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Two linear orders $A$ and $B$ have starting points $a_0$ and $b_0$, and have cofinalities $\omega_1$. Let $(a_\alpha )_{\alpha<\omega_1}$ and $(b_\alpha )_{\alpha<\omega_1}$ be cofinal sequences. Suppose we also know that for every $\alpha<\omega_1$ there is an order isomorphism $[a_\alpha ,a_{\alpha+1}]\simeq [b_\alpha ,b_{\alpha+1}]$ that maps $a_\alpha$ to $b_\alpha$ and $a_{\alpha+1}$ to $b_{\alpha+1}$.

Can we conclude that $A$ is isomorphic to $B$? How would you write this down?

Thank you for any help.

Please let me know if there is anything I can clarify.

hmakholm left over Monica
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user160663
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1 Answers1

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No, this is not true, because there may be elements of $A$ or $B$ that is in neither of the $[a_\alpha,a_{\alpha+1}]$ intervals. For example, consider $$A = \omega_1, \qquad a_\alpha=\alpha$$ and $$B = \omega_2+\omega_1, \qquad b_\alpha=\begin{cases}\alpha & \alpha<\omega \\ \omega_2+\alpha & \alpha\ge\omega\end{cases}$$

Then clearly $[a_\alpha,a_{\alpha+1}] \simeq [b_\alpha,b_{\alpha+1}]$ simply because both intervals are always two-point sets. However, $A$ and $B$ don't even have the same cardinality, so they can't be order isomorphic.

hmakholm left over Monica
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  • I am very pleased that you found a counterexample! You have no idea how helpful this was. I thought I could just go along with a straight-forward induction routine, but you showed me what can go wrong at the limit ordinals ($\omega$ in your example). Fantastic! – user160663 Jun 29 '14 at 04:24
  • So, I think you would need to ensure that if $\gamma$ is a limit ordinal the $b_\alpha$, $\alpha<\gamma$, limit to $b_\gamma$. Then your example would not work because the $b$'s could not escape the $\omega_2$ segment. – user160663 Jun 29 '14 at 04:33
  • @user160663: Yes, that might work. I think you'd still need some kind of choice principle for combining all of the local isomorphisms into a global one. – hmakholm left over Monica Jun 29 '14 at 14:45