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Can any convexity structure be defined by a partial order $\preceq$ in the sense of the order topology: a given set $A$ is convex if for any $a,b \in A$ and any other element $c$ for which $a\preceq c \preceq b$, we have that $c\in A$? I feel like the answer is no because pretty weird things start happening already for $\mathbb{R}^2$ as in Convexity implies the equivalence of order and subspace topologies.

So, I wondered if a set of partial orders defines any convexity structure. I mean in the sense that we call a set $A$ convex if, for any $a,b \in A$ and any partial order $\preceq$ we have that $a\preceq c\preceq b \implies c \in A$.

For example, for the standard convexity of $\mathbb{R}^n$, one could choose a set of partial orders, where each one contains a linear order on some line in $\mathbb{R}^n$. That is for any unit vector $v$, we define $x\preceq_{v} y \Leftrightarrow (\exists \lambda\geq0) \; y-x = \lambda v$ for any $x,y\in \mathbb{R}^n$.

My attempt: Closure stuctures are completely defined by how a (Moore) closure operator acts on finite sets, which is Theorem 1.3 in the book "Theory of convex structures" by Marcel van de Vel. Since partial orders are determined by pairs of elements, it feels like it should be possible to connect the two. I run into problems when I try to combine the partial orders of multiple sets with each other, where I'm not sure that the result is still a partial order.

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    $\begingroup$ If you generate these "convex" sets by assuming $a < c < b$ for some partial order in your collection, then your characterisation of convexity in $\Bbb{R}^n$ is wrong. Adding more partial orders will expand the convex line segment required to be in $A$, not reduce it. For example, the Euclidean line segment between $(0, 0)$ and $(1, 1)$ is not convex in this sense because $(0, 0) < (0.5, 6) < (1, 1)$ under the usual Lexicographic order, but does not lie in the line segment. $\endgroup$
    – Theo Bendit
    Commented Jun 18 at 15:39
  • $\begingroup$ Thanks Theo, indeed my example contained too many line segments and would make some weird sets convex. I edited the original post and I think this fixes that. $\endgroup$
    – user146125
    Commented Jun 18 at 16:12
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    $\begingroup$ OK, I think I understand you. You have a collection of partial orders $\le_v$, indexed by $v \in \Bbb{R}^n \setminus \{0\}$, where $x \le_v y$ if and only if $y - x$ is a non-negative multiple of $v$. And, a set $A$ is convex (in the traditional sense) if, for every $x, y \in A$ and $v \in \Bbb{R}^n \setminus \{0\}$, $x \le_v z \le_v y \implies z \in A$. So, in general, you're wondering if every convexity can be generated in this way by a family of partial orders. Am I understanding this correctly? $\endgroup$
    – Theo Bendit
    Commented Jun 19 at 5:28
  • $\begingroup$ Yes that is correct. $\endgroup$
    – user146125
    Commented Jun 19 at 7:41
  • $\begingroup$ Thank you for your comments already. After rereading what I wrote last night I saw that it was not that clear and even wrong (even though it looked clear for me yesterday). I edited the question and I hope the idea is now clearer. $\endgroup$
    – user146125
    Commented Jun 19 at 8:57

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For a set $X$, denote by $\mathcal{C}_\leq$ the convexity structure induced by partial order $\leq$.

For any convexity structure $\mathcal{C}$ on $X$ we can consider the intersection of all convexity structures $\mathcal{C}_\leq$ which contain $\mathcal{C}$, denote it by $\text{porder}(\mathcal{C})$. This will again be a convexity structure and will be the smallest generated by some set of partial orders.

That is, $A\in \text{porder}(\mathcal{C})$ iff $A$ is $\leq$-convex with respect to all partial orders $\leq$ such that all members of $\mathcal{C}$ are $\leq$-convex.

The smallest convexity structure is $\mathcal{C} = \{\emptyset, X\}$. But $\text{porder}(\mathcal{C})$ contains all the singletons $\{x\}$ for $x\in X$, so $\mathcal{C}\neq \text{porder}(\mathcal{C})$ for $|X| > 1$.

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