text: # Ordinal is Transitive
  
  ## Theorem
  
  Every ordinal is a transitive set.
  
  ## Proof 1
  
  Let $\alpha$ be an ordinal by Definition 1:
  
  $\alpha$ is an ordinal if and only if it fulfils the following conditions:
  
   $(1)$ $:$ $\alpha$ is a transitive set $(2)$ $:$ $\Epsilon {\restriction_\alpha}$ strictly well-orders $\alpha$
  
  where $\Epsilon {\restriction_\alpha}$ is the restriction of the epsilon relation to $\alpha$.
  
  Thus $\alpha$ is a priori transitive.
  
  $\blacksquare$
  
  ## Proof 2
  
  Let $\alpha$ be an ordinal by Definition 2:
  
  $\alpha$ is an ordinal if and only if it fulfils the following conditions:
  
   $(1)$ $:$ $\alpha$ is a transitive set $(2)$ $:$ the epsilon relation is connected on $\alpha$: $\ds \forall x, y \in \alpha: x \ne y \implies x \in y \lor y \in x$ $(3)$ $:$ $\alpha$ is well-founded.
  
  Thus $\alpha$ is a priori transitive.
  
  $\blacksquare$
  
  ## Proof 3
  
  Let $\alpha$ be an ordinal by Definition 3.
  
  An ordinal is a strictly well-ordered set $\struct {\alpha, \prec}$ such that:
  
  $\forall \beta \in \alpha: \alpha_\beta = \beta$
  
  where $\alpha_\beta$ is the initial segment of $\alpha$ determined by $\beta$:
  
  $\alpha_\beta = \set {x \in \alpha: x \prec \beta}$
  
  That is, $\alpha$ is a transitive set.
  
  $\blacksquare$
  
  ## Proof 4
  
  Let $\alpha$ be an ordinal by Definition 4.
  
  $\alpha$ is an ordinal if and only if:
  
  $\alpha$ is an element of every superinductive class.
  
  The proof proceeds by the Principle of Superinduction.
  
  From Empty Class is Transitive we start with the fact that $0$ is transitive.
  
  $\Box$
  
  Let $x$ be transitive.
  
  $x^+$ is transitive.
  
  $\Box$
  
  We have that Class is Transitive iff Union is Subclass.
  
  Hence the union of a chain of transitive sets is transitive.
  
  $\Box$
  
  Hence the result by the Principle of Superinduction.
  
  $\blacksquare$
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