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 Infinite Sequence Property of Well-Founded Relation, 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