Why do we use von Neumann ordinals, $$ 0 = \emptyset $$ $$ n+1 = n \cup \{n\} $$
and not Zermelo ordinals? $$ 0 = \emptyset $$ $$ n+1 = \{ n \} $$
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3https://math.stackexchange.com/questions/85672/the-history-of-set-theoretic-definitions-of-mathbb-n – Asaf Karagila Oct 29 '18 at 09:42
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There is no real, deep, fundamental reason. You can find a bijection between the set of Von Neumann naturals and Zermelo naturals, so anything you can do with the one set you can do with the other.
However, Von Neumann naturals are more convenient in practice for a lot of reasons. For one, the element we call $n$ also has exactly $n$ elements. That means that we can use the actual set $n$ as a cardinality, defining "the set $A$ has $n$ elements" to mean that there is a bijection between $A$ and $n$. For another, a convenient way to define the ordinals is to say that they are the transitive sets which are linearly ordered by $\in$. Then the Von Neumann naturals are precisely the finite ordinals, which is a natural and important way to think about the finite ordinals.
Mees de Vries
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22The fact that exactly one of these generalizes to the transfinite is pretty deep and fundamental. – Asaf Karagila Oct 29 '18 at 09:46
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if i have n <= m with Von Neuman ordinals I can state that this relation is true iff n is in m, how this changes with Zermelo ordinals? – Maicake Oct 29 '18 at 10:41
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5@Maicake $1 = { \emptyset } \not \in {{ { \emptyset }} } = 3$ and $1 = { \emptyset } \not \subseteq {{ { \emptyset }} } = 3$. – Stefan Mesken Oct 29 '18 at 11:41
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@StefanMesken I think Maicake was asking what the analogue is for Zermelo ordinals. – user76284 Nov 11 '18 at 01:38
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@AsafKaragila I assume you mean that the Von Neumann ordinals generalize into the transfinite. Why would the Zermelo ordinals not generalize? – Mwax Apr 05 '20 at 23:56
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@Mwax: Indeed I do. How would you define $\omega$ and $\omega+1$ in a way that truly captures the essence from Zermelo's integers? – Asaf Karagila Apr 05 '20 at 23:58
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@AsafKaragila. Thanks. I think its hard. This is not a fully worked out answer. I was thinking about taking $\omega = \bigcup n$ where $n$ is a finite Zermelo ordinal. But it seems to me that you'll end up with $\omega = { \emptyset, { \emptyset \ } , { { \emptyset } } \dots }$. and we can let $\omega + 1 = {\omega}$. So your point is that in the case of $\omega$ we have not truly captured the essence of what a Zermelo ordinal should look like according to the recursive definition template. – Mwax Apr 06 '20 at 00:12
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@AsafKaragila Can’t you generalise Zermelo’s ordinals by taking unions at limits? – Vivaan Daga Aug 06 '22 at 18:20
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@AsafKaragila Of the previous ordinals. And then using transfinite induction on well ordering a we can define an order for the zermelo ordinals. – Vivaan Daga Aug 07 '22 at 10:27
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@Shinrin-Yoku: Unions are defined by first having a set over which we take the union. In the case of the von Neumann ordinals, it just happens to be the same set for limit ordinals. So, which set are we taking union of? – Asaf Karagila Aug 07 '22 at 16:27
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I mean Defining the limit as the union of the “previous” ordinals, which could be made rigorous by using transfinite induction on a well ordered set to define a zermelo ordinal. – Vivaan Daga Aug 07 '22 at 16:36
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@Shinrin-Yoku: Yes, you can do that. But you no longer satisfy "the concept" that defined the Zermelo ordinals. Do note, at the end of the day, that pretty much any set can represent any ordinal. It's just a representation. But we do like simple representations, preferably that are somehow coherent with the concept being represented. – Asaf Karagila Aug 08 '22 at 06:27
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A simple motivation for the Von Neumann style is to write it in this way: $$ n := \{m |\ m<n\}. $$ I.e., the ordinal $n$ is the set of all the ordinals up to $n-1$. This is, I'd say, the idea behind the Von Neumann ordinals, though obviously it isn't quite a proper definition ($m$ from what base set? What is $<$?).
Alternative way to express it: $$\begin{align} 0 &:= \{\} \\ n+1 &:= \{0 \ldots n\} \end{align}$$
leftaroundabout
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Von Neumann ordinals arise naturally as an answer to the classification problem asking to classify well-orders on sets. In "naive set theory" we might typically say something along the lines of: given two well-ordered sets $(A, <_A)$ and $(B, <_B)$, we will say they are equivalent if there is an order-preserving isomorphism between $A$ and $B$. Then, an "ordinal" will be defined as an "equivalence class" of this equivalence relation.
In axiomatic set theory, however, we run into the problem that the relation described above is "too big" to be an actual object of ZFC, and so is any individual "equivalence class". However, what we can find is that each "equivalence class" has a canonical representative given by the Mostowski Collapse Lemma:
Suppose we have a set $A$ and a relation $R$ on $A$ which is well-founded and extensional ("extensional" means: for all $x,y\in A$, if $\{ z\in A \mid z \mathrel{R} x \} = \{ z\in A \mid z \mathrel{R} y \}$, then $x = y$ -- and a well-order is automatically extensional). Then there exists a unique transitive set $B$ such that $(A, R) \simeq (B, {\in}|_{B\times B})$.
By restricting to well-orders, we see that each "equivalence class" of well-orders has a unique representative of this type, which is a transitive set $X$ such that ${\in}|_{X \times X}$ is a well-order on $X$. This is precisely the definition of von Neumann ordinals.
So, the von Neumann ordinals get around the issues with "too big" sets which cannot be constructed in ZFC, as a "feature not a bug" of ZFC to avoid contradictions such as Russell's paradox, or the Burali-Forti paradox specifically related to the class of ordinal numbers (while these issues are glossed over in naive set theory). Also, having a representative of each well-order in which the relation is gotten just from a basic term of the language of ZFC has numerous technical advantages.
Daniel Schepler
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