A paradox about cardinality of ALL and arithmetic hierarchies ― Did I just prove that ZFC is inconsistent?

Question

This problem arose when I tried to find the arithmetic hierarchy that $\mathsf{ALL}$, the class of all formal languages over a finite alphabet, corresponds to (like how $\mathsf{R} = \Delta^0_1$ and $\mathsf{RE} = \Sigma^0_1$).

Since the cardinality of $\mathsf{ALL}$ is $\beth_1$, I thought that, using a sufficiently large ordinal number as the subscript should cover all decision problems.

Formally, let $\Omega$ be the smallest ordinal number whose cardinality is $\beth_1$. Existence of such ordinal number is justified by the well-ordering principle (which is equivalent to the Axiom of Choice). By the halting problem, $\Sigma^0_o$ has infinitely many more elements than $\Delta^0_o$ for every non-limit ordinal number $o$. As a consequence, $\Delta^0_\Omega = \Sigma^0_\Omega = \Pi^0_\Omega$ has at least $\aleph_0 \times \beth_1 = \beth_1$ elements.

On the other hand, let $\Omega'$ be the smallest ordinal number whose cardinality is strictly bigger than $\beth_1$. By a similar argument as above, $\Delta^0_{\Omega'}$ has a cardinality strictly bigger than $\beth_1$. This contradicts against that $\Delta^0_{\Omega'}$ is still a subset of $\mathsf{ALL}$. What am I missing?

Answer 1

No, you haven't proved ZFC inconsistent. Short version: try to actually define your generalized arithmetical hierarchy and it will become clear how things break down!

The issue is that you've glossed over how you're making sense of $\Sigma^0_\alpha$ for $\alpha$ an arbitrary ordinal. While successor ordinals pose no problem, limit ordinals take us to the thorny issue of notations; see e.g. Kleene's $\mathcal{O}$. Things are relatively simple up until the first noncomputable ordinal $\omega_1^{CK}$ (which is of course still countable); this gives the hyperarithmetic hierarchy. With more work we can continue up through $\omega_1^L$ which may not be countable, the relevant term being mastercodes.

However, no trick at all will let us continue past $\omega_1$ while preserving the key properties of the arithmetical hierarchy (and so your appeal to the relativized halting problem breaks down here). The issue is that we want $\Sigma^0_{\alpha+1}$ to be $X'$, the Turing jump of some language $X$ which "corresponds to" the previous level of our hierarchy. But the only way such an $X$ can possibly exist is if the previous level of our hierarchy was countable! No matter what definition you use, $\Sigma^0_{\omega_1}$ will not be countable, and so either your hierarchy will stop there or you'll lose the connection with Turing jumps (in which case you won't have your non-collapse result).