Bounded Stationary Reflection Chris Lambie-Hanson Department of Mathematical Sciences Carnegie Mellon University GSCL XV Madison, WI 26 April 2014

Section 1 J´ onsson cardinals

J´ onsson algebras

J´ onsson algebras Definition • Let X be a set and κ a cardinal. [X]κ = {Y ⊆ X | |Y | = κ}.

J´ onsson algebras Definition • Let X be a set and κ a cardinal. [X]κ = {Y ⊆ X | |Y | = κ}. • An algebra is a structure X, fi | i < ω such that, for every i < ω, there is ni < ω such that fi : [X]ni → X.

J´ onsson algebras Definition • Let X be a set and κ a cardinal. [X]κ = {Y ⊆ X | |Y | = κ}. • An algebra is a structure X, fi | i < ω such that, for every i < ω, there is ni < ω such that fi : [X]ni → X. • Equivalently, an algebra is a set X together with a single function f : [X]<ω → X.

J´ onsson algebras Definition • Let X be a set and κ a cardinal. [X]κ = {Y ⊆ X | |Y | = κ}. • An algebra is a structure X, fi | i < ω such that, for every i < ω, there is ni < ω such that fi : [X]ni → X. • Equivalently, an algebra is a set X together with a single function f : [X]<ω → X. • If X, f is an algebra and Y ⊆ X, then Y , f [Y ]<ω is a subalgebra of X, f if f “[Y ]<ω ⊆ Y .

J´ onsson algebras Definition • Let X be a set and κ a cardinal. [X]κ = {Y ⊆ X | |Y | = κ}. • An algebra is a structure X, fi | i < ω such that, for every i < ω, there is ni < ω such that fi : [X]ni → X. • Equivalently, an algebra is a set X together with a single function f : [X]<ω → X. • If X, f is an algebra and Y ⊆ X, then Y , f [Y ]<ω is a subalgebra of X, f if f “[Y ]<ω ⊆ Y . • An algebra X, f is a J´ onsson algebra if it has no proper subalgebras of the same cardinality.

J´ onsson cardinals

J´ onsson cardinals Definition Let κ be an infinite cardinal. κ is a J´ onsson cardinal if there are no J´ onsson algebras X, f such that |X| = κ.

J´ onsson cardinals Definition Let κ be an infinite cardinal. κ is a J´ onsson cardinal if there are no J´ onsson algebras X, f such that |X| = κ. Proposition Let κ be an infinite cardinal. The following are equivalent.

J´ onsson cardinals Definition Let κ be an infinite cardinal. κ is a J´ onsson cardinal if there are no J´ onsson algebras X, f such that |X| = κ. Proposition Let κ be an infinite cardinal. The following are equivalent. 1 κ is a J´ onsson cardinal.

J´ onsson cardinals Definition Let κ be an infinite cardinal. κ is a J´ onsson cardinal if there are no J´ onsson algebras X, f such that |X| = κ. Proposition Let κ be an infinite cardinal. The following are equivalent. 1 κ is a J´ onsson cardinal. 2 For every f : [κ]<ω → κ, there is H ∈ [κ]κ such that f “[H]<ω is a proper subset of κ (i.e. κ → [κ]<ω κ ).

J´ onsson cardinals Definition Let κ be an infinite cardinal. κ is a J´ onsson cardinal if there are no J´ onsson algebras X, f such that |X| = κ. Proposition Let κ be an infinite cardinal. The following are equivalent. 1 κ is a J´ onsson cardinal. 2 For every f : [κ]<ω → κ, there is H ∈ [κ]κ such that f “[H]<ω is a proper subset of κ (i.e. κ → [κ]<ω κ ). 3 For all sufficiently large, regular θ and all x ∈ H(θ), there is M ≺ H(θ) such that: • {κ, x} ∈ M. • |M ∩ κ| = κ. • κ ⊆ M.

The large cardinal hierarchy

The large cardinal hierarchy Proposition Let κ be a cardinal. Then κ is measurable ⇒ κ is Ramsey ⇒ κ is Rowbottom ⇒ κ is J´ onsson.

The large cardinal hierarchy Proposition Let κ be a cardinal. Then κ is measurable ⇒ κ is Ramsey ⇒ κ is Rowbottom ⇒ κ is J´ onsson. Theorem (Kleinberg) Con(ZFC + there is a Rowbottom cardinal) ⇔ Con(ZFC + there is a J´ onsson cardinal)

The large cardinal hierarchy Proposition Let κ be a cardinal. Then κ is measurable ⇒ κ is Ramsey ⇒ κ is Rowbottom ⇒ κ is J´ onsson. Theorem (Kleinberg) Con(ZFC + there is a Rowbottom cardinal) ⇔ Con(ZFC + there is a J´ onsson cardinal) Theorem (Prikry) 1 If κ is a singular limit of measurable cardinals, then κ is J´ onsson.

The large cardinal hierarchy Proposition Let κ be a cardinal. Then κ is measurable ⇒ κ is Ramsey ⇒ κ is Rowbottom ⇒ κ is J´ onsson. Theorem (Kleinberg) Con(ZFC + there is a Rowbottom cardinal) ⇔ Con(ZFC + there is a J´ onsson cardinal) Theorem (Prikry) 1 If κ is a singular limit of measurable cardinals, then κ is J´ onsson. 2 If κ is a measurable cardinal and P is Prikry forcing at κ, then κ remains J´ onsson in V P.

Restrictions on J´ onsson cardinals Proposition ω is not a J´ onsson cardinal.

Restrictions on J´ onsson cardinals Proposition ω is not a J´ onsson cardinal. Proof. Let f : [ω]<ω → ω be such that, for every n < ω, f ({n + 1}) = n. Then ω, f is a J´ onsson algebra.

Restrictions on J´ onsson cardinals Proposition Let κ be an infinite cardinal. If κ is not a J´ onsson cardinal, then κ+ is not a J´ onsson cardinal.

Restrictions on J´ onsson cardinals Proposition Let κ be an infinite cardinal. If κ is not a J´ onsson cardinal, then κ+ is not a J´ onsson cardinal. Proof. Suppose κ is not a J´ onsson cardinal. Then there is a regular θ > κ+ and x ∈ H(θ) such that, for all M ≺ H(θ) with {κ, x} ∈ M and |M ∩ κ| = κ, then κ ⊆ M.

Restrictions on J´ onsson cardinals Proposition Let κ be an infinite cardinal. If κ is not a J´ onsson cardinal, then κ+ is not a J´ onsson cardinal. Proof. Suppose κ is not a J´ onsson cardinal. Then there is a regular θ > κ+ and x ∈ H(θ) such that, for all M ≺ H(θ) with {κ, x} ∈ M and |M ∩ κ| = κ, then κ ⊆ M. Let M ≺ H(θ) be such that {κ+, x} ∈ M and |M ∩ κ+| = κ+. There is α ∈ M ∩ κ+ such that |M ∩ α| = κ. Since M contains a bijection between κ and α, |M ∩ κ| = κ. Thus, κ ⊆ M.

Restrictions on J´ onsson cardinals Proposition Let κ be an infinite cardinal. If κ is not a J´ onsson cardinal, then κ+ is not a J´ onsson cardinal. Proof. Suppose κ is not a J´ onsson cardinal. Then there is a regular θ > κ+ and x ∈ H(θ) such that, for all M ≺ H(θ) with {κ, x} ∈ M and |M ∩ κ| = κ, then κ ⊆ M. Let M ≺ H(θ) be such that {κ+, x} ∈ M and |M ∩ κ+| = κ+. There is α ∈ M ∩ κ+ such that |M ∩ α| = κ. Since M contains a bijection between κ and α, |M ∩ κ| = κ. Thus, κ ⊆ M. For every β ∈ M ∩ κ+, since M contains a bijection between κ and β, we have β ⊆ M. Since M ∩ κ+ is unbounded in κ+, we have κ+ ⊆ M.

Restrictions on J´ onsson cardinals Theorem (Keisler-Rowbottom) If V = L, then there are no J´ onsson cardinals.

Restrictions on J´ onsson cardinals Theorem (Keisler-Rowbottom) If V = L, then there are no J´ onsson cardinals. Theorem (Erd˝ os-Hajnal-Rado) If 2κ = κ+, then κ+ is not J´ onsson.

Restrictions on J´ onsson cardinals Theorem (Keisler-Rowbottom) If V = L, then there are no J´ onsson cardinals. Theorem (Erd˝ os-Hajnal-Rado) If 2κ = κ+, then κ+ is not J´ onsson. Theorem (Shelah) If κ is a singular cardinal and κ is not a limit of regular J´ onsson cardinals, then κ+ is not J´ onsson.

Stationary reflection Definition Let κ be an uncountable, regular cardinal. 1 S ⊆ κ is stationary in κ if, for every closed, unbounded C ⊆ κ, S ∩ C = ∅.

Stationary reflection Definition Let κ be an uncountable, regular cardinal. 1 S ⊆ κ is stationary in κ if, for every closed, unbounded C ⊆ κ, S ∩ C = ∅. 2 If S ⊆ κ is stationary and α < κ, then S reflects at α if cf(α) > ω and S ∩ α is stationary in α.

Stationary reflection Definition Let κ be an uncountable, regular cardinal. 1 S ⊆ κ is stationary in κ if, for every closed, unbounded C ⊆ κ, S ∩ C = ∅. 2 If S ⊆ κ is stationary and α < κ, then S reflects at α if cf(α) > ω and S ∩ α is stationary in α. S reflects if there is α < κ such that S reflects at α.

Stationary reflection Definition Let κ be an uncountable, regular cardinal. 1 S ⊆ κ is stationary in κ if, for every closed, unbounded C ⊆ κ, S ∩ C = ∅. 2 If S ⊆ κ is stationary and α < κ, then S reflects at α if cf(α) > ω and S ∩ α is stationary in α. S reflects if there is α < κ such that S reflects at α. Theorem (Tryba, Woodin) Suppose κ is a regular J´ onsson cardinal. Then every stationary subset of κ reflects.

Stationary reflection Definition Let κ be an uncountable, regular cardinal. 1 S ⊆ κ is stationary in κ if, for every closed, unbounded C ⊆ κ, S ∩ C = ∅. 2 If S ⊆ κ is stationary and α < κ, then S reflects at α if cf(α) > ω and S ∩ α is stationary in α. S reflects if there is α < κ such that S reflects at α. Theorem (Tryba, Woodin) Suppose κ is a regular J´ onsson cardinal. Then every stationary subset of κ reflects. Remark If κ is a regular cardinal and S ⊆ κ+ ∩ cof(κ), then S cannot reflect. Thus, if κ is regular, then κ+ is not a J´ onsson cardinal.

The proof of Tryba and Woodin’s theorem actually yields the following, apparently stronger statement:

The proof of Tryba and Woodin’s theorem actually yields the following, apparently stronger statement: Theorem Suppose κ is a regular J´ onsson cardinal, S ⊆ κ is stationary, and λ < κ is a regular cardinal. Then there is α ∈ κ ∩ cof(≥ λ) such that S reflects at α.

The proof of Tryba and Woodin’s theorem actually yields the following, apparently stronger statement: Theorem Suppose κ is a regular J´ onsson cardinal, S ⊆ κ is stationary, and λ < κ is a regular cardinal. Then there is α ∈ κ ∩ cof(≥ λ) such that S reflects at α. In other words, every stationary subset of κ reflects at ordinals of arbitrarily high cofinality.

The proof of Tryba and Woodin’s theorem actually yields the following, apparently stronger statement: Theorem Suppose κ is a regular J´ onsson cardinal, S ⊆ κ is stationary, and λ < κ is a regular cardinal. Then there is α ∈ κ ∩ cof(≥ λ) such that S reflects at α. In other words, every stationary subset of κ reflects at ordinals of arbitrarily high cofinality. Theorem (Eisworth) Suppose µ is a singular cardinal and µ+ → [µ+]2 µ+ . Then every stationary subset of µ+ reflects at ordinals of arbitrarily high cofinality.

The proof of Tryba and Woodin’s theorem actually yields the following, apparently stronger statement: Theorem Suppose κ is a regular J´ onsson cardinal, S ⊆ κ is stationary, and λ < κ is a regular cardinal. Then there is α ∈ κ ∩ cof(≥ λ) such that S reflects at α. In other words, every stationary subset of κ reflects at ordinals of arbitrarily high cofinality. Theorem (Eisworth) Suppose µ is a singular cardinal and µ+ → [µ+]2 µ+ . Then every stationary subset of µ+ reflects at ordinals of arbitrarily high cofinality. Question (Eisworth) Suppose µ is a singular cardinal, κ = µ+, and every stationary subset of κ reflects. Must it be the case that every stationary subset of κ reflects at ordinals of arbitrarily high cofinality?

Section 2 Approachability and forcing

Approachability

Approachability Definition Let µ be a singular cardinal, and let κ = µ+.

Approachability Definition Let µ be a singular cardinal, and let κ = µ+. 1 Let a = aα | α < κ be a sequence of bounded subsets of κ. A limit ordinal β < κ is approachable with respect to a if there is an unbounded A ⊆ β of order type cf(β) such that, for every γ < β, there is α < β such that A ∩ γ = aα.

Approachability Definition Let µ be a singular cardinal, and let κ = µ+. 1 Let a = aα | α < κ be a sequence of bounded subsets of κ. A limit ordinal β < κ is approachable with respect to a if there is an unbounded A ⊆ β of order type cf(β) such that, for every γ < β, there is α < β such that A ∩ γ = aα. 2 Let S ⊆ κ. S ∈ I[κ] if there is a sequence a and a club C ⊆ κ such that, for every β ∈ S ∩ C, β is approachable with respect to a.

Approachability Definition Let µ be a singular cardinal, and let κ = µ+. 1 Let a = aα | α < κ be a sequence of bounded subsets of κ. A limit ordinal β < κ is approachable with respect to a if there is an unbounded A ⊆ β of order type cf(β) such that, for every γ < β, there is α < β such that A ∩ γ = aα. 2 Let S ⊆ κ. S ∈ I[κ] if there is a sequence a and a club C ⊆ κ such that, for every β ∈ S ∩ C, β is approachable with respect to a. 3 The approachability property holds at µ (written APµ) if κ ∈ I[κ].

Approachability Definition Let µ be a singular cardinal, κ = µ+, and θ a sufficiently large, regular cardinal. If A is a countable expansion of H(θ), ∈, <θ and β < κ is a limit ordinal, then β is approachable with respect to A if there is an unbounded A ⊆ β of order type cf(β) such that every initial segment of A is in SkA(β).

Approachability Definition Let µ be a singular cardinal, κ = µ+, and θ a sufficiently large, regular cardinal. If A is a countable expansion of H(θ), ∈, <θ and β < κ is a limit ordinal, then β is approachable with respect to A if there is an unbounded A ⊆ β of order type cf(β) such that every initial segment of A is in SkA(β). Fact Let µ be a singular cardinal, and let κ = µ+.

Approachability Definition Let µ be a singular cardinal, κ = µ+, and θ a sufficiently large, regular cardinal. If A is a countable expansion of H(θ), ∈, <θ and β < κ is a limit ordinal, then β is approachable with respect to A if there is an unbounded A ⊆ β of order type cf(β) such that every initial segment of A is in SkA(β). Fact Let µ be a singular cardinal, and let κ = µ+. 1 S ∈ I[κ] if and only if there is a sufficiently large, regular θ, a countable expansion A of H(θ), ∈, <θ , and a club C ⊆ κ such that, for every β ∈ S ∩ C, β is approachable with respect to A.

Approachability Definition Let µ be a singular cardinal, κ = µ+, and θ a sufficiently large, regular cardinal. If A is a countable expansion of H(θ), ∈, <θ and β < κ is a limit ordinal, then β is approachable with respect to A if there is an unbounded A ⊆ β of order type cf(β) such that every initial segment of A is in SkA(β). Fact Let µ be a singular cardinal, and let κ = µ+. 1 S ∈ I[κ] if and only if there is a sufficiently large, regular θ, a countable expansion A of H(θ), ∈, <θ , and a club C ⊆ κ such that, for every β ∈ S ∩ C, β is approachable with respect to A. 2 I[κ] is a normal ideal on κ.

Approachability Definition Let µ be a singular cardinal, κ = µ+, and θ a sufficiently large, regular cardinal. If A is a countable expansion of H(θ), ∈, <θ and β < κ is a limit ordinal, then β is approachable with respect to A if there is an unbounded A ⊆ β of order type cf(β) such that every initial segment of A is in SkA(β). Fact Let µ be a singular cardinal, and let κ = µ+. 1 S ∈ I[κ] if and only if there is a sufficiently large, regular θ, a countable expansion A of H(θ), ∈, <θ , and a club C ⊆ κ such that, for every β ∈ S ∩ C, β is approachable with respect to A. 2 I[κ] is a normal ideal on κ. 3 If κ<κ = κ, then there is a maximal (modulo the non-stationary ideal) set in I[κ]. This set is called the set of approachable points.

Approachability and forcing Theorem (Shelah) Let λ < κ be cardinals, with κ regular. Suppose that S ⊆ κ ∩ cof(< λ) is stationary, S ∈ I[κ], and P is a µ-closed forcing poset. Then S remains stationary in V P.

Approachability and forcing Theorem (Shelah) Let λ < κ be cardinals, with κ regular. Suppose that S ⊆ κ ∩ cof(< λ) is stationary, S ∈ I[κ], and P is a µ-closed forcing poset. Then S remains stationary in V P. Theorem Let µ be a singular cardinal, let κ = µ+, and suppose κ<κ = κ. Let a = aα | α < κ be an enumeration of all bounded subsets of κ.

Approachability and forcing Theorem (Shelah) Let λ < κ be cardinals, with κ regular. Suppose that S ⊆ κ ∩ cof(< λ) is stationary, S ∈ I[κ], and P is a µ-closed forcing poset. Then S remains stationary in V P. Theorem Let µ be a singular cardinal, let κ = µ+, and suppose κ<κ = κ. Let a = aα | α < κ be an enumeration of all bounded subsets of κ. Let Q be a forcing poset whose conditions are closed, bounded subsets t of κ such that, for every β ∈ t, β is approachable with respect to a, ordered by end-extension.

Approachability and forcing Theorem (Shelah) Let λ < κ be cardinals, with κ regular. Suppose that S ⊆ κ ∩ cof(< λ) is stationary, S ∈ I[κ], and P is a µ-closed forcing poset. Then S remains stationary in V P. Theorem Let µ be a singular cardinal, let κ = µ+, and suppose κ<κ = κ. Let a = aα | α < κ be an enumeration of all bounded subsets of κ. Let Q be a forcing poset whose conditions are closed, bounded subsets t of κ such that, for every β ∈ t, β is approachable with respect to a, ordered by end-extension. Then Q is a cardinal-preserving forcing poset and V Q |= APµ.

Supercompact cardinals

Supercompact cardinals Definition Let κ < λ be cardinals. κ is λ-supercompact if there is an elementary embedding j : V → M, where M is a transitive class, such that: • crit(j) = κ. • j(κ) > λ. • λM ⊆ M.

Supercompact cardinals Definition Let κ < λ be cardinals. κ is λ-supercompact if there is an elementary embedding j : V → M, where M is a transitive class, such that: • crit(j) = κ. • j(κ) > λ. • λM ⊆ M. κ is supercompact if it is λ-supercompact for all λ. Proposition κ is λ-supercompact if and only if there is a normal, fine, κ-complete measure on Pκ(λ).

Theorem (Solovay) Suppose µ is a singular limit of supercompact cardinals. Then every stationary subset of µ+ reflects. Proof. Let µ be a singular limit of supercompact cardinals, and let S ⊆ µ+ be stationary. We will show that S reflects.

Theorem (Solovay) Suppose µ is a singular limit of supercompact cardinals. Then every stationary subset of µ+ reflects. Proof. Let µ be a singular limit of supercompact cardinals, and let S ⊆ µ+ be stationary. We will show that S reflects. By shrinking S if necessary, we may assume that there is λ < µ such that S ⊆ µ+ ∩ cof(λ). Fix a supercompact cardinal κ such that λ < κ < µ, and let j : V → M witness that κ is µ+-supercompact.

Theorem (Solovay) Suppose µ is a singular limit of supercompact cardinals. Then every stationary subset of µ+ reflects. Proof. Let µ be a singular limit of supercompact cardinals, and let S ⊆ µ+ be stationary. We will show that S reflects. By shrinking S if necessary, we may assume that there is λ < µ such that S ⊆ µ+ ∩ cof(λ). Fix a supercompact cardinal κ such that λ < κ < µ, and let j : V → M witness that κ is µ+-supercompact. Let γ = sup(j“µ+). Note that, since j“µ+ ∈ M and M |= “j(µ+) is regular ”, we have γ < j(µ+).

Proof ctd. In M, j(S) is a stationary subset of j(µ+) ∩ cof(λ).

Proof ctd. In M, j(S) is a stationary subset of j(µ+) ∩ cof(λ). Claim M |= “j(S) ∩ γ is stationary in γ”.

Proof ctd. In M, j(S) is a stationary subset of j(µ+) ∩ cof(λ). Claim M |= “j(S) ∩ γ is stationary in γ”. Proof. Suppose not, and let C ∈ M be a club in γ disjoint from j(S). Since j is continuous at ordinals of cofinality < κ, D = C ∩ j µ+ is a < κ-club in γ disjoint from j(S). Thus, E = j−1“D is a < κ-club in µ+ disjoint from S, contradicting the assumptions that S is a stationary subset of µ+ ∩ cof(λ) and λ < κ.

Proof ctd. In M, j(S) is a stationary subset of j(µ+) ∩ cof(λ). Claim M |= “j(S) ∩ γ is stationary in γ”. Proof. Suppose not, and let C ∈ M be a club in γ disjoint from j(S). Since j is continuous at ordinals of cofinality < κ, D = C ∩ j µ+ is a < κ-club in γ disjoint from j(S). Thus, E = j−1“D is a < κ-club in µ+ disjoint from S, contradicting the assumptions that S is a stationary subset of µ+ ∩ cof(λ) and λ < κ. Thus, M |= “j(S) reflects at γ”, so, by elementarity, V |= “S reflects at some ordinal < µ+”.

Supercompacts, forcing, and reflection Fact Suppose n < ω and every stationary subset of ℵω·n+1 reflects. Then APℵω·n holds.

Supercompacts, forcing, and reflection Fact Suppose n < ω and every stationary subset of ℵω·n+1 reflects. Then APℵω·n holds. Theorem (Magidor) Let κn | n < ω be an increasing sequence of cardinals, with κ0 = ω and κn supercompact for all other n. Let P be the full-support iteration of Coll(κn, κn+1) for n < ω. Then, in V P, every stationary subset of ℵω+1 reflects.

Supercompacts, forcing, and reflection Fact Suppose n < ω and every stationary subset of ℵω·n+1 reflects. Then APℵω·n holds. Theorem (Magidor) Let κn | n < ω be an increasing sequence of cardinals, with κ0 = ω and κn supercompact for all other n. Let P be the full-support iteration of Coll(κn, κn+1) for n < ω. Then, in V P, every stationary subset of ℵω+1 reflects. Theorem (Chayut) Let κn | n < ω be as above, and let µ = sup({κn | n < ω}). Let λn | n < ω be a sequence of regular cardinals such that, for all n < ω, κn ≤ λn < κn+1. Let P be the full-support iteration of Coll(λn, κn+1) and, in V P, let Q be the poset to force APµ. Then, in V P∗Q, every stationary subset of µ+ reflects.

Section 3 Bounded stationary reflection

Bounded stationary reflection Definition Let µ be a singular cardinal, and let κ = µ+. Bounded stationary reflection holds at κ if every stationary subset of κ reflects, but there is a stationary S ⊆ κ and a λ < µ such that S does not reflect at any ordinals in κ ∩ cof(≥ λ).

Bounded stationary reflection Definition Let µ be a singular cardinal, and let κ = µ+. Bounded stationary reflection holds at κ if every stationary subset of κ reflects, but there is a stationary S ⊆ κ and a λ < µ such that S does not reflect at any ordinals in κ ∩ cof(≥ λ). Fact Bounded stationary reflection cannot hold at ℵω+1. Proof. For T ⊆ ℵω+1, let T∗ = {α < ℵω+1 | T reflects at α}. Note that, under our assumptions, if T is stationary, then T∗ is also stationary.

Bounded stationary reflection Definition Let µ be a singular cardinal, and let κ = µ+. Bounded stationary reflection holds at κ if every stationary subset of κ reflects, but there is a stationary S ⊆ κ and a λ < µ such that S does not reflect at any ordinals in κ ∩ cof(≥ λ). Fact Bounded stationary reflection cannot hold at ℵω+1. Proof. For T ⊆ ℵω+1, let T∗ = {α < ℵω+1 | T reflects at α}. Note that, under our assumptions, if T is stationary, then T∗ is also stationary. Now, let S ⊆ ℵω+1 be stationary. Define Sn | n < ω as follows: S = S0. Given Sn, find mn such that Tn = Sn ∩ Sℵω+1 ℵmn is stationary, and let Sn+1 = T∗ n . The result follows from the observations that mn | n < ω is strictly increasing and, for every 0 < n < ω, S reflects at every ordinal in Sn.

Bounded stationary reflection Theorem (Cummings, L-H) Assume there are ω · 2-many supercompact cardinals. Then there is a forcing extension in which bounded stationary reflection holds at ℵω·2+1.

Bounded stationary reflection Theorem (Cummings, L-H) Assume there are ω · 2-many supercompact cardinals. Then there is a forcing extension in which bounded stationary reflection holds at ℵω·2+1. Proof sketch. Let κi | i ≤ ω · 2 + 1 be an increasing, continuous sequence of cardinals such that: • κ0 = ω. • If 0 ≤ i < ω or ω < i < ω · 2, then κi+1 is supercompact. • κω+1 = κ+ ω and κω·2+1 = κ+ ω·2 .

Bounded stationary reflection Theorem (Cummings, L-H) Assume there are ω · 2-many supercompact cardinals. Then there is a forcing extension in which bounded stationary reflection holds at ℵω·2+1. Proof sketch. Let κi | i ≤ ω · 2 + 1 be an increasing, continuous sequence of cardinals such that: • κ0 = ω. • If 0 ≤ i < ω or ω < i < ω · 2, then κi+1 is supercompact. • κω+1 = κ+ ω and κω·2+1 = κ+ ω·2 . Let P = P0 ∗ ˙ P1, where P0 is the full-support iteration of Coll(κi , κi+1) for i < ω and, in V P0 , P1 is the full-support iteration of Coll(κi , κi+1) for ω + 1 ≤ i < ω · 2.

Proof sketch ctd. In V P, we have κi = ℵi for all i ≤ ω · 2 + 1. Let κ = κω·2+1. We also have κ<κ = κ. Fix an enumeration a of all bounded subsets of κ in order type κ, and let Q be the forcing to shoot a club through the set of ordinals that are approachable with respect to a.

Proof sketch ctd. In V P, we have κi = ℵi for all i ≤ ω · 2 + 1. Let κ = κω·2+1. We also have κ<κ = κ. Fix an enumeration a of all bounded subsets of κ in order type κ, and let Q be the forcing to shoot a club through the set of ordinals that are approachable with respect to a. In V P∗ ˙ Q, every stationary subset of κ reflects. However, bounded stationary reflection necessarily fails at κ. Thus, let S be the forcing poset whose conditions are bounded subsets s ⊆ κ ∩ cof(ω) such that, for all α ∈ κ ∩ cof(≥ κω+1), S ∩ α is not stationary in α. S is ordered by end-extension.

Proof sketch ctd. In V P, we have κi = ℵi for all i ≤ ω · 2 + 1. Let κ = κω·2+1. We also have κ<κ = κ. Fix an enumeration a of all bounded subsets of κ in order type κ, and let Q be the forcing to shoot a club through the set of ordinals that are approachable with respect to a. In V P∗ ˙ Q, every stationary subset of κ reflects. However, bounded stationary reflection necessarily fails at κ. Thus, let S be the forcing poset whose conditions are bounded subsets s ⊆ κ ∩ cof(ω) such that, for all α ∈ κ ∩ cof(≥ κω+1), S ∩ α is not stationary in α. S is ordered by end-extension. S is a cardinal-preserving forcing poset that adds a stationary subset of κ that does not reflect at any ordinals in κ ∩ cof(≥ κω+1). With a bit of work, one can show that, in V P∗ ˙ Q∗˙ S, it is still the case that every stationary subset of κ reflects. Thus, bounded stationary reflection holds at κ = ℵω·2+1.

Bounded stationary reflection Theorem (Cummings, L-H) Suppose there is a proper class of supercompact cardinals. Then there is a class forcing extension in which, for every singular cardinal µ > ℵω that is not a cardinal fixed point, bounded stationary reflection holds at µ+.

Thank you!