Let $L$ denote a linear order that is unbounded. Then it may or may not satisfy:
- Globally homogeneous. For all $x,y \in L \cup \{-\infty,\infty\}$, if $x < y$ then the interval $(x,y)$ is isomorphic to $L$.
- Locally homogeneous. For all $x,y,x',y' \in L,$ if $x<y$ and $x'<y'$, then the intervals $(x,y)$ and $(x',y')$ are isomorphic.
Please comment if you know the standard terminology for these conditions and/or where I can learn more.
Examples and non-examples.
- $\mathbb{Z}$ satisfies neither condition.
- $\mathbb{R}$ and $\mathbb{Q}$ satisfy both.
- The empty (unbounded) linear order also satisfies both.
Now clearly, globally homogeneous implies locally homogeneous.
Question 0. Does the converse hold (for unbounded linear orders)?
Question 1. Other than $\mathbb{R}$ and $\mathbb{Q}$, what are some examples of unbounded linear orders that are homogeneous (in either sense)?
Bonus Question. Given unbounded linear orders $L$ and $M$ that are globally homogeneous, if we can embed $L$ into $M$ and vice versa, are they necessarily isomorphic?
Another example of a locally homogeneous unbounded order would be $$\mathbb{L} := ( \omega_1 \times ( [0,1) \cap \mathbb{Q} ) ) \setminus \{ \langle 0,0 \rangle \}$$ with the lexicographic order (where, as usual, $\omega_1$ denotes the least uncountable ordinal). (I throw away $\langle 0,0 \rangle$ just to ensure that there is no minimum element.) This example shows that the two notions are not equivalent.
This is locally homogeneous because for any $\langle \alpha , p \rangle < \langle \beta , q \rangle$ the interval $( \, \langle \alpha , p \rangle , \langle \beta , q \rangle \, )$ is a countable dense linear order without endpoints, and is therefore order isomorphic to $\mathbb{Q}$.
It is not globally homogeneous because the entire order is uncountable (yet all intervals $( \, \langle \alpha , p \rangle , \langle \beta , q \rangle \, )$ are countable).