What's the infimum and supremum of countably infinite ordinals?

Question

I'd say the infimum is $\omega$ and the supremum is $\omega_1$, but I am not too sure - especially about the supremum.

Answer 1

Dave Renfro's comments have fully answered your question. However, I think the following will still be helpful to you - let me go back to the definition of $\omega_\alpha$ in the first place.

Actually, there are multiple equivalent definitions; I'm just picking one of them here. In some respects it's not the most efficient one, but I do think it's the most intuitive one, especially for this type of question.

We're most familiar with the cases when $\alpha$ is $0$ or $1$, but the notation "$\omega_\alpha$" makes sense for any ordinal $\alpha$. The ordinal $\omega_\alpha$ is defined recursively as follows: for $\alpha$ an ordinal, $\omega_\alpha$ is the least infinite ordinal which does not inject into any $\omega_\beta$ for $\beta<\alpha$. It's a good exercise to show that this is in fact well-defined, that is, that $\omega_\alpha$ exists for each ordinal $\alpha$; the proof is via transfinite induction.

Now we can say:

• $\omega_0$ is the least infinite ordinal which is greater than $\omega_\alpha$ for each $\alpha<0$. Since there aren't any $\alpha<0$, this means $\omega_0$ is simply the least infinite ordinal - namely, $\omega_0=\omega$.

• Meanwhile, since the only ordinal $<1$ is $0$, this tells us that $\omega_1$ is the least infinite ordinal which doesn't inject into $\omega_0$ - that is, $\omega_1$ is the least uncountable ordinal.

The next key fact connects minima at one level with suprema at the previous level:

• For all $\alpha$, $\omega_\alpha$ is a limit ordinal. Put another way: if $\mu$ is an infinite ordinal, then there is a bijection between $\mu$ and $\mu+1$ (think about Hilbert's hotel), so $\mu+1$ can never be one of the $\omega_\alpha$s.

So, for example, this tells us that $\omega_1$ is both the least uncountable ordinal (that's how it's defined) and the supremum of the countable ordinals (everything less than it is a countable ordinal, so the only way it could fail to be their supremum is if it were a successor ordinal). This also tells us the following analogue to your question:

The least ordinal of cardinality $\aleph_1$ (= in bijection with $\omega_1$) is $\omega_1$, and the supremum of the ordinals of cardinality $\aleph_1$ is $\omega_2$.

Note that in your comment, you asked a different question - not about "cardinality $\aleph_1$," but about "uncountable" ordinals. The problem there is that the uncountable ordinals don't form a set. More generally, the class of ordinals of cardinality $>\kappa$ is never a set, for any $\kappa$; a set of ordinals has to be bounded in cardinality. The point is that "uncountable" is not a "higher-order analogue" of "countable" - the right analogue instead is something like "cardinality $\aleph_{17}$" (or perhaps "cardinality $\le\aleph_{17}$," depending on context).