A "Cantor-Schroder-Bernstein" theorem for partially-ordered-sets

Solution 1:

One does not have to go to partially ordered sets for a counterexample. Linear orders are enough: Consider the rationals in the open interval $(0,1)$ and the rationals in the closed interval $[0,1]$.

The two obviously embed into one another but are not isomorphic due to minimum/maximum considerations.

My example shows that linear orders need not have this property; and Brian's example shows that well-founded partially ordered sets need not have this property.

However if we consider the intersection, well-founded linear orders (i.e. well orders) then indeed this is true. Namely, if $(A,<)$ and $(B,\prec)$ are two well ordered sets and $(A,<)$ embeds into $(B,\prec)$ and vice versa then there exists an isomorphism.

Solution 2:

No. Let $P$ be the disjoint union of chains of every odd length, and let $Q$ be the disjoint union of chains of every positive even length. Clearly each can be embedded in the other, but the two are not isomorphic.

More formally, let $$P=\{\langle 2n+1,k\rangle:n\in\Bbb N\text{ and }1\le k\le 2n+1\}$$ and $$Q=\{\langle 2n,k\rangle:n\in\Bbb Z^+\text{ and }1\le k\le 2n\}\;,$$

each ordered by the relation $\langle m,k\rangle\preceq\langle n,\ell\rangle$ iff $m=n$ and $k\le\ell$. The map $$f:P\to Q:\langle 2n+1,k\rangle\mapsto\langle 2n+2,k\rangle$$ embeds $P$ into $Q$, and the map $$g:Q\to P:\langle 2n,k\rangle\mapsto\langle 2n+1,k\rangle$$ embeds $Q$ into $P$, but $\langle P,\preceq\rangle$ and $\langle Q,\preceq\rangle$ are clearly not isomorphic, as they contain maximal chains of different lengths.

This shows that there is no such result even for well-founded partial orders.