How to demonstrate that the class of ordinal numbers is isomorphic to the class of initial ordinal ones?

Question

In the 11th chapter of the book Set theory for the mathematicians by Jean E. Rubin I've read that: $\omega_\alpha = \omega_\beta \iff \alpha = \beta$ and $\omega_\alpha < \omega_\beta \iff \alpha < \beta$. I've already tried to demonstrate that through transfinite induction, but, apart from the most banal cases, it seems complicated: can someone help me by any chance?

However I think we can proceed in this way

First of all we remember that through transfinite induction it's easily possible to demonstrate that $\omega_\alpha<\omega_{\alpha+1}$.

Then we observe that if it's $0=\alpha< \beta \Rightarrow \omega = \omega_0 = \omega_\alpha < \omega_\beta$, one could try to demonstrate that through transfinite induction; so we suppose that $\forall \gamma < \beta$ | $\alpha < \gamma \Rightarrow \omega_\alpha < \omega_\gamma$

$\begin{cases}\alpha < \beta=(\overline\gamma + 1)\ \\ \alpha < \beta=\bigcup_{\gamma<{\beta}}\gamma \ \end{cases}\Rightarrow \begin{cases}\alpha\le\overline\gamma \rightarrow\begin{cases}\alpha<\overline\gamma \\ \alpha=\overline\gamma\end{cases} \\ \alpha = \overline\gamma < (\overline\gamma + 1)\le\bigcup_{\gamma<{\beta}}\gamma=\beta\ \end{cases}\Rightarrow\begin{cases} \omega_\alpha<\omega_{\overline\gamma} < \omega_{\overline\gamma+1}=\omega_\beta \\ \omega_\alpha=\omega_{\overline\gamma}<\omega_{\overline\gamma+1}=\omega_\beta \\ \omega_\alpha=\omega_{\overline\gamma}<\omega_{\overline\gamma+1}\le\bigcup_{\gamma<\beta}\omega_\gamma=\omega_\beta \end{cases}\Rightarrow\omega_\alpha<\omega_\beta$

Then -trivially- if $\alpha=\beta\Rightarrow\omega_\alpha=\omega_\beta$, we observe that if $\omega_\alpha=\omega_\beta$ it's also $\alpha=\beta$, since if it were $\alpha\neq\beta$, as far as proved, it would be $\omega_\alpha<\omega_\beta$ or $\omega_\beta<\omega_\alpha$ and it's impossible.

What do you think about this demonstration?

Answer 1

It is relatively easy to establish that the property bellow holds for all ordinals $\beta$ by transfinite induction on $\beta$ :

$$P(\beta) := \forall \alpha \in \operatorname{Ord}, \left\{ \begin{array}{ccc} \alpha < \beta & \Longrightarrow & \omega_\alpha < \omega_{\beta} \\ \alpha > \beta & \Longrightarrow & \omega_\alpha > \omega_{\beta} \end{array} \right.$$ Edit : Simplified $P$.

Clearly, $P(0)$ holds.

Let $0 \neq \beta \in \operatorname{Ord}$, assume $P(\beta')$ holds for every $\beta' < \beta$.

• successor : assume $\beta = \gamma + 1$. Then one has $\omega_\gamma < \omega_\beta$ since $\omega_\beta = \omega_\gamma^{\ +}$.
  Let $\alpha \in \operatorname{Ord}$.
  $\qquad$ Clearly, if $\alpha < \gamma$, one has $\omega_\alpha < \omega_\beta$.
  $\qquad$ If $\alpha = \gamma$, then $\omega_\alpha = \omega_\gamma < \omega_\beta$.
  $\qquad$ What remains is the case $\alpha > \beta$, we show that $\omega_\alpha > \omega_\beta$ by transfinite induction on $\alpha$, starting with $\alpha = \beta + 1$.

proof that $\alpha > \beta \Rightarrow \omega_\alpha > \omega_\beta$ :
Assume $\alpha = \beta + 1$, then $\omega_{\alpha} = \omega_{\beta}^{\ +} > \omega_\beta$
Assume $\alpha = \kappa + 1$, where $\omega_\kappa> \omega_\beta$, then $\omega_\alpha = \omega_\kappa^{\ +} > \omega_\kappa > \omega_\beta$.
Assume $\alpha$ is a limit ordinal, then $\omega_\alpha = \bigcup_{\kappa < \alpha} \omega_\kappa \geqslant \omega_{\beta + 1} > \omega_\beta$ since $\alpha > \beta + 1$.

• limit : Here, $\omega_\beta = \bigcup_{\kappa < \beta} \omega_\kappa$.
  Let $\alpha < \beta$. Since $\beta$ is limit, we also have $\alpha + 1 < \beta$. Hence, $\omega_\alpha < \omega_{\alpha + 1} \leqslant \bigcup_{\kappa < \beta} \omega_\kappa = \omega_\beta$.
  Let $\alpha > \beta$, then we get $\omega_\alpha > \omega_\beta$ as above.