If $X_1$,$X_2$ are wosets isomorphic to ordinals $\alpha_1,\alpha_2$ then $X_1\times X_2$ is isomorphic to $\alpha_2\cdot \alpha_1$

Question

I want to prove the following:

Let $X_1$,$X_2$ be wosets, isomorphic to ordinals $\alpha_1,\alpha_2$ respectively. Then $X_1\times X_2$, with the lexicographic order, is isomorphic to $\alpha_2\cdot\alpha_1$.

My attempt:

I will use the notation $X[y]:=\{x\in X:x<y\}$ for a woset $X$ and $y\in X$. I assume one has already shown the following:

Let $X_1$,$X_2$ be disjoints wosets, isomorphic to ordinals $\alpha_1,\alpha_2$ respectively. Then $X_1\cup X_2$, with the linear sum order, is isomorphic to $\alpha_1+\alpha_2$.

Let $a$ denote the smallest element of $X_2$. We proceed by induction on $\alpha_1$. The case $\alpha_1=0$ is trivial.

Suppose it holds for some $\alpha_1$ and let $f:\alpha_1+1\to X_1$ be an isomorphism. Then $f\restriction_{\alpha_1}:\alpha_1\to X_1[f(\alpha_1)]$ is an isomorphism, and so the induction hypothesis implies that $X_1[f(\alpha_1)]\times X_2=(X_1\times X_2)[(f(\alpha_1),a)]$ is isomorphic to $\alpha_2\cdot \alpha_1$. Since $\{f(\alpha_1)\}\times X_2$ is isomorphic to $X_2$ we get that $(X_1\times X_2)[(f(\alpha_1),a)] \cup \{f(\alpha_1)\}\times X_2=X_1\times X_2$

is isomorphic to $\alpha_2\cdot\alpha_1+\alpha_2=\alpha_2\cdot(\alpha_1+1)$.

Suppose it holds for all $\gamma<\alpha_1$ for some nonzero limit ordinal $\alpha_1$, and let $f:\alpha_1\to X_1$ be an isomorphism. Then, for each $\gamma<\alpha_1$, $f\restriction_{\gamma}:\gamma\to X_1[f(\gamma)]$ is an isomorphism, and so the induction hypothesis implies that $X_1[f(\gamma)]\times X_2=(X_1\times X_2)[(f(\gamma),a)]$ is isomorphic to $\alpha_2\cdot \gamma$. Let $g=\cup_{\gamma < \alpha_1}g_\gamma$, where $g_\gamma$ is the unique isomorphism between $(X_1\times X_2)[(f(\gamma),a)]$ and $\alpha_2\cdot \gamma$. Note that $\gamma_1<\gamma_2$ implies $g_{\gamma_1}\subset g_{\gamma_2}$ (because a woset is isomorphic to a unique ordinal and the isomorphism is unique). Then $g:\alpha_2\cdot\alpha_1 \to X_1\times X_2$ is an isomorphism.

Is this correct? Specifically, I am unsure about the limit case (in bold).

1. How can we show that $X_1[f(\gamma)]\times X_2$ and $(X_1\times X_2)[(f(\gamma),a)]$ are the same wosets?

2. Why does $\gamma_1<\gamma_2$ implies $g_{\gamma_1}\subset g_{\gamma_2}$?

Answer 1

• The two (lexicographically ordered) subsets $X_1[f(\alpha_1)]\times X_2$ and $(X_1\times X_2)[(f(\alpha_1),a)]$ of $X_1\times X_2$ are equal, by definition of the notation $X[y]$ and of the lexicographic order: $\forall(u,v)\in X_1\times X_2$, $$(u,v)<(f(\alpha_1),a)\iff\left(u<f(\alpha_1)\text{ or }(u=f(\alpha_1)\text{ and }v<a)\right)\iff u<f(\alpha_1).$$
• If $\gamma_1<\gamma_2$ then $(X_1\times X_2)[(f(\gamma_1),a)]$ is sent isomorphically by $g_{\gamma_2}$ onto the ordinal $(\alpha_2\cdot\gamma_2)[g_{\gamma_2}(f(\gamma_1),a)]$, and by $g_{\gamma_1}$ onto the ordinal $(\alpha_2\cdot\gamma_1)$. Therefore (by the uniqueness you recalled), $g_{\gamma_1}\subset g_{\gamma_2}$.