F(A)$ (canonical basis of $\mathbb{R}^n$)
$3) F(A_1 \cup A_2) = F(A_1) \cup F(A_2)$ if $A_1 \subset A_2$ or $A_2 \subset A_1$
But I don't know how and if to put those facts together.
Could you help me a bit?