Induced total order on equivalence classes from partial order Given a poset $\langle S, \leq \rangle$, we can define an equivalence relation on elements such that $a \sim b$ if $a \nleq b$ and $b \nleq a$, and extend via transitivity and reflexivity.
Put plainly, this says any two "incomparable" elements are in the same equivalence class. So if you have a "diamond" in your poset, the minimum and maximum of the diamond are preserved, but the rest of the elements are grouped into one equivalence class.
If we do so, there is a natural total order on the resulting set of equivalence classes such that for two classes $A, B$, we define $A \leq B$ iff $a \leq b$ for all $a \in A, b \in B$.
My two questions:


*

*Does there exist a name for this induced linear ordering?

*In what conditions is the induced linear ordering a well ordering?


Note this is different than this question about extending a partial order to a linear order - this isn't about extending, but rather creating a linear order on equivalence classes of subsets.
 A: Sadly, this Answer does not definitively resolve your Question. That said, it is too long for a comment, so here we go!

First of all, I'm not familiar with any terminology referring to the induced total order to which you refer.
Second of all, your definition of said "total order" defines a relation that isn't necessarily total! For example, consider a the poset $S=\{a,b,c,d\}$ with the relation $$\le:=\left\{\langle a,a\rangle,\langle a,b\rangle,\langle a,c\rangle,\langle a,d\rangle,\langle b,b\rangle,\langle b,d\rangle,\langle c,c\rangle,\langle c,d\rangle,\langle d,d\rangle\right\}.$$ Put another way, $\langle S,\le\rangle$ is the "diamond" with $d$ at the top and $a$ at the bottom, having both $b$ and $d$ between them. Now, putting $$A=\{a\},B=\{b,c\},C=\{d\},$$ it's clear that $\{A,B,C\}$ is the set of "incomparability" equivalence classes. However, as defined, we have $B\not\le B,$ because neither $b\le c$ nor $c\le b.$
Fortunately, it is fixable! Instead, we define $A<B$ if $a<b$ for all $a\in A$ and $b\in B.$ This is readily irreflexive and transitive, and totality isn't too tricky to prove. The reflexive closure of this relation would be the one you're looking for.

As for when the order will be a well-order, I'm not sure there is any consistent criterion. Any poset with an isolated element will trivially induce a well-order, since there will be only one equivalence class. The set of positive integers under the divisibility partial order will likewise have only one equivalence class, and that partial order is connected and well-founded! On the other hand, consider $\{0,1\}\times\Bbb R$ with $\langle x_1,y_1\rangle\le\langle x_2,y_2\rangle$ if $y_1<y_2$ or ($x_1=x_2$ and $y_1=y_2$). The equivalence classes will be two-element sets of ordered pairs with the same first coordinate, and the induced total order will be isomorphic to the real order, so not a well-order.
