Counter-example for existence of an isomorphism between ordered sets.

Question

Take two totally ordered sets $(A,\leq_A)$ and $(B,\leq_B)$. Assume that none of them have neither a maximum nor a minimum. Assume that $A$ is isomorphic to some final segment of $B$ and $B$ is isomorphic to some final segment of $A$. How to choose $A$ and $B$ so that they are not isomorphic ? I don't think it is a difficult question but I can see it and I need a new point of view.

I assume $A$ and $B$ to have no extrema since I don't want to take $A=\mathbb{R}$ and $B=[0;1[$ for instance.

I would be very happy if the counter example involves dense orders.

Answer 1

The ordinal sum $Z$ of two (disjoint) partial orders $X$ and $Y$, written as $Z=X\oplus Y$, is defined by $a\leq b$ if one of the following holds:

1. $a,b\in X$ and $a\leq_X b$
2. $a,b\in Y$ and $a\leq_Y b$
3. $a\in X$ and $b\in Y$

Intuitively, it's like stacking the Hasse diagrams of two partial orders on top of each other. An infinite ordinal sum can be defined similarly. Then we can define $A=\mathbb{Z}\oplus\mathbb{R}\oplus\mathbb{Z}\oplus\mathbb{R}\oplus\cdots$ and $B=\mathbb{R}\oplus\mathbb{Z}\oplus\mathbb{R}\oplus\mathbb{Z}\oplus\cdots$. Then each order is clearly isomorphic to a final segment of the other, and $A$ and $B$ cannot themselves be isomorphic, since if we take any element in the first occurrence of $\mathbb{R}$ in $B$, say $0$, then that element has the property that the subposet of all elements below it is a dense order, while no element of $A$ has that property.

Answer 2

Assuming a final segment is { x : x > a }, let A be the reals ordered by magnitude and B = (-oo,0] $\cup$ = [1,oo).