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.
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.