Is $\pmb{\eta}\cdot\pmb{\omega_1} = (\pmb{\eta} + \pmb{1})\cdot\pmb{\omega_1}$? 
*

*$\pmb{\eta}$ - order type of $\mathbb{Q}$.

*$\pmb{1}$ - order type of a singleton set.

*$\pmb{\omega_0}$ - order type of $\mathbb{N}$.

*$\pmb{\omega_1}$ - order type of the first uncountable ordinal. 


It is easy to see that $\pmb{\eta}\cdot\pmb{\omega_0} = (\pmb{\eta} + \pmb{1})\cdot\pmb{\omega_0}$, in fact, both sides are $\pmb{\eta}$.
Question:
Is $\pmb{\eta}\cdot\pmb{\omega_1} = (\pmb{\eta} + \pmb{1})\cdot\pmb{\omega_1}$?
 A: Here is a proposed solution, I have not verified all the details, but I believe this should work.
Let $\mathbb Q^\ast$ be the rational numbers plus an endpoint, let $A,B$ be a partition of this set to intervals such that $A$ has order type $\eta$. In $\mathbb Q$ fix some partition into two parts $X,Y$ such that both are intervals and $X$ is of order type $\eta+1$.
For $\alpha<\omega_1$ we write $A_\alpha,B_\alpha,X_\alpha,Y_\alpha$ to be the corresponding parts in the $\alpha$-th copies of $\mathbb Q,\mathbb Q^\ast$.
Now we define by induction:


*

*For $\alpha=0$ simply send $A_0+B_0$ into $X_0$, and $A_1$ into $Y_0$.

*If $\alpha$ is a limit ordinal, do the same. Namely $A_\alpha+B_\alpha$ into $X_\alpha$ and $A_{\alpha+1}$ into $Y_\alpha$.

*If $\alpha=\beta+1$, send $B_\alpha$ into $X_\alpha$ and $A_{\alpha+1}$ into $Y_\alpha$.


It is clear that arriving at any limit ordinal $\alpha$ we have an isomorphism of $(\eta+1)\cdot\alpha$ into $\eta\cdot\alpha$, so the step taken at the limit ordinal itself is well-defined (we do not need to worry about embedding $A_\alpha$ in a prior step).
It is clearly an order isomorphism, and it is a bijection for obvious reasons too.
