A common action in set theory is making a large cardinal axiom "recursive", i.e. turning it from a large uncountable cardinal to a large countable ordinal. For example:

  • Recursively regular = $\alpha$ is admissible, i.e. $L_\alpha \vDash \text{KP}$.
  • Recursively inaccessible = $\alpha$ is admissible and the admissible ordinals are unbounded in $\alpha$.
  • Recursively Mahlo = for every function $f: \alpha \to \alpha$ definable in $L_\alpha$ by a $\Delta_1$ formula, there is an admissible $\beta < \alpha$ closed under $f$.
  • Recursively weakly compact = $\Pi_3$-reflecting, i.e. for any $\Pi_3$ formula $\phi(x)$ and any $a \in L_\alpha$, $(L_\alpha, \in) \vDash \phi(a)$ implies $\exists \beta (\beta \in \alpha \land a \in L_\beta \land (L_\beta, \in) \vDash \phi(a))$.
  • And so on...

I was wondering how we could define a "recursively greatly Mahlo" ordinal. The original greatly Mahlo cardinals are defined like so:

A cardinal $\kappa$ is greatly Mahlo iff it is inaccessible and there is a normal $\kappa$-complete filter on $\mathcal{P}(\kappa)$ which is closed under the function $X \mapsto \{\alpha \in S: \text{cof}(\alpha) \geq \omega_1 \land X \cap \alpha \text{ is stationary in } \alpha\}$.

The resulting large countable ordinal should likely be in between recursively Mahlo and $\Pi_3$-reflecting, as the original was between Mahlo and weakly compact.

However, some of my friends say that recursively greatly Mahlo is $\Pi_3$-reflecting. I don't believe this, as Stegert's PhD says that the recursive analogue of $\Pi^1_n$-indescribability is $\Pi_{n+2}$-reflection, but they may be correct after all and that recursive weak compactness is after all stronger, maybe on the level of stable or nonprojectible ordinals.


1 Answer 1


Let me preface this by stating the obvious in that there is no injection from a given uncountable cardinal to any countable ordinal, thus any system of describing "analogous properties" of large countable ordinals relative to an uncountable set of ordinals will be flawed in infinitely many ways. The connection between large cardinal axioms and large unrecursive countable ordinal axioms is based on intuition, not that there is anything wrong with that. The bottom line is, "recursively greatly Mahlo" ordinals can all be stable, or nonprojectible or $\Pi_3$-reflecting if you define them as such, but they absolutely don't have to be.

I suggest reading "Generalizing the Mahlo hierarchy, with applications to the Mitchell models" by Baldwin, where he defines a very structured hierarchy of Mahloness in analogy to the Mitchel rank for measurability of cardinals. I myself have not read more than a few pages but we only need to use the outline from the paper, so the descriptions I will give here are very oversimplified and do not constitute definitions, so keep that in mind. $m(\kappa)$ is a rank of Mahloness that essentially tells the that a filter on the stationary subsets of $\kappa$, if applied less than $m(\kappa)$ times to an equivalence class on $\kappa$ based on stationarity, will result in a set that is neither 0 nor the set of all nonstationary subsets of $\kappa$. This ranking does not provide much new insight for the structure of Mahloness as $m(\kappa)\leq\kappa^+$ is always the case by definition and for every greatly Mahlo $\kappa$ we already have $m(\kappa)=\kappa^+$ and vice versa, thus this notation is useless for trying to understand cardinals that go beyond greatly Mahlos in stationary properties that are themselves not on the level of the hierarchy of weakly and supercompacts. Further down the paper in section 3, functions $\bar m(\kappa), m^*(\kappa)$ are defined with more complex definitions that provide further structure to the different types of Mahlo cardinals and the different levels of iterated stationarity. According to Baldwin, for these to work as desired, we must assume a particular type of failure of GCH in $2^\kappa=2^{\kappa^+}$ at least for greatly Mahlo $\kappa$, that is, $|\mathcal P(\kappa)|$, which is the size of the domain and codomain of $m(), \bar m(), m^*()$, must be greater or equal to a limit of regular cardinals above $\kappa$. There is no analogy of this that can be translated down to the constructible universe of countable sets $L_{\omega_1}$, which is where we want the least "recursively greatly Mahlo" to fall.

Let $M[\beta](0)$ be the first recursively $\beta$-Mahlo ordinal and let $M[\beta](\lambda+\alpha)$ be the $\alpha$th recursively $\beta$-Mahlo ordinal above $M[\beta](\lambda)$ whenever $\alpha < M[\beta+1](0)$ and let $\beta$-Mahlo ordinals be exactly those ordinals $\kappa$ such that $L_\kappa\models$KP+"$\forall\gamma<\beta:$ the $\gamma$-Mahlo ordinals form a stationary subset of the ordinals".

We could define a collapse of $M[\beta+1](\alpha)$ inside $M[\beta](\text{___})$ to form an evaluation for the hierarchy of $\beta$-Mahlos, limits of such, inaccessible limits of such, hyper-inaccessible limits of such, hyper-hyper-inaccessible limits of such and so on, and keep going from there. If we use, say, the least $\Pi_3$-reflecting ordinal $K$ to collapse inside of $M[\text{___}](0)$ such that $M[K+\gamma](\beta)$ is the $(1+\beta)$th ordinal $\alpha$ that is $\alpha$-Mahlo and for all $\delta_0\in\gamma\land\delta_1\in\alpha:$ the ordinals of the form $M[K+\delta_0](\delta_1)$ form a stationary subset of $\alpha$ with respect to clubs definable in $L_{\omega_1}$. We can have an ordinal representation system $M[K+K\alpha+\beta](\gamma)$ being the $(1+\gamma)$th $\beta$-hyper$^\alpha$-Mahlo ordinal and so on. Then finally, we can define the recursively greatly Mahlo ordinals as exactly those that are of the form $M[\omega^{\rm CK}_{K+1}](\alpha)$ for some $\alpha$ as $\omega^{\rm CK}_{K+1}$ is admissible and above $K$ and thus not reacheable from below by some recursive well-ordering on $K+1$ and thus the least recursively greatly Mahlo will be greater than the $M$-collapse of any ordinal in $[K,K^+)$ or even simply in $K^+$.

If $\alpha^+$ is the least regular above $\alpha$, that means that there is no surjection from $\alpha$ to $\alpha^+$. Analogously, if $\alpha^+$ is the least admissible above $\alpha$ then that means there is no surjection from $\alpha$ to $\alpha^+$ that consitutes a recursive or hyperarithmetical reordering of $\alpha$. This is not a definition but it suffices for making the analogy.

I believe this way of defining the recursively greatly Mahlos fits the bill for placing them in the hierarchy of recursively Mahlos in such a way that they always have the recursively $\beta$-hyper$^\alpha$-Mahlo ordinals and all kinds of limits of such be unbounded in them, yet without the greatly Mahlos necessarily having $\Pi_3$-reflecting or stronger properties.

  • $\begingroup$ I've been asked to comment that $M[\omega_{K+1}^{CK}](0)$ can be equal to $K$ $\endgroup$
    – C7X
    Apr 12, 2022 at 10:58
  • $\begingroup$ And why is that? $\endgroup$ Apr 13, 2022 at 11:35
  • $\begingroup$ If you did something similar but with $I[\beta]$ enumerating rec. $\beta$-inaccessibles, and with the least rec. mahlo $M$ collapsing similarly inside $I[\text{___}](\alpha)$, then $I[\omega_{M+1}^{CK}](0)$ would simply be $M$. There just aren't enough ordinals below $M$ for such collapses, and the same would happen with actual mahlos if we removed all the "recursively". It should also apply to iterating $\Pi_2$-reflection below $\Pi_3$-reflectings, but not to iterating the mahlo operation below weakly compacts, which is why i think $\Pi_3$-reflectings are rec. greatly mahlo. $\endgroup$
    – Yto
    Apr 14, 2022 at 10:25
  • $\begingroup$ The idea that $\Pi_{n+2}$-reflecting is recursively $\Pi^1_n$-indescribable seems to only come from convenience for proof theory, but i haven't found any evidence that this idea is consistent with the details such as greatly mahlos. Some people argue that greatly mahlos should just be skipped for this reason, but i don't think so, as they aren't barely above "recursive" iterations of anything. Also, Taranovsky mentioned that "$\Pi^1_1$-reflection for $\kappa$ can be iterated up to (informally) $\kappa^{++}$ times as opposed to up to $\kappa^+$ times for lower reflection levels" $\endgroup$
    – Yto
    Apr 14, 2022 at 10:35
  • 1
    $\begingroup$ I would say that beyond convenience, the reason greatly Mahlos are skipped is because there simply isn't a nice way to project the notion of a filter down to a level where cardinalities are not increased at each step, as filters on $F$ use $\mathcal P(F)$ in their definition. Maybe you could define a countable filter by replacing $\mathcal P(F)$ with the set of $\Delta^1_1$-definable subsets of $F$ analogously to the "limit of hyperarithmetic" property of $\omega^{CK}_1$ relative to the recursive ordinals, though I am not sure if that would work. I'm a little dubious of Taranovsky's claims. $\endgroup$ Apr 14, 2022 at 21:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.