# Motivation of the von Neumann definition of ordinals

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. Apr 16 '19 at 7:05
• "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. Apr 16 '19 at 9:17
• @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. Apr 16 '19 at 21:17
• 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 $(A,<)$. Apr 16 '19 at 22:49
• That is literally the definition of "smaller", as I said. Apr 17 '19 at 20:49

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...