# How large is the supremum of minimal $V$-heights of all first-order set theories formulated in a particular language of FOST?

Fix a language $$\mathcal{L}$$ of first-order set theory. For this question, we can assume that $$\mathcal{L}$$ is the language described in Chapter 1 of “An introduction to set theory” [William A. R. Weiss | October 2, 2008].

Assuming that the complexity of a formula is not restricted and every formula has a finite length, construct an infinite list $$Z_{\mathcal{L}}$$ of all syntactically valid statements, so that every such statement is assigned an unique natural number (i.e. enumerate all syntactically valid statements).

Consider an infinite binary sequence $$s$$. We can assume that any such sequence corresponds to a set of axioms of a particular set theory $$T$$: an $$i$$-th bit of $$s$$ is non-zero if and only if an $$i$$-th statement of $$Z_{\mathcal{L}}$$ is an axiom of $$T$$. Let $$t(s)$$ denote a theory encoded by $$s$$. (For example, there will be some $$s$$ such that $$t(s) = \text{ZFC}$$.)

Then $$f(s) = \alpha+1$$ if and only if there exists the smallest ordinal $$\alpha$$ such that $$V_{\alpha} \models t(s)$$; otherwise, $$f(s) = 0$$. For example, if $$t(s)=\text{ZFC},$$ then $$f(s)$$ is equal to the successor of the initial ordinal of the smallest worldly cardinal.

Note that the phrase “any binary sequence corresponds to a set of axioms of a particular set theory $$T$$” does not make any assumptions about $$T$$: the theory may be empty (if all bits of $$s$$ are zero) or inconsistent. That is, almost all sequences will not correspond to a consistent theory, but we are not interested in such sequences: for any such sequence $$s$$ we will have $$f(s) = 0$$.

The ordinal $$\beta_{\mathcal{L}}$$ is defined as follows: $$\beta_{\mathcal{L}} = \sup \{ f(s) : s \in {2^\omega }\}.$$

Here “$$s \in {2^\omega}$$” means that we take into account all infinite binary sequences (subsets of $$\omega$$).

Question: is $$\beta_{\mathcal{L}}$$ a well-defined ordinal? If no, why? If yes, how large is $$\beta_{\mathcal{L}}$$ in the hierarchy of large cardinals?

• This question could be shortened substantially: there's no need to bring $2^\omega$ into things, just define $f$ directly on the set of $\{\in\}$-theories. For that matter we don't even need $f$: say that an ordinal $\alpha$ is fresh iff $V_\alpha\not\equiv V_\gamma$ for any $\gamma<\alpha$, then you're asking for the supremum of the fresh ordinals (and I think this makes it easier to see that the ordinal in question is perfectly well-defined). Finally, note that your example is incorrect: "inaccessible" should be replaced by "worldly." Jan 26, 2022 at 5:37

Your ordinal $$\beta_\mathcal{L}$$ is perfectly well-defined: in my opinion it's more easily thought of as $$\sup\{\alpha: \forall \beta<\alpha(V_\beta\not\equiv V_\alpha)\},$$ and this definition should be clearly unproblematic (note that Tarski notwithstanding there is no problem in talking about truth relative to a set-sized structure like the $$V_\alpha$$s). Moreover, this definition avoids any reference to coding of theories by reals, which adds a lot of unnecessary length to the question.

As to how big $$\beta_\mathcal{L}$$ is, the key observation is that levels of the cumulative hierarchy are correct about "local" phenomena. For example, $$\beta_\mathcal{L}$$ is greater than the least measurable cardinal $$\mu$$ if the latter exists, since the measurability of $$\mu$$ is visible in $$V_{\mu+2}$$. To get past $$\beta_\mathcal{L}$$, you need to look at large cardinal properties which more significantly reach up the cumulative hierarchy - e.g. we trivially have that $$\beta_\mathcal{L}$$ is less than the least supercompact if the latter exists.

• It seems that I need an explanation for the claim that $\beta_{\mathcal{L}}$ is less than the least supercompact. Consider the theory T = ZFC + "there exists a supercompact cardinal", which corresponds to some infinite binary sequence (existence of a supercompact cardinal is $\Sigma_3$ in the Lévy hierarchy). What is the smallest ordinal $\alpha$ such that $V_{\alpha} \models \text{T}?$ Note that $\alpha$ must be much smaller than $\beta_{\mathcal{L}}.$ Aug 20, 2022 at 7:04
• More precisely, $\beta_{\mathcal{L}}$ is less than or equal to the least $\Sigma_2$-correct cardinal. Aug 20, 2022 at 9:43
• @lyricallywicked If there is no cardinal that is supercompact up to the next worldly cardinal, there is no $\alpha$ such that $V_\alpha \vDash T$, even if there is a cardinal that is supercompact in $V$, for that supercompact cardinal may be the greatest worldly cardinal. Aug 20, 2022 at 9:50
• @ArvidSamuelsson: thank you for the explanation. Yes, my statement "$\alpha$ must be much smaller than $\beta_{\mathcal{L}}$" should be "$\alpha$ (if it exists) must be much smaller than $\beta_{\mathcal{L}}$"... Aug 20, 2022 at 12:50
• @lyricallywicked I think that $\alpha$ (assuming it exists) has no definition snappier than "The smallest ordinal $\alpha$ such that $V_\alpha\models T$." If you really want a different type of definition, you should specify what you're looking for (and note that it may not exist). Aug 21, 2022 at 21:03