Ordinal equals its Initial Segment

Theorem

Let $\On$ denote the class of all ordinals.

Let $<$ denote the (strict) usual ordering of $\On$.

Let $\alpha$ be an ordinal.

Then $\alpha$ is equal to its own initial segment:

$\alpha = \set {\beta \in \On: \beta < \alpha}$

Proof 1

Ordinal equals its Initial Segment/Proof 1

Proof 2

Ordinal equals its Initial Segment/Proof 2

Proof 3

Ordinal equals its Initial Segment/Proof 3

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.

From Strict Ordering of Ordinals is Equivalent to Membership Relation:

$\forall \alpha, \beta \in \On: \beta < \alpha \iff \alpha \in \beta$

Hence the statement of the result is equivalent to:

$\alpha = \set {\beta \in \On: \beta \in \alpha}$

which is trivially true by definition of a set.

$\blacksquare$

Work In Progress In particular: One proof per definition of ordinal You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by completing it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{WIP}} from the code.