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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?

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  • $\begingroup$ Joel, Thanks and I agree with your comments. I'll revise the question. $\endgroup$ – Haidar Sep 5 '16 at 21:01
  • $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Sep 5 '16 at 21:26
  • $\begingroup$ 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. $\endgroup$ – Haidar Sep 5 '16 at 21:36
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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 $|Y| = 2^{\kappa} > \kappa^+ = |X|$.

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  • $\begingroup$ Very nice. This produces an example under weaker hypotheses. $\endgroup$ – Joel David Hamkins Sep 5 '16 at 22:46
  • $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Sep 6 '16 at 0:28
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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 $X$.

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  • $\begingroup$ Very nice! +1.${}$ $\endgroup$ – Noah Schweber Sep 5 '16 at 21:46
  • $\begingroup$ Clever! This looks nice but I need to work through the details. Thanks Joel $\endgroup$ – Haidar Sep 5 '16 at 21:55

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