Whether partially ordered sets $(Z,\subseteq )$ and $(Q ,\subseteq )$ are isomorphic?

Question

Let $Z= \left\{[k,l]:k,l \in \mathbb Z \wedge k \le l\right\}$ and $Q= \left\{[p,q]:p,q \in \mathbb Q \wedge p \le q\right\}$. Whether partially ordered sets $(Z,\subseteq )$ and $(Q ,\subseteq )$ are isomorphic?

If it is truth by definition I have bijection $f: Q \rightarrow Z$ and $q\subseteq q' \Leftrightarrow f(q) \subseteq f(q')$. I think I can find this bijection because $|\mathbb Z|=|\mathbb Q|$ but I don't have idea how to do it and I don't know if the condition $q\subseteq q' \Leftrightarrow f(q) \subseteq f(q')$ will be met.

I thought about function $f(x)=|x|$ but it is not injective function. At the same time, I do not know how to obtain all integers from rational numbers so that it is a bijection

Have you any ideas?

Answer 1

Exactly one of these partial orders is well-founded. More specifically, in one of them all the intervals are finite, and therefore have only finitely many subsets (in interval form), whereas in $\Bbb Q$ the only finite intervals are singletons.

Answer 2

Suppose they are isomorphic and let $f\colon Z\to Q$ be an isomorphism.

The minimal elements in either $Z$ or $Q$ have the form $[a,a]$.

Since the element $[0,0]\in Z$ is minimal, we have $f([0,0])=[p,p]$, for some $p\in\mathbb{Q}$. By assumption, $p\in f([0,1])=[q,r]$, with $q\ne r$. Suppose $p<r$: then $$ f([0,0])=[p,p]\subseteq[p,r]\subseteq[q,r]=f([0,1]) $$ which leads to a contradiction (hint: there is no $z\in Z$ such that $[0,0]\subsetneq z\subsetneq[0,1]$). Similarly for $q<p$.

Your idea is good as well: bounded subsets in $Z$ have a supremum, but the same is not true for $Q$. However this is a “second order property”, a simpler one might be preferred.