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?

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    $\begingroup$ 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." $\endgroup$ Jan 26, 2022 at 5:37

1 Answer 1


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.

  • $\begingroup$ 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 $\beta_{\mathcal{L}}.$ $\endgroup$ Aug 20, 2022 at 7:04
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    $\begingroup$ More precisely, $\beta_{\mathcal{L}}$ is less than or equal to the least $\Sigma_2$-correct cardinal. $\endgroup$ Aug 20, 2022 at 9:43
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    $\begingroup$ @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. $\endgroup$ Aug 20, 2022 at 9:50
  • $\begingroup$ @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 $\beta_{\mathcal{L}}$"... $\endgroup$ Aug 20, 2022 at 12:50
  • $\begingroup$ @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). $\endgroup$ Aug 21, 2022 at 21:03

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