Showing ordinal addition and ordinal multiplication are defined

I've been reading absoluteless results in Kunens't latest Set Theory text. After talking about $\Delta_0$ formulas and absoluteness, he mentions that certain concepts are absolute, but not $\Delta_0$. Then, he shows that ordinal addition $\alpha + \beta$ and ordinal multiplication $\alpha \cdot \beta$ are absolute for transitive models of $ZF - P$.

In his text, he defines the sum and product of ordinals as follows: $$\alpha + \beta = \mbox{type}(\{0\} \times \alpha \cup \{1\} \times \beta),\\ \alpha \cdot \beta = \mbox{type}(\beta \times \alpha),$$ where "type" is the order type: If $R$ well-orders $A$, then type$(A;R)$ is the unique ordinal $\alpha$ such that $(A;R) \cong (\alpha ; \in)$.

After proving $\alpha + \beta$ and $\alpha \cdot \beta$ are absolute, he moves to $V_\gamma$ for a limit ordinal $\gamma$. Then, he presents an interesting exercise:

Let $\gamma$ be a limit ordinal such that $\forall \alpha < \gamma \thinspace [\alpha^2 < \gamma]$. Show that ordinal sum and product are defined in $V_\gamma$ and are absolute for $V_\gamma$.

I've been looking at this exercise, trying to figure out how it works. Since we don't have Replacement, we are unable to define ordinal addition and multiplication in $V_\gamma$. I'm fairly certain that showing existence will show absoluteness since being an isomorphism and having a lexicographical ordering are $\Delta_0$. I keep getting stuck on how the statement $\forall \alpha < \gamma \thinspace [\alpha^2 < \gamma]$ is used in the proof of absoluteness in $V_\gamma$.

Any help/hints would be greatly appreciated.

• Well, if $\alpha < \gamma$ and $\alpha^2 \ge \gamma$, then in $V_\gamma$, multiplication won't be defined on the pair $(\alpha,\alpha)$ so it wouldn't even make sense to ask whether multiplication is absolute to $V_\gamma$. – Trevor Wilson Oct 31 '13 at 19:37
• In regard to your recent edit, you don't need the entire replacement schema to prove the existence of the isomorphisms you want. If you "run out of room" in $V_\gamma$ when trying to define the operations, perhaps you can show that violates the hypothesis on closure of $\gamma$. – Trevor Wilson Oct 31 '13 at 20:02

$\cdot$ and $+$ on $Ord$ are defined using the fact that there are unique ordinals isomorphic to $(\beta\times\alpha,\leq_{\beta\times\alpha})$ and $((\{0\}\times\alpha)\cup(\{1\}\times\beta),\leq_{\alpha\cup\beta}).$
It is easy to prove by induction on $\alpha$ that for all ordinals $\alpha$, $\alpha\cdot 2\leq \alpha^2$, hence $\forall\alpha,\beta<\gamma$, $\alpha+\beta<\gamma$.
By the hypothesis on $\gamma$, in particular we have that $\gamma$ is limit, so if $\alpha,\beta<\gamma$, i.e., $\alpha,\beta\in R(\gamma)$, $(\beta\times\alpha,\leq_{\beta\times\alpha})$ and $((\{0\}\times\alpha)\cup(\{1\}\times\beta),\leq_{\alpha\cup\beta})$ lie at $R(\gamma)$; as the ranks of both sets are less than $\max\{\alpha,\beta\}+n$ for some $n<\omega.$ Also there are unique ordinals in $R(\gamma)$; $\alpha+\beta$ and $\alpha\cdot\beta$ respectively, isomorphic to both well-orders, and such isomorphisms are easy to show to be in $R(\gamma)$; using that $\gamma$ is limit.
Thus $\cdot$ and $+$ can be defined for ordinals in $R(\gamma)$ and they are absoulute.