# 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.

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$\blacksquare$

## 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.

$\blacksquare$