8
$\begingroup$

Question: What is the consistency strength of existence of a $(κ^{++})^L$-Mahlo cardinal $κ$?

I am particularly interested in how the strength compares to weakly compact cardinals (and other levels of indescribability), and I am looking for equiconsistencies between an $α$-Mahlo $κ$ (for some sense of $α$-Mahlo) and existence of weakly compact $κ$.

The motivation for the question is that in set theory, strong levels of one axiom schema naturally morph into weak versions of the next schema. For example:
* The replacement schema corresponds to Ord being inaccessible, and a strong limit $κ$ is inaccessible iff for every $U⊂V_κ$, $V_κ$ satisfies replacement for formulas in $(V_κ,∈,U)$.
* A schema $∃(\text{regular }λ) \, V_λ ≺_{Σ_n} V$ corresponds to Ord being Mahlo, and a strong limit $κ$ is Mahlo iff $∀U⊂V_κ \, ∃(\text{regular }λ<κ) \, (V_λ,∈,U∩V_λ) ≺ (V_κ,∈,U)$.
* After iterations of Mahloness, the next step is weak compactness, prompting the question: What is the correspondence between strong levels of Mahloness and weak compactness?

The definition of $α$-Mahlo $κ$ for $α≤κ^+$ is standard (or see below). Beyond $(κ^+)^L$, we have different choices:
a. require $α≤κ^+$, so $(κ^+)^L < κ^+$
b. use the definition for $α≤κ^+$ beyond $κ^+$ (allowing GCH to fail at $κ$)
c. use ordinal representation systems for diagonalization beyond $κ^+$ (not sure how far this works)
d. $κ$ being $α$-Mahlo in a sufficiently large generic collapse $\mathrm{Coll}(κ,λ)$
(and for more choices, the above also makes sense for weakly Mahlo cardinals).

My guess is that these possibilities have the same strength, but in any case, any of these will be fine for an answer.

Definition of $α$-Mahlo cardinals

If the reader is unfamiliar with greatly Mahlo cardinals, here is a definition and some properties.

A strong limit cardinal $κ$ is 0-Mahlo iff it is regular uncountable.
$κ$ is $α+1$-Mahlo iff the set of $α$-Mahlo cardinals is stationary below $κ$.
For limit $α≤κ$, $κ$ is $α$-Mahlo iff it is $α'$ Mahlo for every $α'<α$.

Beyond $κ$-Mahlo, we rely on diagonalization, and it turns out that the iteration is well-defined up to $κ^+$.

Define $M_α$ modulo the nonstationary ideal on $κ$:
$M_0$ = $κ$
$M_{α+1}$ = {$λ<κ$: $\mathrm{cf}(λ)>ω$ and $M_α$ is stationary below $λ$}
$M_α$ for limit $α<κ^+$ is greatest lower bound of $(M_{α'}:α'<α)$ (so $M_α \setminus M_{α'}$ is nonstationary).

A strong limit $κ$ is $α$-Mahlo ($α≤κ^+$) iff for every $α'<α$, $M_{κ+α}$ is stationary below $κ$.

Also, the greatest lower bound above is defined modulo the nonstationary ideal (which suffices). If desired, given a choice of fundamental sequences, we can pick a canonical representative of $M_α$ as follows:
$\mathrm{cf}(α)<κ$ -- $M_α$ is the intersection of $M_{α'}$ for $α'$ on the fundamental sequence of $α$.
$\mathrm{cf}(α)=κ$ -- $M_α$ is the diagonal intersection of $M_{α'}$ for $α'$ on the fundamental sequence of $α$. For example, $M_κ$ consists of (modulo a nonstationary set) regular cardinals below $κ$.

Weakly compact cardinals are greatly Mahlo (i.e. $κ^+$-Mahlo) and more. For example, the property that every stationary subset of $κ$ reflects (i.e. is stationary below some $λ<κ$) is consistency-wise (and in terms of the least such cardinal in L) strictly between greatly Mahlo and weakly compact.

$\endgroup$
  • $\begingroup$ Do you know if weakly compact cardinals in $L$ are $\kappa^{++}$-Mahlo (in any one of those definitions above)? $\endgroup$ – Yair Hayut Dec 26 '18 at 10:30
  • $\begingroup$ @YairHayut No, I do not. My attempts at diagonalization have not yielded $κ^{++}$, and I do not know what happens in $\mathrm{Coll}(κ,κ^+)$. However, one negative result is that in $L$ there is no canonical stationary set of rank $κ^+$. $\endgroup$ – Dmytro Taranovsky Dec 26 '18 at 17:46
  • $\begingroup$ Maybe the following papers be related (though I'm not sure): A Lower Bound on the Mahlo Rank of a Weakly Compact Cardinal and Normal Ultrafilters and Mahlo Rank $\endgroup$ – Mohammad Golshani Dec 27 '18 at 5:35
1
$\begingroup$

I have an idea for diagonalization. First, separate ordinals into three categories. $\alpha$ is a successor, $\alpha$ is a limit with $cf\alpha\lt\kappa$, or $\alpha$ is a limit with $cf\alpha\ge\kappa$. Then let $M(X)$ be the standard Mahlo operation, and take $M^{\alpha+1}(X)=M(M^\alpha(X))$, and at limits of the first type $M^\alpha(X)=\cap_{\beta\lt\alpha} M^\beta(X)$, and at limits of the second type $M^\alpha(X)=\Delta_{\beta\lt\alpha} M^\beta(X)$ (The diagonal intersection). Then $\kappa$ is $\alpha-$Mahlo if and only if $\kappa\in M^{1+\alpha}(R)$, where $R$ is the class of regular cardinals. Then for $\alpha\lt\kappa^+$ this system agrees with the normal system. Eureka! (Unless I have made a terrible mistake)

Then, for instance, the hyper$^\kappa-$Mahlo cardinals are precisely the $\kappa\cdot\kappa-$cardinals. Now with this system, we can take $\alpha$ beyond $\kappa^{++}$; to $Ord$ even. I claim that if $\kappa$ is weakly compact, $\kappa$ is $\alpha-$Mahlo for every $\alpha$. Let $F$ be the $\Pi^1_1-$indescribable filter on $\kappa$. It is normal, $\kappa-$complete, and closed under $M(X)$. Therefore, it follows that if $\{\alpha\lt\kappa|\alpha\text{ is Mahlo}\}\in F$, then $\kappa$ is $\alpha-$Mahlo for every $\alpha$. But if $\kappa$ is weakly compact, then there exists many $(V_\lambda,\in,S)\prec (V_\kappa,\in,S)$, such that $\lambda$ is Mahlo, and so the proof is complete and no amount of degrees of Mahloness can approach weak compactness.

If you want exactly the consistency strength of a $(\kappa^{++})^L-$Mahlo cardinal, then that is same as a $\kappa^{++}-$Mahlo cardinal, because if $M\vDash(\kappa\text{ is }(\kappa^{++})^L\text{-Mahlo})$, then $L^M\vDash(\kappa\text{ is }\kappa^{++}\text{-Mahlo})$, because stationarity, regular cardinals, intersection, and diagonal intersection are preserved when switching to an inner model.

$\endgroup$
  • $\begingroup$ I'm not sure why this would be the third or fourth time people tell you to use \text{...} when writing text inside math mode... $\endgroup$ – Asaf Karagila Jul 6 at 10:44
  • $\begingroup$ I'm sorry. I will fix it. I thought you meant for particularly long text blocks. Is this better? $\endgroup$ – Master Jul 6 at 16:07
  • $\begingroup$ Diagonal intersection is only generally defined (modulo the nonstationary ideal) up to $κ^+$, so unfortunately your construction does not work beyond that. $\endgroup$ – Dmytro Taranovsky Jul 6 at 22:14
  • $\begingroup$ I mean this definition of diagonal intersection. It has nothing to due with Mahloness (And can be applied to inaccessibility): en.wikipedia.org/wiki/Diagonal_intersection $\endgroup$ – Master Jul 6 at 22:28
  • $\begingroup$ @Master For the construction to work as intended beyond $M_κ$, you would need fundamental sequences (the "modulo the nonstationary ideal" makes the choice of sequences irrelevant), and even then $M_{κ^+}$ would need a different definition. $\endgroup$ – Dmytro Taranovsky Jul 6 at 23:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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