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Question: Suppose $(A,<)$ and $(B,<)$ are two dense linear orders without end points, of size $2^{\aleph_0}$, and both $\aleph_1$-saturated (every type over a countable subset has a realization). Are they isomorphic?

I know that under the continuum hypothesis, one can use a back & forth argument because it is true that any two $\kappa$-saturated models of size $\kappa$ have to be isomorphic. However, I don't know what happen in the general case.

Are there examples assuming $\neg CH$?

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    $\begingroup$ I have heard that the negation of the CH proves there are two non-isomorphic ultrapower of $\mathbb{R}$. $\endgroup$ – Hanul Jeon Oct 26 '16 at 19:20
  • $\begingroup$ @HanulJeon Well, there are always if we consider the ultrafilter to be either principal or non-principal, but of course you mean (I guess) two ultrapowers of $\mathbb{R}$ with respect to non-principal ultrafilters. Do you have a reference for that? $\endgroup$ – Darío G Oct 26 '16 at 19:46
  • $\begingroup$ I don't have a reference for that, but the question in that site seems to be relevant. $\endgroup$ – Hanul Jeon Oct 26 '16 at 19:50
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    $\begingroup$ Although, probably even the existence of two non-isomorphic ultrapowers of the field $\mathbb{R}$ does not answer the question: It is still possible that the linear orderings induced by them are isomorphic, isn't it? $\endgroup$ – Darío G Oct 26 '16 at 19:53
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    $\begingroup$ Sorry, I was checking the paper in your link and it seems to answer the problem: Theorem 1 in the link below is that if $CH$ fails and $A$ is unstable with $|A|\leq 2^{\aleph_0}$ then there are $2^{\aleph_0}$ non-isomorphic ultraprowers of $A$. So, we can simply use $A=(\mathbb{Q},<)$ and apply the theorem. Interesting! math.yorku.ca/~ifarah/Ftp/2009i19-ultrapowers.pdf $\endgroup$ – Darío G Oct 26 '16 at 20:00
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Suppose we have $2^{\aleph_0}>\aleph_1$.

Hausdorff showed that there is an $\eta_1$-order without end points of size $2^{\aleph_0}$, that is, a totally ordered set $\Bbb A=(A,<)$ having the following properties:

  • $\Bbb A$ has neither a coinitial nor a cofinal subset of size $<\aleph_1;$
  • For any $B,C$ subsets of $A$ both of size less than $\aleph_1$ with $B<C$, there is some $a\in A$ with $B<a<C$.

$\eta_1$-orders without end points are $\aleph_1$-saturated models; see section $5.4$ of Chang & Keisler's Model Theory, third edition.

Let $<_1$ be the lexicographical order on $\omega_1\times A$, and set $\Bbb A_1=(\omega_1\times A,<_1)$.

Build another total order $\Bbb A_2$ just like $\Bbb A_1$, but instead of using $\omega_1$, use $\omega_2$.

As $A$ has no end points, it is easy to see both $\Bbb A_1$ and $\Bbb A_2$ are $\eta_1$-orders. These orders have no end points, thus they are $\aleph_1$-saturated, and have size $2^{\aleph_0}$. However, they cannot be isomorphic as $\Bbb A_1$ has a cofinal subset of size $\aleph_1$, while $\Bbb A_2$ does not.

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  • $\begingroup$ By the first property, do you mean "Every countable set is bounded from above and below", right? $\endgroup$ – Asaf Karagila Nov 23 '16 at 19:09
  • $\begingroup$ Hi @AsafKaragila, yes, that's indeed what I mean. $\endgroup$ – Camilo Arosemena-Serrato Nov 23 '16 at 19:12

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