Other ways of proving that the set of all countable ordinals is uncountable

Question

I know that the standard way of proving that the set of all countable ordinals is uncountable is by stating that if the set is countable, then it incurs Burali-Forti paradox.

Is there other ways of proving this?

Answer 1

The set of all countable ordinals is the supremum of all countable ordinals, which is simply their unions. If $\omega_1$ were countable, $\omega_1+1$ would be countable too and hence $\omega_1+1<\omega_1$. Hence $\omega_1\in\omega_1+1\in\omega_1$, which shows that the set $\{\omega_1,\omega_1+1\}$ has no $\in$-minimum, which is impossible since the ordinals are well-ordered by $\in$.

Answer 2

I think that all proofs will be similar in some way to the Burali-Forti paradox. Here is a proof that is slightly different than other proofs on this site, so maybe someone will find it useful. Note that the Axiom of Regularity is not used anywhere.

Definition: An ordinal is a transitive set well-ordered by $\in$.

Fact: The class of ordinals is transitive and well-ordered by $\in$.

Definition: $\omega_1$ is the class of countable ordinals.

Fact: $\omega_1$ is a set.

To see that $\omega_1$ is an ordinal using the above facts, it remains to observe that every element of a countable ordinal is a subset of that ordinal and is therefore countable also.

Now suppose toward a contradiction that $\omega_1$ is countable. Then $\omega_1 \in \omega_1$ by definition. This contradicts the fact that the class of ordinals is well-ordered by $\in$.