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 ($\alpha$).

  • 1
    $\begingroup$ That's okay. You're only using the fact that it's an order-preserving mapping, which is necessarily injective - that's enough. $\endgroup$ – David Jan 3 '16 at 20:32
  • $\begingroup$ @David: Are you claiming that it is enough to assume $f$ is an injective function? $\endgroup$ – Asaf Karagila Jan 3 '16 at 21:02
  • $\begingroup$ No, I'm saying that this follows from $f$ being an order isomorphism, which is part of the assumption. However, the hypothesis could be weakened to saying that $f$ is an isomorphism from $\beta$ onto a subset of $\alpha$, in which case $f$ would be injective, but not bijective. The same proof would carry through. $\endgroup$ – David Jan 3 '16 at 21:11
  • $\begingroup$ Your proof could be streamlined a bit by letting $\gamma$ be the smallest ordinal such that $f(\gamma) < \gamma$. $\endgroup$ – David Jan 3 '16 at 21:37
  • $\begingroup$ You only use $f$ is injective because you only show the proof of the $\alpha\in\beta$ case. Implicitly, to get the proof for the case $\beta\in \alpha$, one would exchange the symbols $\beta$ and $\alpha$ and write $f^{-1}$ in place of $f$. That is, you used bijectivity in the line "without loss of generality". $\endgroup$ – Milo Brandt Jan 3 '16 at 21:47

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.