# Ordinals are Well-Ordered/Proof 2

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## Theorem

The ordinals are well-ordered.

## Proof

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.

$\blacksquare$

## Sources

- 1993: Keith Devlin:
*The Joy of Sets: Fundamentals of Contemporary Set Theory*(2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals