Regressive order on subsets of regular cardinals anchez Terraf 1 Y. Peng P . S´ W. Weiss University of Toronto CIEM-FaMAF — Universidad Nacional de C´ ordoba International Congress of Mathematicians Rio de Janeiro, 2018 / 08 / 02 1 Supported by CONICET and SeCyT-UNC.
Summary Intro: The club filter on ω 1 1 Club sets Stationary sets Pressing-down lemma The regressive order on [ ω 1 ] ℵ 1 2 Problems and examples Lower bounds Characterization of < R for ω 1 Generalizations 3 < β -to-one regressive maps Many maximal elements ] ω 1 Regressive order on [ [ [ ω 1 ] ] Y. Peng, PST, W. Weiss (UofT, UNC) ICM2018. Rio, 2018/08/02 2 / 15
The club filter on ω 1 Let ω 1 be the first uncountable ordinal. C ⊆ ω 1 is club ( closed unbounded ) if it is unbounded in ω 1 and it contains all of its limit points. Analogy : Borel sets of measure 1 in [ 0 , 1 ] . Example The set Lim of limit ordinals in ω 1 : { ω , ω · 2 , ω · 3 ,..., ω 2 , ω 2 + ω ,... } Given g : ω 1 → ω 1 , C g : = { β ∈ ω 1 : ∀ α < β ( g ( α ) < β ) } Lemma Clubs are closed under countable intersections. Hence subsets containing a club form a filter, the club filter. Analogy : Lebesgue measurable sets of measure 1 in [ 0 , 1 ] . ] ω 1 Regressive order on [ [ [ ω 1 ] ] Y. Peng, PST, W. Weiss (UofT, UNC) ICM2018. Rio, 2018/08/02 3 / 15
The club filter on ω 1 Let ω 1 be the first uncountable ordinal. C ⊆ ω 1 is club ( closed unbounded ) if it is unbounded in ω 1 and it contains all of its limit points. Analogy : Borel sets of measure 1 in [ 0 , 1 ] . Example The set Lim of limit ordinals in ω 1 : { ω , ω · 2 , ω · 3 ,..., ω 2 , ω 2 + ω ,... } Given g : ω 1 → ω 1 , C g : = { β ∈ ω 1 : ∀ α < β ( g ( α ) < β ) } Lemma Clubs are closed under countable intersections. Hence subsets containing a club form a filter, the club filter. Analogy : Lebesgue measurable sets of measure 1 in [ 0 , 1 ] . ] ω 1 Regressive order on [ [ [ ω 1 ] ] Y. Peng, PST, W. Weiss (UofT, UNC) ICM2018. Rio, 2018/08/02 3 / 15
The club filter on ω 1 Let ω 1 be the first uncountable ordinal. C ⊆ ω 1 is club ( closed unbounded ) if it is unbounded in ω 1 and it contains all of its limit points. Analogy : Borel sets of measure 1 in [ 0 , 1 ] . Example The set Lim of limit ordinals in ω 1 : { ω , ω · 2 , ω · 3 ,..., ω 2 , ω 2 + ω ,... } Given g : ω 1 → ω 1 , C g : = { β ∈ ω 1 : ∀ α < β ( g ( α ) < β ) } Lemma Clubs are closed under countable intersections. Hence subsets containing a club form a filter, the club filter. Analogy : Lebesgue measurable sets of measure 1 in [ 0 , 1 ] . ] ω 1 Regressive order on [ [ [ ω 1 ] ] Y. Peng, PST, W. Weiss (UofT, UNC) ICM2018. Rio, 2018/08/02 3 / 15
The club filter on ω 1 Let ω 1 be the first uncountable ordinal. C ⊆ ω 1 is club ( closed unbounded ) if it is unbounded in ω 1 and it contains all of its limit points. Analogy : Borel sets of measure 1 in [ 0 , 1 ] . Example The set Lim of limit ordinals in ω 1 : { ω , ω · 2 , ω · 3 ,..., ω 2 , ω 2 + ω ,... } Given g : ω 1 → ω 1 , C g : = { β ∈ ω 1 : ∀ α < β ( g ( α ) < β ) } Lemma Clubs are closed under countable intersections. Hence subsets containing a club form a filter, the club filter. Analogy : Lebesgue measurable sets of measure 1 in [ 0 , 1 ] . ] ω 1 Regressive order on [ [ [ ω 1 ] ] Y. Peng, PST, W. Weiss (UofT, UNC) ICM2018. Rio, 2018/08/02 3 / 15
The club filter on ω 1 Let ω 1 be the first uncountable ordinal. C ⊆ ω 1 is club ( closed unbounded ) if it is unbounded in ω 1 and it contains all of its limit points. Analogy : Borel sets of measure 1 in [ 0 , 1 ] . Example The set Lim of limit ordinals in ω 1 : { ω , ω · 2 , ω · 3 ,..., ω 2 , ω 2 + ω ,... } Given g : ω 1 → ω 1 , C g : = { β ∈ ω 1 : ∀ α < β ( g ( α ) < β ) } Lemma Clubs are closed under countable intersections. Hence subsets containing a club form a filter, the club filter. Analogy : Lebesgue measurable sets of measure 1 in [ 0 , 1 ] . ] ω 1 Regressive order on [ [ [ ω 1 ] ] Y. Peng, PST, W. Weiss (UofT, UNC) ICM2018. Rio, 2018/08/02 3 / 15
Stationary sets N ⊆ ω 1 is nonstationary if its complement contains a club. They form an ideal, dual to the club filter. Analogy : sets of outer measure 0 in [ 0 , 1 ] . Example N 0 : = � { ( δ , δ + ω ] : δ ∈ ω 1 limit } ⊆ ω 1 \{ ω α : 2 ≤ α ∈ ω 1 } . S ⊆ ω 1 is stationary if it is not nonstationary. Equivalently, S intersects every club. Analogy : sets of positive outer measure in [ 0 , 1 ] . Some properties Every stationary set is unbounded in ω 1 (intersects every [ α , ω 1 ) ). Every stationary set contains (many) limit ordinals. ] ω 1 Regressive order on [ [ [ ω 1 ] ] Y. Peng, PST, W. Weiss (UofT, UNC) ICM2018. Rio, 2018/08/02 4 / 15
Stationary sets N ⊆ ω 1 is nonstationary if its complement contains a club. They form an ideal, dual to the club filter. Analogy : sets of outer measure 0 in [ 0 , 1 ] . Example N 0 : = � { ( δ , δ + ω ] : δ ∈ ω 1 limit } ⊆ ω 1 \{ ω α : 2 ≤ α ∈ ω 1 } . S ⊆ ω 1 is stationary if it is not nonstationary. Equivalently, S intersects every club. Analogy : sets of positive outer measure in [ 0 , 1 ] . Some properties Every stationary set is unbounded in ω 1 (intersects every [ α , ω 1 ) ). Every stationary set contains (many) limit ordinals. ] ω 1 Regressive order on [ [ [ ω 1 ] ] Y. Peng, PST, W. Weiss (UofT, UNC) ICM2018. Rio, 2018/08/02 4 / 15
Stationary sets N ⊆ ω 1 is nonstationary if its complement contains a club. They form an ideal, dual to the club filter. Analogy : sets of outer measure 0 in [ 0 , 1 ] . Example N 0 : = � { ( δ , δ + ω ] : δ ∈ ω 1 limit } ⊆ ω 1 \{ ω α : 2 ≤ α ∈ ω 1 } . S ⊆ ω 1 is stationary if it is not nonstationary. Equivalently, S intersects every club. Analogy : sets of positive outer measure in [ 0 , 1 ] . Some properties Every stationary set is unbounded in ω 1 (intersects every [ α , ω 1 ) ). Every stationary set contains (many) limit ordinals. ] ω 1 Regressive order on [ [ [ ω 1 ] ] Y. Peng, PST, W. Weiss (UofT, UNC) ICM2018. Rio, 2018/08/02 4 / 15
Stationary sets N ⊆ ω 1 is nonstationary if its complement contains a club. They form an ideal, dual to the club filter. Analogy : sets of outer measure 0 in [ 0 , 1 ] . Example N 0 : = � { ( δ , δ + ω ] : δ ∈ ω 1 limit } ⊆ ω 1 \{ ω α : 2 ≤ α ∈ ω 1 } . S ⊆ ω 1 is stationary if it is not nonstationary. Equivalently, S intersects every club. Analogy : sets of positive outer measure in [ 0 , 1 ] . Some properties Every stationary set is unbounded in ω 1 (intersects every [ α , ω 1 ) ). Every stationary set contains (many) limit ordinals. ] ω 1 Regressive order on [ [ [ ω 1 ] ] Y. Peng, PST, W. Weiss (UofT, UNC) ICM2018. Rio, 2018/08/02 4 / 15
Stationary sets N ⊆ ω 1 is nonstationary if its complement contains a club. They form an ideal, dual to the club filter. Analogy : sets of outer measure 0 in [ 0 , 1 ] . Example N 0 : = � { ( δ , δ + ω ] : δ ∈ ω 1 limit } ⊆ ω 1 \{ ω α : 2 ≤ α ∈ ω 1 } . S ⊆ ω 1 is stationary if it is not nonstationary. Equivalently, S intersects every club. Analogy : sets of positive outer measure in [ 0 , 1 ] . Some properties Every stationary set is unbounded in ω 1 (intersects every [ α , ω 1 ) ). Every stationary set contains (many) limit ordinals. ] ω 1 Regressive order on [ [ [ ω 1 ] ] Y. Peng, PST, W. Weiss (UofT, UNC) ICM2018. Rio, 2018/08/02 4 / 15
The pressing-down lemma Intuition Sets in the club filter have “density 1 at infinity.” Stationary sets have “positive density at infinity” We can’t bring a stationary set from infinity in a 1-1 fashion. Fodor’s Lemma Let S ⊆ ω 1 be stationary and f : S → ω 1 such that f ( α ) < α for all α ∈ S . Then there exists β ∈ ω 1 such that f − 1 ( β ) is stationary (viz., uncountable). ] ω 1 Regressive order on [ [ [ ω 1 ] ] Y. Peng, PST, W. Weiss (UofT, UNC) ICM2018. Rio, 2018/08/02 5 / 15
The pressing-down lemma Intuition Sets in the club filter have “density 1 at infinity.” Stationary sets have “positive density at infinity” We can’t bring a stationary set from infinity in a 1-1 fashion. Fodor’s Lemma Let S ⊆ ω 1 be stationary and f : S → ω 1 such that f ( α ) < α for all α ∈ S . Then there exists β ∈ ω 1 such that f − 1 ( β ) is stationary (viz., uncountable). ] ω 1 Regressive order on [ [ [ ω 1 ] ] Y. Peng, PST, W. Weiss (UofT, UNC) ICM2018. Rio, 2018/08/02 5 / 15
The pressing-down lemma Intuition Sets in the club filter have “density 1 at infinity.” Stationary sets have “positive density at infinity” We can’t bring a stationary set from infinity in a 1-1 fashion. Fodor’s Lemma Let S ⊆ ω 1 be stationary and f : S → ω 1 such that f ( α ) < α for all α ∈ S . Then there exists β ∈ ω 1 such that f − 1 ( β ) is stationary (viz., uncountable). ] ω 1 Regressive order on [ [ [ ω 1 ] ] Y. Peng, PST, W. Weiss (UofT, UNC) ICM2018. Rio, 2018/08/02 5 / 15
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