aa r X i v : . [ m a t h . L O ] F e b A note on Woodin’s HOD dichotomy
Gabriel GoldbergEvans HallUniversity DriveBerkeley, CA 94720February 19, 2021
The purpose of this note is to prove a version of Woodin’s HOD dichotomy [4] from a stronglycompact cardinal. A cardinal κ is strongly compact if every κ -complete filter extends toa κ -complete ultrafilter. The class of hereditarily ordinal definable sets, denoted HOD, isthe minimum class M that contains all the ordinals and is definably closed: if S ⊆ M isdefinable (in the universe of sets V ) by a set theoretic formula with parameters in M , then S ∈ M . HOD is a proper class transitive model of ZFC. Theorem 1.1.
Suppose κ is strongly compact. Then one of the following holds:(1) For every singular strong limit cardinal λ ≥ κ , λ is singular in HOD and λ +HOD = λ + .(2) All sufficiently large regular cardinals are measurable in HOD . If (1) holds, (2) cannot (since there are arbitrarily large successor cardinals that are notinaccessible in HOD), so Theorem 1.1 is truly a dichotomy.We will actually prove a stronger dichotomy. An inner model M is said to have the λ -cover property if every set of ordinals of cardinality less than λ is covered by (that is,included in) a set of ordinals in M of cardinality less than λ . Theorem 1.2.
Suppose κ is strongly compact. Then one of the following holds:(1) For every strong limit cardinal λ ≥ κ , HOD has the λ -cover property.(2) All sufficiently large regular cardinals are ω -strongly measurable in HOD . The definition of ω -strong measurability is deferred until the beginning of the next sec-tion, but let us note here that any cardinal that is ω -strongly measurable in HOD is indeedmeasurable in HOD. Note that Theorem 1.2 (1) strengthens Theorem 1.1 (1), while Theo-rem 1.2 (2) strengthens Theorem 1.1 (2).Assuming instead that κ is HOD-supercompact, Woodin [4] proved an even strongerdichotomy theorem, for example establishing that if (either) condition (1) above holds, then κ is supercompact in HOD. On the other hand, Cheng-Friedman-Hamkins [1] producea model of ZFC with a supercompact cardinal κ in which (1) holds, yet κ is not weaklycompact in HOD, which shows that Woodin’s theorem cannot be proved from the hypotheseswe assume here. 1 The proof
Suppose δ is a regular cardinal and S ⊆ δ is a stationary set. We say an inner model M splits S if S ∈ M and for all γ such that (2 γ ) M < δ , there is a partition of S into γ -manystationary sets. A cardinal is ω -strongly measurable in HOD if HOD does not split the setof ordinals less than δ with countable cofinality, which is denoted by S δω . Theorem 2.1 (Woodin, [4, Lemma 3.37]) . If S is an ordinal definable stationary subset ofa regular cardinal δ and HOD does not split S , then ( C δ ↾ S ) ∩ HOD is atomic in
HOD .In particular, if δ is ω -strongly measurable in HOD , then δ is measurable in HOD and δ contains an ω -club of cardinals inaccessible in HOD . The last sentence is supposed to highlight that the existence of ω -strongly measurablecardinals entails a massive failure of the cover property even for countable sets. Actuallythere is an ω -club of cardinals measurable in HOD below any cardinal that is ω -stronglymeasurable in HOD in the natural sense. Proposition 2.2.
Suppose κ is strongly compact, δ ≥ κ is regular, and HOD splits S δω .Then for any γ such that (2 γ ) HOD < δ , there is a κ -complete fine ultrafilter U on P κ ( γ ) such that U concentrates on P κ ( γ ) ∩ HOD .Proof.
Let j : V → M be an elementary embedding from the universe into an inner modelsuch that crit( j ) = κ and j [ δ ] is contained in some set S ∈ M with | S | M < j ( κ ). Inparticular, the ordinal δ ∗ = sup j [ δ ] has cofinality less than j ( κ ) in M . Let C ⊆ δ ∗ be aclosed unbounded set of ordertype cf M ( δ ∗ ).Fix a cardinal γ such that (2 γ ) HOD < δ and let h S α i α<γ witness that HOD splits S δω .Then let σ = { ξ < j ( γ ) : T ξ ∩ δ ∗ is stationary in M } where ~T = j ( ~S ). Thus σ ∈ HOD M . Notice that j [ γ ] ⊆ σ : j [ S ξ ] ⊆ T j ( ξ ) , so in fact T j ( ξ ) ∩ δ ∗ is truly stationary (not just in M ). For all ξ ∈ σ , T ξ ∩ C = ∅ , so let f ( ξ ) = min( T ξ ∩ C ).Then f : σ → C is an injection. So | σ | = cf M ( δ ∗ ). In particular, σ ∈ j ( P κ ( γ )).Finally let U be the ultrafilter on P κ ( γ ) derived from j using σ . That is, let U = { A ⊆ P κ ( γ ) : σ ∈ j ( A ) } . Since σ ∈ HOD M , σ ∈ j ( P κ ( γ ) ∩ HOD), and hence HOD ∩ P κ ( γ ) ∈ U .Since j [ γ ] ⊆ σ , U is fine. Since crit( j ) = κ , U is κ -complete.The main observation involved in the proof above is that the stationary splitting argu-ment from [4] (which Woodin calls “Solovay’s Lemma” although it is a bit different fromthe related lemma in [2]) can be adapted to strongly compact cardinals. Usuba [3] madethe same observation independently and earlier. Lemma 2.3.
Suppose κ is strongly compact. Then one of the following holds:(1) HOD has the κ -cover property.(2) All sufficiently large regular cardinals are ω -strongly measurable in in HOD .Proof.
Assume that there are arbitrarily large regular cardinals δ that are not ω -stronglymeasurable in HOD, or in other words, HOD splits S δω . Applying Proposition 2.2 to suffi-ciently large such δ , for all γ ≥ κ , there is a κ -complete fine ultrafilter U on P κ ( γ ) such that P κ ( γ ) ∩ HOD ∈ U . For each σ ∈ P κ ( γ ), { τ ∈ P κ ( γ ) : σ ⊆ τ } ∈ U . Since U is a filter, itfollows that { τ ∈ P κ ( γ ) ∩ HOD : σ ⊆ τ } ∈ U , and in particular this set is nonempty. Thisyields τ ∈ P κ ( γ ) ∩ HOD covering σ , as desired.2e now extend the cover property of HOD to all strong limit cardinals greater than orequal to the first strongly compact cardinal. Proof of Theorem 1.2.
Suppose (2) fails, so by Lemma 2.3, HOD has the κ -cover property.Given this, it suffices to show that for all δ ≥ κ , for some δ ′ < i ω ( δ ), every set of ordinalsof cardinality at most δ is covered by a set of ordinals in HOD of cardinality at most δ ′ .Let U be a fine κ -complete ultrafilter on P κ ( δ ). Fix A ⊆ κ such that V κ ⊆ HOD A .We will show that for every set S of ordinals of cardinality at most δ , there is a set T ofcardinality at most 2 δ such that T ∈ HOD A, U and S ⊆ T . By Vopenka’s theorem, HOD A, U is a forcing extension of HOD by a forcing in HOD of cardinality less than i ω ( δ ), so thetheorem follows.Note that HOD A, U is closed under <κ -sequences since HOD A, U has the κ -cover propertyand V κ ⊆ HOD A, U . As a consequence, M U satisfies that j U (HOD A, U ) is closed under By Lemma 2.3 (and Theorem 2.1), we can assume that HOD hasthe κ -cover property. Fix a singular strong limit cardinal λ > κ . Theorem 1.2 easily impliesthat λ is singular in HOD. Let γ = λ +HOD . Since γ is regular in HOD, Theorem 1.2 impliescf( γ ) ≥ λ . Since λ is singular, cf( γ ) > λ , and so γ = λ + . References [1] Yong Cheng, Sy-David Friedman, and Joel David Hamkins. Large cardinals need notbe large in HOD. Annals of Pure and Applied Logic , 166(11):1186 – 1198, 2015.[2] Robert M. Solovay, William N. Reinhardt, and Akihiro Kanamori. Strong axioms ofinfinity and elementary embeddings.