Jech's proof of canonical well-ordering of $\alpha\times\alpha$.

Question

I'm reading Jech's Set Theory. The canonical well-ordering of $\mathrm{Ord}\times\mathrm{Ord}$ is defined as $$( \alpha ,\beta ) < ( \gamma ,\theta ) :\begin{cases} \max\{\alpha ,\beta \} < \max\{\gamma ,\theta \} & \text{or}\\ \max\{\alpha ,\beta \} =\max\{\gamma ,\theta \} \land \alpha < \gamma & \text{or}\\ \max\{\alpha ,\beta \} =\max\{\gamma ,\theta \} \land \alpha =\gamma \land \beta < \theta & \end{cases}$$

The well-ordering is indeed simple. Let's define $\Gamma:\mathrm{Ord}^2\to\mathrm{Ord}$ by $$\Gamma(\alpha,\beta):=\text{the order-type of}\,\{(\xi,\eta):(\xi,\eta)<(\alpha,\beta)\}$$

He gives a "concise" proof of $\Gamma(\omega_\alpha\times\omega_\alpha)=\omega_\alpha$. This can proof $\aleph_\alpha\cdot\aleph_\alpha=\aleph_\alpha$.

Let's define $\gamma (\alpha)=\Gamma(\alpha\times\alpha)$.

I have already understand why $\gamma(\alpha)\geqslant \alpha$ . Maybe I can do Transfinite Induction by $\alpha$ from $\alpha=\emptyset$.

But his proof of $\Gamma(\omega_\alpha\times\omega_\alpha)=\omega_\alpha$ is too fast that I cannot understand.

• The frist question is why $\Gamma(\omega_0\times\omega_0)=\omega_0$. He don't give a proof of this and just state this.

• The second question is about:

Thus let $\alpha$ to be the least ordinal such that $\Gamma(\omega_\alpha\times\omega_\alpha)\neq \omega_\alpha$.Let $\beta,\gamma<\omega_\alpha$ be such that $\Gamma(\beta,\gamma)=\omega_\alpha$.

What makes me confused is that how to ensure that such a $\beta ,\gamma$ always exist?

• And more, I even can't understand why $\gamma(\alpha)$ is continuous. He stated this before the "Proof of Theorem 3.5.

This is his book's link and my question is at page 31

https://link.springer.com/book/10.1007/3-540-44761-X

Answer 1

1. This question is equivalent to asking why $\Gamma$ is a bijection between $\omega \times \omega$ and $\omega$. You can do this by induction, but as an example here are some examples: $$ \Gamma(0,0) = 0, \; \Gamma(1,0) = 1, \; \Gamma(1,1) = 2, \; \Gamma(2,0) = 3, \; \Gamma(2,1) = 4, \; \Gamma(2,2) = 5, \dots $$

2. If $\alpha$ is the least ordinal such that $\Gamma(\omega_\alpha \times \omega_\alpha) \neq \omega_\alpha$, then since $\Gamma(\omega_\alpha \times \omega_\alpha) = \gamma(\omega_\alpha) \geq \omega_\alpha$, we must have $\Gamma(\omega_\alpha \times \omega_\alpha) > \omega_\alpha$. In other words, $\omega_\alpha \in \Gamma(\omega_\alpha \times \omega_\alpha)$ (recall that $\alpha < \beta$ iff $\alpha \in \beta$) so there exists some $\beta,\gamma < \omega_\alpha$ such that $\Gamma(\beta,\gamma) = \omega_\alpha$.

3. If $\alpha = \sup_{\beta < \alpha} \beta$, then: \begin{align*} \gamma(\alpha) &= \Gamma(\alpha \times \alpha) \\ &= \Gamma\left(\bigcup_{\beta < \alpha} \beta \times \beta\right) \\ &= \bigcup_{\beta < \alpha} \Gamma(\beta \times \beta) \\ &= \bigcup_{\beta < \alpha} \gamma(\beta) \\ &= \sup_{\beta < \alpha} \gamma(\beta) \end{align*} so $\gamma$ is continuous.