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

Vladimir Reshetnikov
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  • You define the product as the order type of the lexicographic ordering, right? Namely, $\eta\cdot\omega_1$ is the order type of replacing every rational by a copy of $\omega_1$. – Asaf Karagila Jul 22 '12 at 23:18
  • @AsafKaragila I was assuming it was the "reverse" lexicographic ordering like the one used for ordinal multiplication. – William Jul 22 '12 at 23:21
  • Oh, makes sense. So $\eta\cdot\omega_1$ is stacking up $\omega_1$ many copies of $\mathbb Q$? – Asaf Karagila Jul 22 '12 at 23:23
  • @AsafKaragila He claims that $\eta \cdot \omega_0 = \eta$. If you replace every rational by a copy of $\omega_0$, then you are no longer dense. It can not be isomorphic to $\eta$. – William Jul 22 '12 at 23:25
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    The product is defined so that $\pmb{\eta}\cdot\pmb{\omega_1}$ is dense. – Vladimir Reshetnikov Jul 22 '12 at 23:26
  • I have an idea, but I can't put it into formality due to a notational nightmare. – Asaf Karagila Jul 22 '12 at 23:52

1 Answers1

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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.

Asaf Karagila
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