TITLE: Is $(\pmb{1} + \pmb{\eta})\cdot\pmb{\omega_1} = \pmb{1} + \pmb{\eta}\cdot\pmb{\omega_1}$?
QUESTION [17 upvotes]: $\pmb{\eta}$ - 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 $(\pmb{1} + \pmb{\eta})\cdot\pmb{\omega_1} = \pmb{1} + \pmb{\eta}\cdot\pmb{\omega_1}$?

REPLY [16 votes]: 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.