Ngtjyty

I have encountered myself with the following exercise:

Let $langle A, <_Rrangle$ and $langle B, <_Srangle$ 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:

$$langle A,<_Rrangleconglangle B',<_Scap(B'times B')rangleqquad&qquadlangle B,<_Srangleconglangle A',<_Rcap(A'times A')rangle$$

Is it necessarily true that $langle A,<_Rrangleconglangle B,<_Srangle$?

The original statement talked about well-ordered sets, but that case is pretty easy, because supposing that $langle B',<_Scap(B'times B')rangle$ is not isomorphic to $langle B,<_Srangle$, the theorem of comparison between well-ordered sets assures that $B'$ with its restricted relation $<_S$ is isomorphic to an initial section of $langle B,<_Srangle$. 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 many examples between 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? Or does this property actually characterize when two ordered structures are isomorphic? In case the second part holds, why is that the case?

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, if true?

This should work: the real intervals $I = (-1, 1)$ and $J = [-1, 1]$ 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 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.