# Ordinals are Well-Ordered

## Theorem

The ordinals are well-ordered.

### Corollary

Let $\Epsilon {\restriction_A}$ denote the epsilon restriction on $A$.

Then $A$ is strictly well-ordered by $\Epsilon {\restriction_A}$.

## Proof 1

Recall that the Ordinals are Totally Ordered.

Let $A$ be a non-empty set of ordinals.

Let $\alpha \in A$.

Let $B = \alpha^+ \cap A$, where $\alpha^+$ denotes the successor set of $\alpha$.

Recall that $\alpha^+$ is an ordinal.

Note that $\alpha \in B$, so $B$ is non-empty.

By Intersection is Subset, $B \subseteq \alpha^+$.

It follows that there exists a smallest element $\kappa$ of $B$.

We claim that $\kappa$ is the smallest element of $A$.

So let $\beta \in A$. We need to show that $\kappa = \beta$ or $\kappa \in \beta$.

By Ordinal Membership is Trichotomy, either $\beta \in \alpha^+$, $\alpha^+ = \beta$, or $\alpha^+ \in \beta$.

If $\beta \in \alpha^+$, then $\beta \in B$; it follows by the definition of $\kappa$ that $\kappa = \beta$ or $\kappa \in \beta$.

If $\alpha^+ = \beta$ or $\alpha^+ \in \beta$, then it follows by the transitivity of $\beta$ that $\alpha^+ \subseteq \beta$.

Since $\kappa \in \alpha^+$, it follows that $\kappa \in \beta$.

$\blacksquare$

## Proof 2

By Ordinals are Totally Ordered, the ordinals are totally ordered by $\subseteq$:

\(\ds X\) | \(\subsetneqq\) | \(\ds Y\) | ||||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds \exists a \in Y: \ \ \) | \(\ds X\) | \(=\) | \(\ds Y_a\) | where $Y_a$ denotes the initial segment of $Y$ determined by $a$ | |||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds X\) | \(=\) | \(\ds a\) | since $Y_a = a$ | ||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds X\) | \(\in\) | \(\ds Y\) |

Thus:

- the strict ordering $\subsetneqq$ on ordinals

and

- the strict ordering $\in$ on ordinals

are the same.

Aiming for a contradiction, suppose the ordinals were not well-ordered by $\subsetneqq$.

Then we could find a sequence $\sequence {X_n}_{n \mathop = 0}^\infty$ of ordinals such that:

- $X_0 \supsetneqq X_1 \supsetneqq X_2 \cdots$

So for all $n > 0$:

- $X_n \subsetneqq X_0$

so:

- $X_n \in X_0$

Thus $\sequence {X_{n + 1} }_{n \mathop = 0}^\infty$ is a decreasing sequence under $\subsetneqq$ of elements of $\sequence {X_n}$.

But since $X_0$ is an ordinal it is well-ordered by $\subsetneqq$.

From Condition for Well-Foundedness, this demonstrates a contradiction.

$\blacksquare$