Motivation of the von Neumann definition of ordinals

Question

The von Neumann ordinals are defined in such a way that each ordinal is exactly the set of all smaller ordinals. I am wondering about the origin/motivation for this definition of ordinals (that is, how one got to this definition from the goal of choosing a representative for each equivalence class of well-orderings). I read that the motivation was the fact that each well-ordering is isomorphic to the set of all smaller well-orderings. But when I looked for a proof of this fact, I saw that this proof contained ordinals as a tool to prove it. Now this seemed circular to me (not in the logical sense, but in the historical sense).

Is there also an ordinal-free proof of the fact that each well-ordering is isomorphic to the set of all smaller well-orderings?

Also, I wonder: The definition "An ordinal is the set of all smaller ordinals" would be somehow circular. But would it work rigorously? (Maybe it's some kind of recursive/inductive definition -- these things also seem "circular" but are ok -- also, for example, hereditary sets are defined as sets whose elements are hereditary sets, and this definition also works rigorously.)

Furthermore: How did one get from the slogan "an ordinal is the set of all smaller ordinals" to the definition that an ordinal is a transitive set that is a well-ordering under $\in$?

Fix a well-ordering $(A,<)$ and you look at the obvious map $a\mapsto (A_{<a},<)$. Well, it's very easy to prove it's an isomorphism and we haven't used the ordinals. — Asaf Karagila, Apr 16 '19 at 07:05
"The definition "An ordinal is the set of all smaller ordinals" would be somehow circular. " This is not the def: "A set S is an ordinal if and only if S is strictly well-ordered with respect to set membership and every element of S is also a subset of S" where the notion of "well-order" is defined previously and thus independently from that of ordinal. — Mauro ALLEGRANZA, Apr 16 '19 at 09:17
The context is the Set-theoretic definition of natural numbers : originals attempts of Frege and Russell were not suitable for axiomatic set theory and thus Zermelo provided a new one. von Neumann's def is an improvement of Zermelo's and has an "intuitive" aspect: the numebr $n$ has exactly $n$ elements. — Mauro ALLEGRANZA, Apr 16 '19 at 09:22
By the way, when you see seemingly circular definitions, they are either wrong, or there is a recursive definition hiding underneath. — Asaf Karagila, Apr 16 '19 at 09:22
@Mauro: Also on this site, and probably a few other questions and answers over the years. — Asaf Karagila, Apr 16 '19 at 09:27
@AsafKaragila Please give a link to your previous answer to these questions, otherwise it doesn't help me at all that you feel like you answered this before. Also, it would be quite a coincidence if someone asked exactly the three questions I asked here. — , Apr 16 '19 at 20:57
@AsafKaragila Okay, the map $a\mapsto (A_{<a}, <)$ is an isomorphism between the fixed well-ordering and the set of its principal ideals. But I asked about an isomorphism between the well-ordering and the set of all smaller well-orderings. (Technically, this is not a set, but one could consider the set of all well-orderings with domain $\subseteq A$ quotient by $\sim$.) Or are the principal ideals corresponding exactly the the smaller well-orderings? If yes, then at least for me this is not something that is easy to see. — , Apr 16 '19 at 21:02
@MauroALLEGRANZA Of course I know that "An ordinal is the set of all smaller ordinals" is not the common definition. But I ask if this would formally work as a definition. — , Apr 16 '19 at 21:03
@AsafKaragila Would "An ordinal is the set of all smaller ordinals" formally work as a definition? Is there some kind of recursion underneath it that could justify this as a definition? — , Apr 16 '19 at 21:04
@user7280899: What is the "set of all well-orderings"? You agree it's not a set. So in what sense is there an isomorphism between a set and a proper class? — Asaf Karagila, Apr 16 '19 at 21:10
Yes. There is a definition that works. For example, an ordinal is a transitive set which is well-ordered by $\in$. Which you can then prove that by recursion the class of ordinals is the class such $\varnothing$ is an ordinal, if $x$ is an ordinal then $x\cup{x}$ is an ordinal, and if $x$ is a set of ordinals, then $\bigcup x$ is an ordinal. — Asaf Karagila, Apr 16 '19 at 21:12
@AsafKaragila As I wrote, "set of all well-orderings smaller than $A$" is of course just abuse of language and means the set of all well-orderings with domain $\subseteq A$ quotient by $\cong$ (without the equivalence class of $A$). — , Apr 16 '19 at 21:16
@user7280899: If it's abuse of language to the set of all initial segments of $A$, I don't see how my above comment fails to satisfy you. Please decide which version of your question you want answered. — Asaf Karagila, Apr 16 '19 at 21:17
@AsafKaragila I asked if "An ordinal is the set of all smaller ordinals" would formally work as a definition and you answer this by saying "Yes. There is a definition that works." I asked whether this specific sentence would formally work as a definition and not whether there is a definition that works. Do you read what a write? — , Apr 16 '19 at 21:18
@AsafKaragila I want all three parts of the question to be answered. It is common that a math.stackexchange question consists of various aspects. — , Apr 16 '19 at 21:20
@AsafKaragila So why is the set of principal ideals of $A$ isomorphic to the set of all well-orderings with domain $\subseteq A$ quotient by $\cong$ (and without the equivalence class of $A$)? — , Apr 16 '19 at 21:22
Because every well-ordering is isomorphic to some initial segment. — Asaf Karagila, Apr 16 '19 at 21:26
@AsafKaragila To some initial segment of what? Can you please give a proof of your claim? — , Apr 16 '19 at 21:57
When people say something like "each well-ordering is isomorphic to the set of all smaller well-orderings", they usually mean exactly what Asaf referred to in his second comment: a "smaller" well-ordering than $(A,<)$ is defined to be a well-ordering that is (isomorphic to) a proper initial segment of $(A,<)$. — Eric Wofsey, Apr 16 '19 at 22:49
@EricWofsey Thanks! I see that for each $a\in A$, $A_{<a}$ is a well-ordering. But how can one prove that each smaller well-ordering than $A$ is (isomorphic to a principal ideal) of the form $A_{<a}$ for some $a$? — , Apr 17 '19 at 20:07
@EricWofsey Do you think "An ordinal is the set of all smaller ordinals" would formally work as a definition? And is there some recursion principle with which one could define "ordinal" in some "seemingly circular" way? — , Apr 17 '19 at 21:02
@EricWofsey Why not define ordinals by recursing on well orders, why isn’t that definition more intuitive? — Vivaan Daga, Mar 11 '23 at 13:57
@EricWofsey Or is there a sort of historical inertia against it? — Vivaan Daga, Mar 12 '23 at 05:01

Alexander Kuleshov · Answer 1 · 2019-09-19T22:00:11.190

Blockquote

I think "an ordinal is the set of all smaller ordinals" can be formulated as follows: consider an arbitrary class $\mathrm{No}$ with a property $(1)$: $\mathrm{No} = \{s: s=\{x\in \mathrm{No}: x\subset s\}\}$, where $\subset$ denotes a strict embedding. If $y\in x\in s$ then $y\subset x\subset s$ which means $y\in s$ by definition, thus each $s \in No$ is transitive and $\in$ is a partial order on $\mathrm{No}$. Now let's consider an arbitrary $A\subseteq \mathrm{No}$ and $m:=\bigcap A$. If $m\subset a$ for all $a\in A$ then $m \in \mathrm{No}$ $(*)$, so $m \in a$ and $m \in \bigcap A = m \Rightarrow m \subset m$ - contradiction. So $m\in A$ and $m \in a$ for all $a \in A: a \ne m$, which means that $\mathrm{No}$ is totally and well-ordered by $\in$. So we came to a classical von Neumann definition of ordinals.

Question: can we replace property $(1)$ with property $(2)$: $\forall s\in \mathrm{No} \Rightarrow s=\{x\in \mathrm{No}: x\subset s\}$? I feel that the answer is yes, but I cannot prove $(*)$ by now...

Edit: the question has been positively answered here https://mathoverflow.net/questions/341868/compact-definition-of-ordinals#comment854834_341900

And I guess you suggested some reverse reasoning here: Definition of Ordinals in Set Theory in Layman Terms

Motivation of the von Neumann definition of ordinals

1 Answers1