- $\pmb{\eta}$ - 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 $\pmb{\eta}\cdot\pmb{\omega_0} = (\pmb{\eta} + \pmb{1})\cdot\pmb{\omega_0} = \pmb{\eta}$ and $(\pmb{1} + \pmb{\eta})\cdot\pmb{\omega_0} = \pmb{1} + \pmb{\eta}\cdot\pmb{\omega_0} = \pmb{1} + \pmb{\eta}$. As it was shown in the answer to my last question, we also have $(\pmb{\eta} + \pmb{1})\cdot\pmb{\omega_1} = (\pmb{\eta}+\pmb{1})\cdot\pmb{\omega_0}\cdot\pmb{\omega_1}= \pmb{\eta}\cdot\pmb{\omega_1} $.
Question: Is $(\pmb{1} + \pmb{\eta})\cdot\pmb{\omega_1} = \pmb{1} + \pmb{\eta}\cdot\pmb{\omega_1}$?