Counterexample: a pair of linearly ordered sets that are isomorphic to subsets of the other, but not isomorphic between them

Question

I have encountered myself with the following exercise:

Let $\langle A, <_R\rangle$ and $\langle B, <_S\rangle$ be two linearly ordered sets so that each one is isomorphic to a subset of the other, that is, there exists $A'\subseteq A$ and $B'\subseteq B$ such that:

$$\langle A,<_R\rangle\cong\langle B',<_S\cap(B'\times B')\rangle\qquad\&\qquad\langle B,<_S\rangle\cong\langle A',<_R\cap(A'\times A')\rangle$$

Is it necessarily true that $\langle A,<_R\rangle\cong\langle B,<_S\rangle$?

The original statement talked about well-ordered sets, but that case is pretty easy, because supposing that $\langle B',<_S\cap(B'\times B')\rangle$ is not isomorphic to $\langle B,<_S\rangle$, the theorem of comparison between well-ordered sets assures that $B'$ with its restricted relation $<_S$ is isomorphic to an initial section of $\langle B,<_S\rangle$. But then, composing the unique isomorphism between $A$ and $B'$ with the restriction to $B'$ of the only isomorphism between $B$ and $A'$, we find that $A$ is isomorphic to a subset of $A$ with strict upper bounds (namely, the image of the element of $B$ that defines the initial section of $B'$ via the isomorphism between $B$ and $A'$) in the sense of $<_R$, which is absurd.

However, when the sets are not well-ordered, can we find a couple of linearly ordered sets that, although verifying the stated property, are not isomorphic to each other?

I have tried with some subsets of $\mathbb{R}$ and other subsets of $\mathbb{R}$, but it seems that they are all isomorphic to each other.

I ran out of ideas. Are there any counterexamples to this statement?

Thanks in advance for your time.

P.S.: I have thought, for instance, that a closed interval of $\mathbb{R}$ shouldn't be isomorphic to the whole $\mathbb{R}$. How can we prove this assertion?

Answer 1

This should work: the real intervals $I = (-1, 1) \subset [-1, 1] = J$ with the usual ordering are order isomorphic to subsets of one another:

• $I$, being a subset of $J$, is obviously order isomorphic to itself
• $J$ is order isomorphic to $I' = [-\frac{1}{2}, \frac{1}{2}] \subset I$ by the scaling map $\omega : J \ni j \mapsto \frac{1}{2}j \in I'$

However, $I$ and $J$ are not order isomorphic to each other because $J$ has a least element $-1$ and a greatest element $1$ while $I$ has neither.

Indeed, order isomorphisms must preserve the existence of least and greatest elements.

For if $j_\text{min}$ were the least element of $J$ and $J$ were order isomorphic to $I$ by an order isomorphism $\omega$, then $\omega(j_\text{min})$ must be the least element of $I$. To see this, let $i \in I$ be any element. Then the unique preimage $j = \omega^{-1}(i) \in J$ of $i$ must be greater than $j_\text{min}$ by definition of $j_\text{min}$ being the least element of $J$. That is, $j_\text{min} \leq_J j$. So as $\phi$ is order preserving, this implies $\omega(j_\text{min}) \leq_I \omega(j) = i$. So, $\omega(j_\text{min})$ is indeed smaller than every element of $I$.

P.S.: I have thought, for instance, that a closed interval of $\mathbb{R}$ shouldn't be isomorphic to the whole $\mathbb{R}$. How can we prove this assertion, if true?

Using the exact same idea as before, you can show this to be true. Any closed interval of $\mathbb{R}$ must have a least element; however, $\mathbb{R}$ does not have a least element.