Infinite strictly decreasing sequence of ordinals

Question

I know that on a linearly ordered set $X$, $\leq$ is a well order if and only if $X$ contains no strictly decreasing sequence. However this conflicts in my head with the following problematic 'intuition' which I do not know where it fails.

Let $\alpha$ be a limit ordinal such that $\alpha\geq \omega_1$. All ordinals strictly less than $\alpha$ are a proper set and therefore we should not be able to find an infinite strictly decreasing sequence of ordinals. However since there uncountably many ordinals lesser than $\alpha$, it seems like we should be able to choose a strictly decreasing sequence in some complicated way. Aside from this proposition, can someone explain to me why a decreasing sequence must stablize?

I know there is probably some simple point which has not fully clicked for me.

Answer 1

Well. Choose $\alpha_0$, then maybe it was finitely many points below $\alpha$, but maybe it was infinitely many points below $\alpha$, i.e. there was a limit ordinal between $\alpha_0$ and $\alpha$.

Repeat the process. After finitely many steps you must have reached $\omega_1$, or went below it. Otherwise we had an infinite set of ordinals without a minimal element (that would be $\{\alpha_n\mid n<\omega\}$). Say that $\alpha_5$ was $\omega_1$. Now you have to choose a point below it, $\alpha_6$. But that's a countable ordinal, so you've discarded "most" ordinals already.

Rinse and repeat, and we hit $\omega$, say at $\alpha_{52}$, and then you have to choose $\alpha_{53}$ which is a natural number, and then there's only finitely many points left anyway. So your sequence had to be finite.

You might read my answer and object that I didn't really use the fact that it was $\omega_1$. And that's true. Because it's true for every ordinal. And it is one of those situations where "uncountable" is a red herring. Start with $\alpha_0=\omega$, there are infinitely many points below it, so principle dictates that we can find a decreasing sequence. But we can't, since after choosing just one we have only finitely many points to work with.

Answer 2

This is a direct consequence of the definition of a well-ordered set and the definition of an ordinal number.
A linearly ordered set $(X, \leq)$ is well-ordered if and only if every nonempty subset $S \subseteq X$ has a least element.
Contrapositively, if we could come up with some infinite decreasing sequence $$x_1 > x_2 > x_3 > ... $$ of elements of $X$, then this would correspond to a nonempty, infinite subset $$S := \{ x_1, x_2, x_3, ... \} \subseteq X$$ with no least element, so $X$ would not be well-ordered.

Ordinal numbers, meanwhile, are typically defined to be certain well-ordered sets. In Von Neumann's formulation, for instance, any given ordinal is defined to be the well-ordered set of all smaller ordinals:

A set S is an ordinal if and only if S is strictly well-ordered with respect to set membership and every element of S is also a subset of S.

So if I try to build my infinite descending sequence of ordinals starting from some infinite--even some uncountably infinite--ordinal number $\alpha$, I can't do it, because that sequence necessarily lives inside the well-ordered set $\alpha$, and as a subset of a well-ordered set, it has a least element.