Ordinal is Transitive

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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.

From Successor Set of Transitive Set is 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.

$\blacksquare$