# Ordinal arithmetic exercise in Kunen

I'm self studying Kunen (2013 edition) right now, and got stuck on this exercise - I generally lack the tools and/or intuition for these kind of ordinal arithmetic exercises.

Exercise. Let $\alpha\geq\omega$. Then $type(\alpha\times\alpha,\vartriangleleft)=\alpha$ iff $\alpha=\omega^\mu$ for $\mu=1$ or $\mu$ infinite indecomposable ordinal.

As for the definitions:

Definition. A limit ordinal $\gamma$ is indecomposable if any (hence, all) of the following equivalent statements hold:

1. $\forall\alpha,\beta<\gamma: \alpha+\beta<\gamma$
2. $\forall\alpha<\gamma: \alpha+\gamma=\gamma$
3. $\forall X\subseteq\gamma: type(X)=\gamma\lor type(\gamma\backslash X)=\gamma$
4. $\exists\delta: \gamma=\omega^\delta$

Definition. $(\xi_1,\xi_2)\vartriangleleft(\eta_1,\eta_2)$ iff $\max(\xi_1,\xi_2)<\max(\eta_1,\eta_2)$ or $\max(\xi_1,\xi_2)=\max(\eta_1,\eta_2)$ and $(\xi_1,\xi_2)\prec(\eta_1,\eta_2)$, where $\prec$ is the lexicographical order.

I've tried the "$\Leftarrow$" direction, where the $\mu=1$ case seems straight forward, by grouping the elements into "blocks" of equal max value $n$, noticing these blocks have $2n+1$ elements and thus seeing that it's type is $\omega$. On the indecomposable case, I've tried to show inductively that $type(\omega^\delta\times\omega^\delta,\vartriangleleft)<\omega^\mu$ for all $\delta<\mu$, which therefore implies the wanted result (since $type(\alpha\times\alpha)\geq\alpha$ always hold). I can't see how I use indecomposability though, since I only use $\omega$ as base case, exploit $\mu$ being a limit in the successor case and that $\bigcup\{\omega^\zeta\mid\zeta<\gamma\}\leq\omega^\gamma$ for $\gamma$ limit.

I briefly tried the "$\Rightarrow$" direction, but I was completely lost as to how to even begin the argument.

Hints would be greatly appreciated.

For the "$\Leftarrow$" use the fact that $\mathbf{type}((\alpha+1)\times(\alpha+1),\lhd)=\mathbf{type}(\alpha\times\alpha,\lhd)+\alpha\cdot 2+1$, to prove by transfinite induction that $\mathbf{type}(\alpha\times\alpha,\lhd)\leq \alpha^3$ for all ordinals $\alpha$; then if $\gamma=\omega^\mu$ with $\mu$ indecomposable, we have $\mathbf{type}(\gamma\times\gamma,\lhd)=\sup\{\mathbf{type}(\alpha\times\alpha,\lhd):\alpha<\gamma\}$ as $\gamma$ is a limit ordinal, then by the bound given above we obtain $\mathbf{type}(\gamma\times\gamma,\lhd)\leq \gamma$; as $\alpha^3<\gamma$ for all $\alpha<\gamma$, but for any ordinal $\alpha$ we have $\mathbf{type}(\alpha\times\alpha,\lhd)\geq \alpha$.
• If $\gamma=\omega^\theta$ with $\theta$ not indecomposable, there is $\mu<\gamma$ such that $\mu\cdot 2>\theta$; by the equivalent definitions you gave in your question, then prove that if $A=\{(\alpha,\xi):\alpha<\omega^\mu\wedge\omega^\mu\leq\xi<\omega^\mu\cdot 2\}$, $(A,\lhd)\simeq(\omega^{\mu\cdot 2},\in)$, and so $\mathbf{type}(\gamma\times\gamma,\lhd)\geq \mathbf{type}(\omega^\mu\cdot 2\times\omega^\mu\cdot 2,\lhd)\geq \omega^{\mu\cdot 2}>\gamma$.
• If $\gamma=\omega^{\beta_n}\cdot l_n+\cdots+\omega^{\beta_1}\cdot l_1+l_0,$ with $l_n\geq 2$, if we put $m=l_n-1$, we get $\omega^{\beta_n}\cdot m+1<\gamma$, prove that $\mathbf{type}((\omega^{\beta_n}\cdot m+1)\times(\omega^{\beta_n}\cdot m+1),\lhd)\geq \omega^{\beta_n}\cdot 3\cdot m+1$, which implies $\mathbf{type}(\gamma\times\gamma,\lhd)> \gamma$.
• If $\gamma=\omega^{\beta_n}\cdot l_n+\cdots+\omega^{\beta_1}\cdot l_1+l_0,$ with $l_n=1$ and $n\geq 1$, $\omega^{\beta_n}+1\leq \gamma$, so proving $\mathbf{type}((\omega^{\beta_n}+1)\times(\omega^{\beta_n}+1),\lhd)\geq\omega^{\beta_n}\cdot 3+1,$ it follows that $\mathbf{type}(\gamma\times\gamma,\lhd)> \gamma$.
• Thanks! I have one quick question though. Say we have $\gamma=\omega^{\beta_n}\cdot l_n+\dots+\omega^{\beta_n}\cdot l_1+l_0$. Does it then always hold that $\omega^{\beta_n+1}>\gamma$ and $\omega^{\beta_n}\cdot(l_n+1)>\gamma$? And if not, then something close to? (Btw, in your first paragraph, I have presumed you meant "but for any ordinal $\alpha$ we have $type(\alpha\times\alpha,\vartriangleleft)\geq\alpha$".) – Dan Saattrup Nielsen Jul 27 '13 at 7:36
• @Leidem: You're welcome!. Yes, indeed, we have both $\omega^{\beta_n+1}>\gamma$ and $\omega^{\beta_n}\cdot(l_n+1)>\gamma$, as $\omega^{\beta_n+1}\geq\gamma$ and $\omega^{\beta_n}\cdot(l_n+1)\geq\gamma$ and uniqueness of the Cantor normal form. – Camilo Arosemena-Serrato Jul 27 '13 at 14:36