Isomorphic ordinals are equal

Question

Let $\alpha$ and $\beta$ be two isomorphic ordinals. Then $\alpha = \beta$.

I want to whether the following proof is correct.

I already know that there are three prossible cases: $\alpha \in \beta, \alpha=\beta$ or $\beta\in\alpha$. Without loss of generality, assume that $\alpha \in \beta$. Let $f: \beta \to \alpha$ be an isomorphism. We consider $f(\alpha)$. Because the range of $f$ is $\alpha$, we have $f(\alpha) < \alpha$. Because $f$ is an isomorphism, it is order preserving, so $f^2(\alpha) < f(\alpha)$. Hence $f^{n+1}(\alpha) < f^n(\alpha)$ for all $n$. So $\{f^n(\alpha): n \in \mathbb{N}\}$ is a strictly decreasing sequence in a well-ordered set. This is impossible, so we get the desired contradiction.

My doubt stems from the fact that I don't seem to use that $f$ is a bijection. I tried to use this argument for special cases of $\alpha$ and $\beta$, but I can't find order-preserving functions from a larger ordinal ($\beta$) to a smaller one ($\alpha$).

Answer 1

Your argument is fine. The assumption that $f$ is an isomorphism has many consequences, among them that $f$ is injective, that $f$ is surjective, that $f$ is bijective, and that $f$ is strictly order-preserving; some are useful here and some are not, so you shouldn’t worry just because you’re not explicitly using one of them. In fact the one that you’re using is that $f$ is a strictly order-preserving function into $\alpha$, which is strictly weaker than the stated assumption that $f$ is an isomorphism onto $\alpha$. Thus, you’ve actually proved that no ordinal admits a strictly order-preserving function into a proper initial segment of itself, which of course implies that no ordinal is isomorphic to a smaller ordinal.