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

*$\pmb{\eta}$ - 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 $\pmb{\eta}\cdot\pmb{\omega_0} = (\pmb{\eta} + \pmb{1})\cdot\pmb{\omega_0} = \pmb{\eta}$ 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 $(\pmb{1} + \pmb{\eta})\cdot\pmb{\omega_1} = \pmb{1} + \pmb{\eta}\cdot\pmb{\omega_1}$?
 A: Here is a somewhat relaxed way to see that the orders $1+\eta\cdot\omega_1$ and $(1+\eta)\cdot\omega_1$ are not isomorphic, by finding a topological feature of one order not exhibited in the other.
First, observe that $(1+\eta)\cdot\omega_1$ has a closed subset of
order type $\omega_1$, namely, the points corresponding to the
initial point of each of the intervals $1+\eta$ used to create it. This suborder contains limit points for all its increasing $\omega$-sequences. 
But meanwhile, $1+\eta\cdot\omega_1$ has no such suborder. This is because this order contains no limits of sequences of points from distinct intervals (the $\eta$ intervals used to build it), but any subset
of order type $\omega_1$ will have increasing sequences of points from distinct intervals.
So they are not isomorphic.
A: $(1+\eta)\cdot\omega_1\ne 1+\eta\cdot\omega_1$. Equivalently, I’ll show that $\eta+(1+\eta)\cdot\omega_1\ne\eta\cdot\omega_1$.
Let $Q_0=\omega_1\times\big(\Bbb Q\cap[0,1)\big)$ and $Q=\omega_1\times\Bbb Q$, ordered lexicographically. Let $X=Q_0\setminus\{\langle 0,0\rangle\}$, and suppose that $h:X\to Q$ is an order-isomorphism. For each limit ordinal $\xi\in\omega_1$ let $\langle\xi_k:k\in\omega\rangle$ be a strictly increasing sequence of ordinals with limit $\xi$. Then 
$$\Big\langle h\big(\langle\xi_k,0\rangle\big):k\in\omega\Big\rangle\nearrow h\big(\langle\xi,0\rangle\big)\;.$$
If $h\big(\langle\xi,0\rangle\big)=\langle f(\xi),q_\xi\rangle$, there is an $n(\xi)\in\omega$ such that $h\big(\langle\xi_k,0\rangle\big)\in\{f(\xi)\}\times(\leftarrow,q_\xi)_{\Bbb Q}$ for $k\ge n(\xi)$. The map $\xi\mapsto\xi_{n(\xi)}$ is a pressing-down function on the cub of limit ordinals in $\omega_1$, so there is a stationary subset $S$ of limit ordinals and an $\alpha\in\omega_1$ such that $\xi_{n(\xi)}=\alpha$ for each $\xi\in S$. Then for each $\xi\in S$ we have $h\big(\langle\alpha,0\rangle\big)\in\{f(\xi)\}\times(\leftarrow,q_\xi)_{\Bbb Q}$, so there is some $\beta\in\omega_1$ such that $f(\xi)=\beta$ for each $\xi\in S$. But then $h\big[S\times\{0\}\big]\subseteq\{\beta\}\times\Bbb Q$, which is absurd, since $|S\times\{0\}|=\omega_1$, $\{\beta\}\times\Bbb Q$ is countable, and $h$ is injective.
