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

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    @joriki: after thinking a little more, I share your concern about the $\omega_1$ stage of the idea I had put in an answer, and I am going to delete that answer. Thanks for your helpful comments.2012-07-23

2 Answers 2

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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.

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$(1+\eta)\cdot\omega_1\ne 1+\eta\cdot\omega_1$. 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.

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    @Vladimir: In the other case, the $1$s are always adjacent to the $\eta$ of the successor ordinal, so they're sandwiched between two $\eta$s and are relatively easy to deal with. In the present case, because they're at the front, we'd need predecessor ordinals to deal with them in a regular fashion, but limit ordinals don't have predecessors, and Brian's proof shows that it's impossible to choose *ersatz* predecessors in a consistent manner.2012-07-23