# Ordinal equals its Initial Segment

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## 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 ProgressIn particular: One proof per definition of ordinalYou 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. |