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 $\subseteq$:

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

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.

$\blacksquare$


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