Non-isomorphic discrete linear order without endpoints

Question

What would be an example of two discrete linear orders without endpoints which are not isomorphic? I'm having trouble understanding how this would be possible.

Answer 1

The simplest example is that $(\mathbb Z, <)$ is not isomorphic with $(\{0,1\} \times \mathbb Z, <')$, where $(0,n) <' (1, m)$ always, and $(i,n) <' (i, m)$ if and only if $n < m$.

Edit: note that they are not isomorphic because one of them has an infinite ascending chain with an upper bound, while the other does not.

More generally, you can take any poset $P$ and consider $P \times \mathbb Z$, with $(p,n) < (q,m) \iff p < q \lor (p = q \land n < m)$. Nonisomorphic $P$ give nonisomorphic discrete unbounded linear orders.