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.
$\blacksquare$