Partially isomorphic but not isomorphic structures of the same cardinality

248 Views Asked by At

I think this should be a standard counterexample but I can't find it anywhere, I'm looking for examples of $L$-structures $\mathfrak A$ and $\mathfrak B$ of the same cardinality such that $\mathfrak A\cong_p \mathfrak B$ but $\mathfrak A\not\cong \mathfrak B$ (an example in a nice language with naturally arising structures would be great, but an ad hoc example would also be appreciated).

Here $\mathfrak A\cong_p\mathfrak B$ means that there is a family $I$ of finite partial isomorphisms $\mathfrak A\to\mathfrak B$ with the back and forth property, that is:

  1. for all $f\in I$ and each $a\in\mathfrak A$ there is $g\in I$ extending $f$ with $a\in\operatorname{dom}(g)$
  2. for all $f \in I$ and each $b\in\mathfrak B$ there is $g\in I$ extending $f$ with $b\in\operatorname{Im}(g)$

I know that $\mathfrak A\cong_p \mathfrak B\implies \mathfrak A\equiv \mathfrak B$ and that by Scott's isomorphism theorem $\mathfrak A\cong_p \mathfrak B\implies \mathfrak A\cong \mathfrak B$ for countable $\mathfrak A,\mathfrak B$ so we need to look at bigger structures.
The source I'm reading uses $\Bbb Q$ and $\Bbb R$ as structures in the language $L=\{<\}$ as an example of structures which are partially isomorphic but not isomorphic but they don't have the same cardinality, so what are some counterexamples?

1

There are 1 best solutions below

2
On BEST ANSWER

Well, take any two dense linear orders without endpoints, then easily they are partially isomorphic.

So we are just looking for two non-isomorphic dense linear orders without endpoints. Well, for example $\Bbb R$ and $\Bbb{R\times R}$ with the lexicographic order.