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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 $\in$?

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    $\begingroup$ 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. $\endgroup$ – Asaf Karagila Apr 16 '19 at 7:05
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    $\begingroup$ "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. $\endgroup$ – Mauro ALLEGRANZA Apr 16 '19 at 9:17
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    $\begingroup$ @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. $\endgroup$ – Asaf Karagila Apr 16 '19 at 21:17
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    $\begingroup$ 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 $(A,<)$. $\endgroup$ – Eric Wofsey Apr 16 '19 at 22:49
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    $\begingroup$ That is literally the definition of "smaller", as I said. $\endgroup$ – Eric Wofsey Apr 17 '19 at 20:49
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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

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