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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!

Arturo Magidin
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UnsinkableSam
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1 Answers1

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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.)

Eric Wofsey
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  • Ah I see, thank you so much Eric! That makes sense! However, my course only defined the ordinal to be well ordered by $\in, =$ and so I don't quite have the notion of strict order. Wondering how could I adapt your argument to suit this definition? – UnsinkableSam Jun 05 '21 at 17:58
  • Oof, that's an awful definition of ordinals. It just doesn't work in the absence of Foundation at all. For instance, in a model where there is a set $x$ such that $x={x}$, then $x$ will be an ordinal by that definition. – Eric Wofsey Jun 05 '21 at 18:00
  • In particular, using that definition in the absence of Foundation, you can't even prove that $\omega_1$ is a set. – Eric Wofsey Jun 05 '21 at 18:01