Studying linear orderings, I learned two theorems.
Suppose two linearly ordered sets A and B satisfy the following: (1) countably infinite, (2) dense, i.e. if x<z then there exists y such that x<y<z, (3) no first and last elements. Then, A and B are order isomorphic.
Suppose two linearly ordered sets A and B satisfy the following: (1) least upper bound property, (2) separable, i.e. there exists countable dense subset, (3) no first and last elements. Then, A and B are order isomorphic.
I thought the conditions of first theorem determine the order type of Q, and the conditions of second theorem determine the order type of R.
What I wonder is the conditions which determine the order type of N and Z.
I guess the conditions that determine the order type of N are the following: (1) countably infinite, (2) well ordered, (3) For all elements, except the first element, there exists immediate predecessor.
And, I guess the conditions that determine the order type of Z are the following: (1) countably infinite, (2) For all elements, there exist immediate predecessor and successor, (3) least upper bound property.
I wonder whether these conditions really determine the order type of N, Z respectively, or whether there are better conditions.
For both $\Bbb N$ and $\Bbb Z$, they are quickly seen to meet the indicated conditions. Conversely:
Suppose $A$ is a infinite well-ordered set where every element $x$ except the least has an immediate predecessor, denoted by $p(x)$. Define $f: \Bbb N \to A$ inductively by
This defines $f$ as an order-preserving injection. Suppose $f$ is not surjective. Let $x = \min A \setminus f(\Bbb N)$. $x \ne \min A$, since $f(0) = \min A$, so $x$ has a predecessor. But $p(x) \notin A \setminus f(\Bbb N)$ since it is smaller than $x$. Thus $p(x) = f(n)$ for some $n$. But then $f(n+1) = \min\{m \in A: m > p(x)\}$. So $f(n+1) > p(x)$, but because $x > p(x), f(n+1) \le x$. If $f(n+1) < x$, then $p(x) < f(n+1) < x$ contradicts that $p(x)$ is the immediate predecessor of $x$, while $f(n+1) = x$ contradicts that $x \in A\setminus f(\Bbb N)$. This cannot be. Thus $f$ must be an order-preserving bijection between $\Bbb N$ and $A$. They have the same order-type.
Now suppose $B$ is a non-empty linearly ordered set for which every $x \in B$ has an immediate predecessor $p(x)$ and immediate successor $s(x)$, and such every non-empty $C \subset B$ which is bounded above has a least upper bound. Note that because of the predecessor condition, this least upper bound $m$ will always be the maximum element of $C$. If $m$ were not an element of $C$, then $p(m)$ would also be an upper bound for $C$. Thus we could have equivalently given the condition that every non-empty subset of $B$ that is bounded above must have a maximum. Another equivalent is that every non-empty subset of $C$ that is bounded below has a minimum, as the set of lower bounds is bounded above, and thus has a maximum, whose successor must be the minimum of $C$.
Define $f: \Bbb Z \to B$ by choosing an arbitrary element of $B$ for $f(0)$, then defining inductively for $n > 0$
This defines $f$ as an order-preserving injection of $\Bbb Z$ into $B$.
Suppose $x \in B \setminus f(\Bbb Z)$. If $x < f(0)$, let $z = \min\{y \in f(\Bbb Z) : x < y\}$. This set is non-empty and bounded below, so $z$ must exist. Since $z \in f(\Bbb Z)$, there is some $m \in\Bbb Z$ with $z = f(m)$. But $f(m-1) < f(m)$ so it must be that $f(m-1) < x$, but $f(m-1) = p(f(m)) = p(z)$, so $p(z) < x < z$, which cannot be. A similar contradiction occurs if $x > f(0)$. So there cannot be such an $x \in B \setminus f(\Bbb Z)$. I.e., $f$ is bijective, and $\Bbb Z$ and $B$ must have the same order-type.
The condition on $B$ could be simplified to $B$ is not empty, has no maximum or minimum, and for every non-empty $C \subseteq B$, if $C$ is bounded below, then $C$ has a minimum, and if $C$ is bounded above, then $C$ has a maximum.