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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?

Tian Vlasic
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Tom Finet
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  • As we generalize the notion of an open set with a topology, a convexity space is just a generalization of the concept of convex set. Note that for the usual definition of convex set you need to know how to sum elements in your set and multiply by scalars. But, following the definition of convexity space, the only thing you need is an abstract set. Think about a topology: if you don't have a metric, is impossible to prove that an open set of the topology is open in the usual sense (by open balls). Is the same idea. – Carlos Jiménez Sep 14 '21 at 17:57
  • I have not studied any topology and so do not understand what you mean. I am trying to read the linked paper which is computer science oriented so just want to intuitively understand this definition without going into too much detail. Is there a more intuitive explanation? – Tom Finet Sep 14 '21 at 18:02
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    The more intuitive explanation is the next: forget all you know about convex spaces. Then, read the article. The autor defined a convexity space as a colecction of sets that have a common propertie. And that's all. Imagine that the "convex" sets that the autor defines are different from the convex sets that you know. – Carlos Jiménez Sep 14 '21 at 18:09

1 Answers1

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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.

Zim
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