# 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.
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# 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.
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The text passage is a discussion about a proof in set theory, specifically about the equality of isomorphic ordinals. It does not contain any mathematical questions or calculations in the form of word problems or numerical computations.

TYPE: NO_QUESTION