# 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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