Countably infinite and monotonically countably infinite

While reviewing Meaning of finite, countably finite, infinite?, I wondered about $\mathbb N$ vs $\mathbb Q$, both countably infinite, but $|\mathbb Q|$ is clearly much nearer to $|\mathbb R|$ than $|\mathbb N|$, given that between any distinct pair $q_1,q_2\in \mathbb Q$ there is an infinite number of $r \in (q_1,q_2) \subset \mathbb R$, and vice versa.

This concept has been explained by this answer: there's cardinality, density and measure; all various comparisons of the size of a set.

But can the difference between $\mathbb N$ and $\mathbb Q$ be described by the mapping of $\mathbb N \to \mathbb Q$ not being monotonic?

Having seen the above answer, I assume I'm asking: Is there is a countably infinite set, $S$, that is nowhere dense, but still has not got a mapping from $\mathbb N, M,$ such that $\forall m,n \in \mathbb N: m\le n \iff M(m)\le M(n)$ (with equality only when $m=n$), for some ordering $\le \text{on } S$?

(EDIT: I changed "not dense" to "nowhere dense".)

By way of example consider mappings of $\mathbb N \to \mathbb Z$. With the normal meaning of $\le$ you can't find a mapping (mostly because you need to change one infinite "end" to "two"). But if you define $x < y$ to mean $|x|<|y|\text{ or } (|x|=|y| \text{ and } x<y)$ again with equality only when $x=y$, you can define a mapping of $n \to (-1)^{n}\left\lfloor{\frac{n+1}{2}}\right\rfloor$.

Every countably infinite set that you like works. Just pick any bijection with the naturals and then transfer the ordering on the naturals over to get an ordering on the set that satisfies your strict monotonicity condition.

One easy way to distinguish between naturals and rationals in a way you are looking for is that rationals are dense in themselves but naturals are not.

• I assume I need to specify some sort of "continuity", but that probably implies "denseness". E.g. Any set similar to $\mathbb Q \backslash (1,2)$ has a non-dense part. – Mark Hurd Jun 22 '15 at 6:30
• @MarkHurd: What do you mean? What is your dissatisfaction with the characterization I gave to distinguish rationals and naturals? In fact, as Stefan noted, the rationals are the essentially unique unbounded dense linear order, where "dense" means "dense in itself". – user21820 Jun 22 '15 at 6:36
• @MarkHurd: If you want a more fine-grained view, the naturals are nowhere dense in itself while the rationals are everywhere dense. Things like $\{ [2n,2n+1] : n \in \mathbb{Z} \}$ are dense somewhere and not dense somewhere else. – user21820 Jun 22 '15 at 6:38
• @MarkHurd: Note that my above comment is not using the standard topological definition of "nowhere dense" but rather I mean that the naturals are not dense in any interval of itself, while every interval of the rationals is dense in any interval of itself. The intermediate example I gave is dense in some intervals of itself but not in others. – user21820 Jun 22 '15 at 6:42
• @Stefan's answer has effectively shown there is no "intermediate" set of the form I'm considering, but, yes, my problem with the current wording of my assumed question is that it should probably say "nowhere dense". – Mark Hurd Jun 22 '15 at 7:00

Having seen the above answer, I assume I'm asking: Is there is a countably infinite set, $S$, that is not dense, but still has not got a mapping from $\mathbb N, M,$ such that $\forall m,n \in \mathbb N: m\le n \iff M(m)\le M(n)$ (with equality only when $m=n$), for some ordering $\le \text{on } S$?

This has a rather trivial answer. Let $S \subseteq \mathbb Q$ be any infinite countable subset that is not dense. Fix a bijection $\pi: \mathbb N \rightarrow S$ and let $\pi(m) \le_\pi \pi(n) :\Leftrightarrow m \le n$. This defines a linear order $(S; \le_S)$. So there cannot be such an $S$ as in your question.

Let me elaborate on this a little more:

When we compare the size of two sets $X$, $Y$, we are only talking about injections / bijections between them - this is what characterizes the size (cardinality) of a set. If $X$,$Y$ carry additional structure (for example $(X, \prec_X), (Y, \prec_Y)$ are total orderings), we may talk about injections / bijections between them, that respect these structures (in your example, we might want to consider order-isomorphisms). But then, we don't compare the "size" of these sets, but rather their underlying structures. Their size is part of this structure, but usually we are interested in additional properties.

Let's consider $(\mathbb Q, \le)$: By the famous back-and-forth method of Georg Cantor, one can show that $(\mathbb Q, \le)$ is (up to order-isomorphism) the unique countable dense linear order without endpoints. (Btw. this is a nice exercise - but it may be pretty challenging, if you're unfamiliar with this method.)

• Oh, I see. Let me edit my answer. – Stefan Mesken Jun 22 '15 at 3:57