During my course on elementary set theory, we defined $\omega_1$ to be the set of all countable ordinals. Now with Foundation we know that $\omega_1$ cannot be countable since that otherwise means $\omega_1\in \omega_1$, which just violates Axiom of Foundation. Now the question is can we prove $\omega_1$ is uncountable without using Foundation?
How should I proceed? This was a past paper question that did not worth so many marks and so I was wondering if there is a really slick way of showing this. I thought about using the criterion that a set is countable if and only if there exists a surjective function from $\omega$ to this set / if and only if there exists an injective function from this set to $\omega$ but I am stuck after some initial thoughts.
Many thanks in advance for any helps!
An ordinal is by definition strictly totally ordered by $\in$. So this means that $\alpha\not\in\alpha$ for any ordinal $\alpha$: if $\alpha\in\alpha$, then $x=\alpha$ is an element of $\alpha$ that satisfies $x\in x$, contradicting antisymmetry of the strict total order $\in$ on $\alpha$.
Now if $\omega_1$ were an element of $\omega_1$, then by definition $\omega_1$ would be a countable ordinal, and in particular it would be an ordinal. But an ordinal cannot be an element of itself, so this is a contradiction.
(Alternatively, if you already know that $\omega_1$ is an ordinal, you can just directly conclude that $\omega_1\not\in\omega_1$. But proving that $\omega_1$ is an ordinal takes a lot of work and is much harder than the argument by contradiction as above. That said, if you have managed to already prove that $\omega_1$ is a set at all then probably you have the machinery needed to quickly conclude that it is an ordinal.)