Is there an infinite set with a discrete cyclic order? Let's call a cyclic order of a set discrete if every cut of the order is a jump.
A cut of a cyclic order is a linear order $<$ such that $x < y < z \implies (x, y ,z)$ for any elements $x$, $y$, $z$ of the set.
A cut of a cyclic order is a jump if it has the least and the greatest elements.
Clearly, the induced cyclic order of integers is not discrete since the natural linear order of integers does not have the least and the greatest elements.
However, there are other ways of ordering integers cyclically, e.g. https://math.stackexchange.com/a/2196717/427611.
I am wondering if it is possible to find a discrete cyclic order of integers or maybe of some other infinite set.
If it is not possible, what would be the easiest way to prove that?
By cyclic order I mean a total strict cyclic order defined in here: https://en.wikipedia.org/wiki/Cyclic_order#The_ternary_relation
 A: Given a cyclic order on $A$ and an element $a\in A$, we can define $<$ as
$$ x<y\iff [x,y,a]\lor x\ne y=a$$
(i.e., we "cut" immediately behind $a$). This obviously has $a$ as a maximal element. Assume that there is also a minimal element, no matter what $a$ we pick. Call it $S(a)$, and we have a successor map on $A$. By the same argument, we obtain a predecessor map and this is clearly inverse to the successor map. Using these (and picking an element $a_0\in A$) we can map $\iota\colon\Bbb Z\to A$ such that no elements of $A$ are between the images of consecutive integers.
If $\iota$ is not injective, then it must be periodic and so $\iota(\Bbb Z)$ finite. In that case $\iota$ must be onto because there is no way to "squeeze" any further elements of $A$ in-between. As we are interested in the case of infinite $A$, we can ignore this case. [Thanks to a comment by Eric Wolsey]
Now we can make a new cut "above $\Bbb Z$", i.e., we define
$$x\prec y\iff \exists n\in\Bbb Z\colon [x,y,\iota(n)]. $$
This does not have a maximal element.
A: Step 1. Let's show that every cut of a discrete cyclic order is discrete.
A linear order is discrete if every cut of it $(A, \overline A)$ is a jump.
A cut $(A, \overline A)$ of a linear order is a jump if $A$ has the greatest and $\overline A$ has the least element.
Assuming a cut $(A, \overline A)$ of a cut $<$ of a discrete cyclic order is not a jump.
Defining the new linear order $<_A$ in the following way:
$x <_A y$ if and only if


*

*$x \in \overline A \land y \in A$, or

*$x \in \overline A \land y \in \overline A \land x < y$, or

*$x \in A \land y \in A \land x < y$.


It is easy to check that $<_A$ is a cut of the cyclic order using the rule:
$(x, y, z) \iff x < y < z \lor y < z < x \lor z < x < y$.
If $A$ does not have the greatest element, then the cut $<_A$ does not have the greatest element.
If $\overline A$ does not have the least element, then the cut $<_A$ does not have the least element.
Therefore, if $(A, \overline A)$ is not a jump of $<$ then $<_A$ is not a jump of the cyclcic order.
Step 2. Any cut of an infinite discrete cyclic order is an infinite discrete linear order.
Any infinite discrete linear order is isomorphic to $\mathbb N$ or $\mathbb Z$.
Therefore, an infinite discrete cyclic order does not have any jump.
