If $\zeta<\omega_\zeta$ how is it possible there exist a cardinal $\alpha$ such that $\alpha=\aleph_\alpha$?

Question

Since using transfinite induction it's possible to demonstrate that for every ordinals $\zeta$, we have $\zeta<\omega_\zeta$, it seems that it isn't possible that there exist an infinite cardinal $\alpha$ (which can also be an initial ordinal) such that $\alpha=\aleph_\alpha$: infact if it's true, it results that there exist two different ordinal ($\alpha$ and $\omega_\alpha$) such that $\alpha<\omega_\alpha$ and $|\alpha|=|\omega_\alpha|$, so while $\omega_\alpha$ is an initial ordinal it must be $\alpha=\omega_\alpha$ and for the hypothesis this is impossible.

Answer 1

Since using transfinite induction it's possible to demonstrate that for every ordinals $\zeta$ it's $\zeta<\omega_\zeta$

No, it isn't. What you can show is that $\zeta\le\omega_\zeta$ for all $\zeta$, but that's quite different.

Any attempted inductive argument to prove the claim above will run into trouble at limit stages. For example, just because $\alpha_n<\beta_n$ for all $n<\omega$ does not mean that $\sup\{\alpha_n:n\in\omega\}<\sup\{\beta_n:n\in\omega\}$; consider $$\alpha_n=n, \beta_n=2n.$$

So when $\lambda$ is a limt you cannot argue that $$\forall\zeta<\lambda(\omega_\zeta<\omega_\lambda)\quad\implies\quad\lambda<\omega_\lambda.$$

Answer 2

What you say is simply false. It's true that for some $\xi$ it is true that $\xi<\omega_\xi$, but that's not true for all ordinals.

Indeed, in a technical sense, one can say that a typical ordinal is of the form $\xi=\omega_\xi$.