How to prove that the set of all countable ordinals, $\omega_1$, is uncountable / has the cardinality $\aleph_1$?

Question

The only reasoning I've seen given for this is that it's uncountable because it can't include itself an element. I'm a little unconvinced and was looking for a more proper formal proof demonstrating the equality: $\omega_1 = \aleph_1$.

Answer 1

Following Henning's suggestion, posting my comment as an answer.

$\omega_1$ is usually defined to be the the least uncountable ordinal number. So it is uncountable by definition. Thus every ordinal in $\omega_1$ is countable. Moreover any countable ordinal $\alpha$ cannot be larger than or equal to $\omega_1$ and so $\alpha \in \omega_1$. Thus $\omega_1$ is the set of countable ordinals.

Answer 2

The reasoning is correct: The set $\omega_1$ of all countable ordinals (here always meant as: including finite ones) is an ordinal, hence we have $\omega_1\notin\omega_1$ (even in set theories that allow Quine atoms), hence it is not a countable ordinal. Since it is definitely an ordinal, it must be an uncountable ordinal.

EDIT: removed bad reason for $\le$ argument

Answer 3

Why is this not a reasonable reason?

Ordinals are sets which are transitive and well-ordered by $\in$. $\omega$ is defined to be the first infinite cardinal, $\omega_1$ is defined to be the first ordinal which is larger than $\omega$ not in bijection with any of its members.

The definition of $\omega_1$, then, implies that $(1)$ There is no bijection between $\omega_1$ and $\omega$, that is to say that $\omega_1$ is uncountable; and $(2)$ that every ordinal below $\omega_1$ has to be countable. Since $\in$ is well-founded we have that $\omega_1\notin\omega_1$.