Is the $\omega_1$'th element of a well ordered set the first with uncountable predecessors?

Question

As far as I understand the topic, ordinal numbers are defined as sets, but their "purpose" is to generalize the concept of indices beyond infinity.

So, can it be said that

1. the element at index $\omega$ is the first with (countably) infinite predecessors

and likewise,

2. the element at index $\omega_1$ is the first with uncountable predecessors?

I find (1) somewhat intuitive, but I can't think of an example that illustrates (2).

Answer 1

Yes, that is correct. $\omega$ is the first ordinal which has infinitely many smaller ordinals; and $\omega_1$ is the first ordinal which has uncountably many smaller ordinals. Similarly $\omega_2$ is the first ordinal which has more than $\aleph_1$ smaller ordinals, and so on.

It seems to me that you're trying to visualize $\omega_1$. And that's not really possible. The reason is that however you might want to imagine $\omega_1$, it's much longer than that. And it's more than just that, it's also the fact that if you think about $\omega_1$ with any additional functions or relation symbols and you try to describe it using a first-order property of that structure, there will be some countable ordinal with the same properties under the same relations.

So it really means that $\omega_1$ is far far beyond our reach, when it comes to examples which illustrate it. It is something that is best to work with formally using the definition until you get a good intuition about it.