- Von Neumann defines an ordinal $\alpha$ as a transitive set whose elements are well-ordered with respect to the membership relation $\in$.
- Meanwhile, in Naive set Theory, Halmos defines an ordinal $\alpha$ as a well-ordered set such that $\forall \zeta \in \alpha$, the initial segment determined by $\zeta$ is $\zeta$ itself. That is : $s_\zeta = \zeta$ with \begin{equation}s_\zeta := \{\beta \in \alpha :~~ \beta < \zeta\}.\end{equation}
Althought my intuition tells me that these definitions are equivalent, I'm not able to prove it in clean and clear way.
Here is what I got so far :
For $1. \Leftarrow 2.$, the transitivity of the set $\alpha$ follows from the condition $s_\zeta = \zeta$ since $s_\zeta \subseteq \alpha$ for all for all $\zeta \in \alpha$.
Also, from the condition $s_\zeta = \zeta$, we have that the predecessors of $\zeta$ in the well-ordered set $\alpha$ are exactly the element of $\zeta$.
All that is lacking to turn this into an actual proof of 1. is a bit of structure...
For $1. \Rightarrow 2.$, since $\alpha$ is well-ordered by membership we have $s_\zeta \supseteq \zeta$. I'm currently working on the reverse inclusion using transitivity but once again the whole thing lacks clarity.
Let $\alpha$ be a Neumann ordinal. By definition,$\alpha$ is well-ordered by $<\;:=\;\in$. Let $\zeta\in \alpha$. Then for $\beta\in\alpha$ we have $\beta<\zeta\iff \beta\in \zeta$. Hence $s_\zeta\subseteq \zeta$ and more precisely $s_\zeta=\zeta\cap\alpha$. But by transitivity, $\zeta\in\alpha$ implies $\zeta\subseteq \alpha$ so that indeed $s_\zeta=\zeta$. We conclude that $\alpha$ is a Halmos ordinal.