Q:

Let $(P,<)$ and $(Q,\prec)$ be order isomorphic. What kind of properties that $P$ and $Q$ have in common by the order isomorphism?

Let $(P,<)$ and $(Q,\prec)$ be order isomorphic. What kind of properties that $P$ and $Q$ have in common by the order isomorphism?

When I do exercises (at basic level) related to order isomorphism, I often encounter such statements as:

Since $P$ is dense, and $(P,<)$ and $(Q,\prec)$ are order isomorphic, $Q$ is dense.
Since $P$ has no endpoints, and $(P,<)$ and $(Q,\prec)$ are order isomorphic, $Q$ has no endpoints.
Since $P$ is complete, and $(P,<)$ and $(Q,\prec)$ are order isomorphic, $Q$ is complete.

These statements are usually in the form: Since $P$ is $.....$, and $(P,<)$ and $(Q,\prec)$ are order isomorphic, $Q$ is $.....$.

Immediately, two questions appear in my mind:

What kind of properties that we can correctly fill in the blank $.....$ in the statement Since $P$ is $.....$, and $(P,<)$ and $(Q,\prec)$ are order isomorphic, $Q$ is $.....$

Of course, this kind of properties must relate to the orderings of $P$ and $Q$. Is it still correct if such properties relate to other things besides the orderings. I can not figure out how to describe such kind of properties formally.

Is it possible to prove statement $1,2,3$ and similar ones in a general way, rather than prove one by one?

Thank you for your help!

A:

All "order properties", i.e. first order set theory logical formulas that only use $< (\le)$ and set theoretic notions and are quantified over the ordered set and its subsets.
The order $P$ is dense can be formulated as $$\forall x,y \in P: (x < y) \to (\exists z \in P: x < y \land y < z)$$
$P$ has no endpoints: 
$$\forall x \in P: \exists a,b \in P: (a < x) \land (x < b)$$
etc. Any order isomorphism must then preserve the truth of such formulas. One could attempt to show this by induction on the complexity of the formula perhaps. 

