How do abstract convexity spaces generalise convex sets?

Question

I am reading this paper where the definition of a convexity space is given as follows (page $3$ of paper).

Def 1: A convexity space on a set $V$ is a collection $C\subseteq2^{|V|}$ satisfying

1. $\emptyset, V\in C$,
2. $A, B\in C$ implies $A\cap B \in C$.

It calls $X\in C$ a convex set. However, I am familiar with the following definition of a convex set.

Def 2: A set $X$ is convex iff for all $x, y \in X$ and $0\leq\lambda\leq1$ we have $\lambda x+(1-\lambda)y\in X$.

Suppose we have a convexity space $C=\{\emptyset,\{1, 2\}\}$ over $V=\{1, 2\}$. Then take $X=\{1, 2\}\in C$, from the paper $X$ is convex. However, the definition I am familiar with for convex sets would say that $X$ is not convex since for $\lambda=\frac{1}{2}$ we have $$\frac{1}{2}\cdot1+\frac{1}{2}\cdot2=\frac{3}{2}\not\in X$$

Is the definition in the paper for a convexity space at all related to convex sets as defined in def 2? If so, I can't understand how the def 1 generalizes def 2. How does it capture the idea of a convex set containing all elements in between any two elements in the set?

The paper then states that any convexity space has the closure operator mapping any set $X\in V$ to a minimal superset $X\subseteq<X>\in C$ called the convex hull of $X$. I can't understand when the convex hull of a set $X$ would differ from $X$, since the minimal superset of $X$ which is in $C$ is $X$ correct?

Answer 1

This definition appears similar to a topology in that we can start with some basic elements, then build up the complicated ones using the rules 1 and 2. However, the one rule (intersection) preserves the traditional notion of convexity. Here's my take on it:

Since all closed/convex sets can be written as an intersection of closed halfspaces, $C$ describes the traditional collection of "convex sets" when it contains all closed halfspaces and their intersections. More generally (under this definition), $C$ does not necessarily need to start with all closed half-spaces; by removing this point, $C$ can sometimes contain nonconvex sets, even though the definition appears motivated by convexity-preserving operations.

Concerning the second point, note that we start with arbitrary $X\in \mathbf{ V}$, not $X\in C$. So, for $X\in V\setminus C$, the convex hull will be different, even for this definition of a convexity space.