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Cantor define an ordinal as the class of well ordered set that are isomorphic each other. In ZF is possible to show that the Von Neumann ordinals are canonical representatives of Cantor ordinals. I know that Cantor ordinals fails to be a set, but i don't know any prove of this. So my question is how to prove in Z (Zermelo set theory) that Cantor ordinals are proper classes?

Or, alternatively: Let a “Cantor ordinal” be defined as an equivalence class of well-ordered sets (w.o. sets $X$ and $Y$ being “equivalent” if there exists an order-preserving bijection between them). Without the Axiom Schema of Replacement, how can one prove that such a “Cantor ordinal” is a proper class?

NikS
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Manuel Bonanno
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    Maybe useful this post. – Mauro ALLEGRANZA Jun 13 '25 at 12:20
  • @MauroALLEGRANZA I have read this post, but in the anwer is stated that: 'This means that an ordinal is not a set, in modern perspective, since for any non-empty well-ordered set, there is a proper class of well-ordered sets which are isomorphic to this given order'. This is exactly my question. – Manuel Bonanno Jun 13 '25 at 12:24
  • See also discussion in this MSO post. – Mauro ALLEGRANZA Jun 13 '25 at 13:42
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    Let $W$ be a nonempty well-ordered set. Suppose, for contradiction, that there exists a set $\mathcal C$ such that $X\in\mathcal C$ if and only if $X$ is order-isomorphic to $W$. Then we can form the set $\mathcal D$ such that $X\in\mathcal D$ if and only if $X$ is the underlying set of a well-ordered set which is order-isomorphic to $W$. But given any set $x$, it is clear that $x\in X$ for some $X\in W$. Hence the union of the sets in $W$ is the entire universe; contradiction. – Joe Jun 13 '25 at 15:25
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    (It is easiest to show that $\mathcal D$ exists using the replacement axiom, but I'm fairly sure that this can be made to work in Zermelo set theory, where we don't have replacement. Ask me if you would like more details.) – Joe Jun 13 '25 at 15:25
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    By the way, I think your question is perfectly fine, and it should be reopened. – Joe Jun 13 '25 at 16:36
  • Thank you very much! I thinks there is some typo on your first commenti: any set $x \in X$ for some $X\in \mathcal{C}$ then the union of $\mathcal{D}$ is the entire universe. – Manuel Bonanno Jun 13 '25 at 17:28
  • If a understand the construction $\mathcal{D}={\mathcal{C}}$ so It can be formed in Zermelo using axiom of pair – Manuel Bonanno Jun 13 '25 at 17:30
  • Yes, to your first comment: I meant that the union of the sets in $\mathcal D$ is the entire universe. For your second comment, I'm not sure what you mean: $\mathcal C$ consists of the well-ordered sets which are order-isomorphic to $W$. So an element of $\mathcal C$ must be a pair $(S,<)$ consisting of a set $S$ and a well-ordering $<$ on $S$. We obtain $\mathcal D$ by extracting the underlying sets in $\mathcal C$. It's easiest to prove that $\mathcal D$ exists using replacement. – Joe Jun 13 '25 at 17:35
  • But if we don't have replacement, then I think we can try taking the union of the sets in $\mathcal C$ (which feels quite unnatural, but I think works). Recall that an element of $\mathcal C$ is a pair $(S,<)$, and $(S,<)={{S},{S,<}}$ in the standard construction of the ordered pair by Kuratowski, so by taking the union of the sets in $\mathcal C$, we get all the underlying sets, plus a bunch of well-orders $<$. I'm not sure how to get $\mathcal D$ from this weird collection, but I think you can get a contradiction anyway by taking the union again. – Joe Jun 13 '25 at 17:38
  • Thanks! Now i have understand – Manuel Bonanno Jun 14 '25 at 11:38
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    I’m rather mystified as to how this question is considered a duplicate of this one. I don’t see how they’re related at all (other than being generally about set theory) – NikS Jun 17 '25 at 22:54

1 Answers1

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The precise way of stating that "Cantor ordinals" do not exist is given by the following theorem.

Theorem. Suppose $(W,<)$ is a well-ordered set, and $W\neq\varnothing$. It is not the case that there exists a set $\mathcal C$ such that $X\in\mathcal C$ iff $X$ is a well-ordered set which is order-isomorphic to $\mathcal C$.

Proof. Suppose, for contradiction, that there does exist such a set $\mathcal C$. Then there exists a set $\mathcal D$ such that $X\in\mathcal D$ iff $X$ is the underlying set of a well-ordered set which is order-isomorphic to $\mathcal C$. Note that $\mathcal D$ contains every set which is equinumerous to $W$, since the order-structure on $(W,<)$ can be pushed along bijections. Thus given any set $x$, we have $x\in X$ for some $X\in\mathcal D$. That is, the union of the sets in $\mathcal D$ is the entire universe, which is absurd. QED.

This proof tacitly uses some of the basic axioms of set theory, such as the axiom of union. The only nontrivial step appears to be in proving the existence of $\mathcal D$. If we have the axiom (schema) of replacement, then we can argue as follows: consider the sentence $S(a,b)$ in the language of set theory corresponding to the English phrase "$a$ is a well-ordered set, and $b$ is the underlying set of $a$". Then, for all $X\in\mathcal C$, there is a unique $Y$ such that $S(X,Y)$ holds; indeed, $Y$ is just the underlying set of $X$. Hence by replacement, we can form the set $\mathcal D$ such that $Y\in\mathcal D$ iff there is an $X\in\mathcal C$ such that $S(X,Y)$; that is, $Y\in\mathcal D$ iff there is an $X\in\mathcal C$ such that $Y$ is the underlying set of $X$.

If we don't have replacement, then I think the following ad hoc approach works, even though it feels quite unnatural. Recall that an element of $\mathcal C$ is a well-ordered set $(E,<)$, and $(E,<)=\{\{E\},\{E,<\}\}$ under the standard definition of an ordered pair due to Kuratowski. This means that if we form the union of the sets in $\mathcal C$, we end up with a strange-looking set $\mathcal U$ consisting of all of the underlying sets in $\mathcal C$, and a bunch of well-orders. The set $\mathcal D$ is given by $\{Y\in\mathcal U\mid\text{$Y$ is equinumerous to $W$}\}$. Alternatively, we could just take the union of the sets in $\mathcal U$, and we would wind up with the entire universe again.

Joe
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