I have had some troubles when trying to rephrase and prove the things I am reading on different books. I would like to prove the equivalence of defining a convex set the usual way, or as a set which contains all convex combinations of its points.
Definition of Convex Combination A point of the form $\alpha_1x_1 + \alpha_2 x_2 + \cdots + \alpha_k x_k$ where $\alpha_1 + \alpha_2 + \cdots + \alpha_k = 1$ and $\alpha_i \geq 0, \; i = 1, 2, \ldots, k$ is called a convex combination of the points $x_1, x_2, \ldots, x_k$
Proposition 1 Let $V$ be a vector space, and $\vec{x}_i \in V$, $i=1,\ldots,k$. Let $C$ be the set of all convex combinations of $\vec{x}_i$, then $C$ is convex
Proposition 2 Let $V$ be a vector space, and $\vec{x}_i \in V$, $i=1,\ldots,k$. Let $C$ be a convex set containing all $\vec{x}_i$ then all convex combinations of $\vec{x}_i$ are contained in $C$
From these propositions, can we say this ?
Proposition 3 Let $V$ be a vector space, and $\vec{x}_i \in V$, $i=1,\ldots,k$. Let $C$ be a set containing all $\vec{x}_i$ then $C$ is convex if and only if it contains every convex combinations of $\vec{x}_i$
I have been able to prove (2) by induction and could do the same for (1). But would such a proof be valid for something like (4) ?
Proposition 4 Let $V$ be a vector space. A set $C \subset V$ is convex if and only if it contains every convex combinations of its points.
And is (3) equivalent (4) ? If not, how can I prove (4) as it is stated, or rephrase it to be able to ?
4) can be proved exactly like 2), using induction. 3) is false. Only one way implication is true. If $C$ contains all convex combinations of the given points it need not be convex. There may be other points in $C$ some of whose convex combinations do not belong to $C$.