Backward Path of Well-Ordering is Finite

Theorem

Let $A$ be a class under a well-ordering $\preccurlyeq$.

Let $S$ be a backward path of $\preccurlyeq$ in $A$.

Then $S$ is finite.

Proof

First suppose that $S$ has no terms.

Then $S$ is finite by default.

Hence let it be assumed that $S$ has at least $1$ term.

Let $T = \set {x: x \in \sequence {x_n} }$ be the range of $S$.

Then by definition $T$ is a subclass of $A$.

Aiming for a contradiction, suppose $S$ is infinite.

By definition of well-ordering, $T$ has a smallest element, which can be referred to as $y$.

Hence:

$\exists n \in \N: y = x_n$

But then by definition of backward path:

$x_{n + 1} \prec x_n$

and so $y$ is not the smallest element of $T$ after all.

Hence $T$ can have no such smallest element.

From this contradiction we conclude that $S$ cannot be an infinite sequence.

Hence the result.

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Exercise $1.4$