Ordinals are Well-Ordered/Proof 2

Theorem

$\ds X$

$\subsetneqq$

$\ds Y$

$\ds \leadstoandfrom \ \ $

$\ds \exists a \in Y: \, $

$\ds X$

$=$

$\ds Y_a$

$\ds \leadstoandfrom \ \ $

$\ds X$

$=$

$\ds a$

since $Y_a = a$

$\ds \leadstoandfrom \ \ $

$\ds X$

$\in$

$\ds Y$

Thus:

and

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 $\subsetneqq$.

$\blacksquare$