text: # Isomorphic ordinals are equal
  
  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 ($\alpha$).
  
  • That's okay. You're only using the fact that it's an order-preserving mapping, which is necessarily injective - that's enough. – David Jan 3 '16 at 20:32
  • @David: Are you claiming that it is enough to assume $f$ is an injective function? – Asaf Karagila Jan 3 '16 at 21:02
  • 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. – David Jan 3 '16 at 21:11
  • Your proof could be streamlined a bit by letting $\gamma$ be the smallest ordinal such that $f(\gamma) < \gamma$. – David Jan 3 '16 at 21:37
  • 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". – 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.
explanation: The document contains a detailed mathematical discussion and a proof related to isomorphic ordinals. It includes a question about the correctness of the proof and a discussion of the assumptions and implications of the proof. The document also includes comments from other users providing feedback and additional insights.
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distilabel_metadata.content: # Isomorphic ordinals are equal
  
  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 ($\alpha$).
  
  • That's okay. You're only using the fact that it's an order-preserving mapping, which is necessarily injective - that's enough. – David Jan 3 '16 at 20:32
  • @David: Are you claiming that it is enough to assume $f$ is an injective function? – Asaf Karagila Jan 3 '16 at 21:02
  • 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. – David Jan 3 '16 at 21:11
  • Your proof could be streamlined a bit by letting $\gamma$ be the smallest ordinal such that $f(\gamma) < \gamma$. – David Jan 3 '16 at 21:37
  • 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". – 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.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a detailed mathematical discussion and a proof related to isomorphic ordinals. It includes a question about the correctness of the proof and a discussion of the assumptions and implications of the proof. The document also includes comments from other users providing feedback and additional insights.
  
  Label: question-and-answer