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Proof

(related to Proposition: Equivalent Notions of Ordinals)

For a better readability, we will write in the following $\in$ instead of $\in_X$ but mean the contained relation $\in_X$ defined on $X.$

$(1) \Rightarrow (2)$

• Let $X$ be an ordinal.
• By definition, $(X,\in)$ is transitive and strictly and totally and well-ordered by $\in_X.$
• Let $u \in v \in w\in X.$
• Because $X$ is transitive, $u\in v\in X.$
• Because $X$ is transitive, $u\in X.$
• For the same reason, $u\in w.$
• Therefore, every element $w\in X$ is itself a transitive set.

$(2) \Rightarrow (1)$

• Let $X$ be a transitive set and all of its elements $w\in X$ be also transitive sets.
• Since $w$ is transitive, then from $u \in v$ and $v\in w$ it follows that $u\in w.$
• This means, by definition, that $(X,\in)$ is a strict total order, since $\in$ is connex.
• It remains to be shown that $\in$ is a well-order.
  • The axiom of foundation postulates that $X$ contains an element that is disjoint from $X.$
  • This means that there is an element $z\in X$ and for no other element $x\in X$ we have $z\in x.$
  • This means that the contained relation $\in$ defined on a set $X$ is well-founded.
  • Therefore, $X$ contains a minimal element with respect to the contained relation $\in.$
• Altogether, we have shown that $(X,\in)$ is transitive, strictly and totally and well-ordered by $\in.$
• Therefore, $X$ is an ordinal number.

$(2) \Rightarrow (3)$

$"\Rightarrow"$:

• Let $w\in X.$
• From $(2)$ it follows that $w\in X$ is itself a transitive set. It remains to be shown that $w\subset X$ (i.d. that $w$ is a proper subset of $X.$)
• If $v\in W$ then $v\in X,$ therefore $w\subseteq X.$
• But $w\neq X$ since otherwise $X\in X,$ in contradiction to the axiom of foundation.

$"\Leftarrow"$:

• Let $w\subset X$ be itself a transitive set.
• Since $w$ is a proper subset of $X,$ the set difference $X\setminus w$ is not empty.
• Since $(2)$ and $(1)$ are equivalent, $\in$ is a strict order and a well-order.
• Therefore, $X\setminus w$ contains a minimal element $z\in X\setminus w.$
• Thus, $z\in X$ and $z\not\in w$ and there is no element $x\in X\setminus w$ with $x\in z.$ We will show that $z=w,$ which means $w\in X.$
  • We have $z\subseteq w.$ * If $y\in z,$ then since $y\in z\in X$ we have $y\in X,$ because $X$ is transitive. * But $z$ is minimal in $X\setminus w$, i.e. there is no other element $x\in X\setminus w$ with $x\in z.$
    * Since $y\in z$ and $y\in X$ and there is no other element $x\in X\setminus w$ with $x\in z,$ we have $y\in w.$
  • We have $w\subseteq z.$ * Let $y\in w.$ * We have $z\not\in y,$ otherwise we would have $z\in w$ because $w$ is transitive, in contradiction to the hypothesis $z\not\in w.$ * Therefore, $y\neq z$ and $z\not\in y$ which means $y\in z,$ since $\in$ is a strict total order.
• Altogether, we have shown, given the equivalence $(1)\Leftrightarrow (2),$ that from $w$ being transitive and $w\subset X$ it follows that $w\in X.$

$(3) \Rightarrow (2)$

• Let $w\in X$ if and only if $w\subset X$ and $w$ is transitive for all elements $w\in X.$
• Therefore, all elements $w\in X$ are transitive sets.
  ∎

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References

Bibliography

1. Hoffmann, Dirk W.: "Grenzen der Mathematik - Eine Reise durch die Kerngebiete der mathematischen Logik", Spektrum Akademischer Verlag, 2011
2. Hoffmann, D.: "Forcing, Eine Einführung in die Mathematik der Unabhängigkeitsbeweise", Hoffmann, D., 2018