Let $L$ be the language of set theory, and for each countable ordinal $\kappa$, define $L_\kappa$ as follows: $L_0 = L$, and $L_{\alpha+1}$ is obtained by adding countably many new class constants $c_\phi$ to the signature of $L_\alpha$ to stand for each formula $\phi(x)$ of one free variable. Define $L_\alpha$ for $\alpha$ a limit ordinal to be the union of $L_\beta$ for $\beta < \alpha$. Given some set theory such as ZF, it is possible to add axioms for the new constants: let $T_0 = ZF$ and $T_{\alpha+1}$ be the theory got from $T_\alpha$ by adding $$x \in c_\phi \iff \phi(x)$$ as an axiom where $\phi(x)$ ranges over the countably many formulas in $L_\alpha$. Let $T_\alpha$ for $\alpha$ a limit ordinal be the union of $T_\beta$ for $\beta<\alpha$. This is ordinary expansion by definitions.

So, for each countable ordinal $\kappa$, $T_\kappa$ is countably axiomatizable. Define $L_\Omega$ and $T_\Omega$ as the union of $L_\kappa$ and $T_\kappa$ as $\kappa$ ranges over countable ordinals.

Is $T_\Omega$ countably axiomatizable?

What relation does MK have to $T_\Omega$, if any?

New question: is every formula of $L_\Omega$ equivalent to a formula of $L_1$ modulo $T_1$? (seems likely) please see: Is every formula of LΩ equivalent to a formula of L1 modulo T1?

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    $\begingroup$ The way you defined it, $T_1$ (and a fortiori all subsequent theories) is inconsistent. You need to either start with a theory of classes rather than ZF, or use unary predicates instead of constants. $\endgroup$ Dec 13, 2016 at 15:55
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    $\begingroup$ Anyway, it’s impossible to countably axiomatize a (consistent) theory in an uncountable language, unless all but countably many of the symbols are left completely arbitrary by the axioms. $\endgroup$ Dec 13, 2016 at 15:59
  • $\begingroup$ Oh, sorry. I meant to say this: add a unary predicate $M x$ for "x is a set" and for each axiom in ZF, relativize it with respect to $M x$, so that the ZF axioms only apply to sets (except for the axiom of extension) and not the new constants. As in MK, $x$ is a set if and only if there exists $y$ such that $x\in y$. $\endgroup$ Dec 13, 2016 at 16:02
  • $\begingroup$ It seems to me that the theory basically amounts to GB+AC, since the naming process is simply ensuring instances of class comprehension. $\endgroup$ Dec 13, 2016 at 19:15
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    $\begingroup$ @JoelDavidHamkins I don't even think it's doing that - note that it only applies to parameter-free $\varphi$. $\endgroup$ Dec 13, 2016 at 21:02

1 Answer 1


As Emil points out, $T_\Omega$ is not countably axiomatizable for trivial reasons: it has an uncountable language, and says non-trivial things about each symbol in that language.

That said, it is equivalent to a countably axiomatizable theory, in a precise sense: namely, you don't really get anything new after $T_\omega$. To see why, think about $T_{\omega+1}$: every formula in $L_\omega$ is also in $L_n$ for some finite $n$, so you've already added a constant symbol naming the relevant class; this means each constant symbol you add at stage $\omega+1$ is - provably in $T_{\omega+1}$ - equivalent to an already-added constant symbol. (Incidentally, this makes it easy to show that MK is much stronger than $T_\omega$, and hence much stronger "morally" than $T_\Omega$.)

In fact, unless I'm missing something your hierarchy collapses right away - $T_1$ is essentially the same as $T_0$. This is because, in forming $T_1$, we can replace every "new" sentence with constant symbols $c_\varphi$ by an "old" sentence where "$-\in c_\varphi$" is replaced with "$\varphi(-)$".

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    $\begingroup$ Unlike the other levels, the construction of $T_1$ from $T_0$ is not conservative. The constants in $T_1$ are not just an expansion by definitions, they actually imply the existence of the corresponding “classes” which is not provable in $T_0$ alone. So, the last part is now incorrect. $\endgroup$ Dec 14, 2016 at 6:41

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