# locally incomparable dense linear orderings extending $\langle \mathbb{R}, < \rangle$

This follows up on Incomparable dense linear orderings extending $\langle \mathbb{R},< \rangle$ and hopefully sparks more discussion.

Where $a<b$, say that the four “types” of nonempty bounded intervals are:

$(a,b)$, $[a,b]$, $(a,b]$, and $[a,b)$.

Let $\langle X,< \rangle$ and $\langle Y,< \rangle$ be dense linear orderings without endpoints such that

1. $\langle \mathbb{R},< \rangle$ is order-embeddable into $\langle X,< \rangle$, and into $\langle Y,< \rangle$,

2. all the nonempty bounded intervals of $\langle X,< \rangle$ of the same type are order-isomorphic,

3. all the nonempty bounded intervals of $\langle Y,< \rangle$ of the same type are order-isomorphic,

4. $\langle X,< \rangle$ is isomorphic to its inverse order, and

5. $\langle Y,< \rangle$ is isomorphic to its inverse order.

Are there examples of $\langle X,< \rangle$ and $\langle Y,< \rangle$ such that

“no nonempty open interval of $\langle X,< \rangle$ is order-embeddable into any nonempty open interval of $\langle Y,< \rangle$, and no nonempty open interval of $\langle Y,< \rangle$ is order-embeddable into any nonempty open interval of $\langle X,< \rangle$”?

Is it independent of $ZFC$ whether there are such pairs of orderings for which the statement in quotes holds?

If $ZFC$ proves that there are such pairs of orderings, then which definitions (if any) of such orderings $\langle X,< \rangle$ and $\langle Y,< \rangle$ are such that it is independent of $ZFC$ whether the statement in quotes holds of them?

• Joel, Thanks and I agree with your comments. I'll revise the question. Sep 5, 2016 at 21:01
• Thanks for the edit; much better! I have deleted my comments. But one further objection: it isn't just the statement in quotation marks that should be independent, but rather the quantifiers over X and Y also should be part of that statement. Sep 5, 2016 at 21:26
• It would be interesting if it is independent of $ZFC$ whether there are such pairs of orderings. But if $ZFC$ does prove that there are such pairs of orderings, I'm interested in the features of those orderings. Let me split the question about independence into two. Sep 5, 2016 at 21:36

Here's a slight variation on Hamkins' example using compactness theorem under the assumption $2^{\kappa} > \kappa^+$ for some $\kappa \geq \mathfrak{c}$. Let $T$ be the (consistent) theory consisting the elementary diagram of $(\mathbb{R}, <)$ and the following sentences:

(1) $c_i < c_j$ for each $i < j < \kappa^{+}$

(2) For all $a < b$, the map $x \mapsto f(a, b, x)$ is an order preserving bijection from $(a, b)$ to the universe

(3) $x \mapsto g(x)$ is a reverse order isomorphism of the universe

By Lowenheim-Skolem, $T$ has a model $X$ of size $\kappa^{+}$. Let $Y = {}^{\kappa}\mathbb{R}$ under lexicographic order. Then neither one of $X, Y$ embeds in the other - $X$ has an increasing $\kappa^+$-sequence while $Y$ has size $|Y| = 2^{\kappa} > \kappa^+ = |X|$.

• Very nice. This produces an example under weaker hypotheses. Sep 5, 2016 at 22:46
• This suggests the question whether the original inquiry is equivalent to the negation of the GCH. But meanwhile, the hypotheses in my answer are consistent with the GCH. Sep 6, 2016 at 0:28

Let me prove at least that it is consistent with ZFC that there are two linear orders as you requested.

Work in ZFC plus the assumption that $2^{\omega_1}>\omega_3^L$, but $\omega_2=\omega_2^L$ and $\mathbb{R}=\mathbb{R}^L$. This situation is easy to arrange by forcing, since you can simply add $\omega_4$ many Cohen subsets of $\omega_1$ over $L$, which preserves cardinals and adds no reals.

Under that assumption, here are two linear orders with the properties you have requested. Let $X=(\mathbb{Q}^{\omega_2})^L$ and $Y=\mathbb{R}^{\omega_1}$, each under the lexical order.

The real line embeds into both of these orders. Both of these orders have the all-intervals-look-alike property. And both of the orders is symmetric, in the sense of being isomorphic to its own inverse order.

Meanwhile, every nontrivial interval of $X$ has an increasing $\omega_2$-sequence, and there is no such sequence in $Y$, so no interval of $X$ embeds into $Y$.

Conversely, every interval of $Y$ has size $2^{\omega_1}$, which is larger than $X$, since $X$ has size merely $(2^{\omega_2})^L=\omega_3^L$, and so no interval of $Y$ embeds into $X$.

• Very nice! +1.${}$ Sep 5, 2016 at 21:46
• Clever! This looks nice but I need to work through the details. Thanks Joel Sep 5, 2016 at 21:55