Covering an uncountable square by countably many continuous - PowerPoint PPT Presentation

Covering an uncountable square by countably many continuous functions Wiesław Kubi´ s Instytut Matematyki Akademia ´ Swie ¸tokrzyska Kielce, POLAND http://www.pu.kielce.pl/˜wkubis/ Measure Theory Edward Marczewski Centennial Conference Be ¸dlewo, 9–15 September 2007 W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 1 / 14

Motivations Theorem (Sierpi´ nski) Let S be a set of cardinality ℵ 1 . Then there exists a sequence of functions { f n : S → S } n ∈ ω , such that � ( f n ∪ f − 1 S × S = n ) . n ∈ ω Proof. We assume that S = ω 1 . For each β ∈ S fix a surjection g β : ω → β + 1. Define f n ( β ) = g β ( n ) . W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 2 / 14

Motivations Theorem (Sierpi´ nski) Let S be a set of cardinality ℵ 1 . Then there exists a sequence of functions { f n : S → S } n ∈ ω , such that � ( f n ∪ f − 1 S × S = n ) . n ∈ ω Proof. We assume that S = ω 1 . For each β ∈ S fix a surjection g β : ω → β + 1. Define f n ( β ) = g β ( n ) . W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 2 / 14

Motivations Theorem (Sierpi´ nski) Let S be a set of cardinality ℵ 1 . Then there exists a sequence of functions { f n : S → S } n ∈ ω , such that � ( f n ∪ f − 1 S × S = n ) . n ∈ ω Proof. We assume that S = ω 1 . For each β ∈ S fix a surjection g β : ω → β + 1. Define f n ( β ) = g β ( n ) . W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 2 / 14

Motivations Theorem (Sierpi´ nski) Let S be a set of cardinality ℵ 1 . Then there exists a sequence of functions { f n : S → S } n ∈ ω , such that � ( f n ∪ f − 1 S × S = n ) . n ∈ ω Proof. We assume that S = ω 1 . For each β ∈ S fix a surjection g β : ω → β + 1. Define f n ( β ) = g β ( n ) . W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 2 / 14

Motivations Theorem (Sierpi´ nski) Let S be a set of cardinality ℵ 1 . Then there exists a sequence of functions { f n : S → S } n ∈ ω , such that � ( f n ∪ f − 1 S × S = n ) . n ∈ ω Proof. We assume that S = ω 1 . For each β ∈ S fix a surjection g β : ω → β + 1. Define f n ( β ) = g β ( n ) . W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 2 / 14

Remark (Sierpi´ nski) If S has the above property then | S | � ℵ 1 . Proof. Fix A ∈ [ S ] ℵ 1 . For each x ∈ A let F x = { f n ( x ): n ∈ ω } . The set � x ∈ A F x has cardinality � ℵ 1 . Suppose p ∈ S is such that p / ∈ F x for x ∈ A . For each a ∈ A there is n ( a ) ∈ ω such that a = f n ( a ) ( p ) . The map a �→ n ( a ) must be one-to-one. A contradiction. W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 3 / 14

Remark (Sierpi´ nski) If S has the above property then | S | � ℵ 1 . Proof. Fix A ∈ [ S ] ℵ 1 . For each x ∈ A let F x = { f n ( x ): n ∈ ω } . The set � x ∈ A F x has cardinality � ℵ 1 . Suppose p ∈ S is such that p / ∈ F x for x ∈ A . For each a ∈ A there is n ( a ) ∈ ω such that a = f n ( a ) ( p ) . The map a �→ n ( a ) must be one-to-one. A contradiction. W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 3 / 14

Remark (Sierpi´ nski) If S has the above property then | S | � ℵ 1 . Proof. Fix A ∈ [ S ] ℵ 1 . For each x ∈ A let F x = { f n ( x ): n ∈ ω } . The set � x ∈ A F x has cardinality � ℵ 1 . Suppose p ∈ S is such that p / ∈ F x for x ∈ A . For each a ∈ A there is n ( a ) ∈ ω such that a = f n ( a ) ( p ) . The map a �→ n ( a ) must be one-to-one. A contradiction. W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 3 / 14

Remark (Sierpi´ nski) If S has the above property then | S | � ℵ 1 . Proof. Fix A ∈ [ S ] ℵ 1 . For each x ∈ A let F x = { f n ( x ): n ∈ ω } . The set � x ∈ A F x has cardinality � ℵ 1 . Suppose p ∈ S is such that p / ∈ F x for x ∈ A . For each a ∈ A there is n ( a ) ∈ ω such that a = f n ( a ) ( p ) . The map a �→ n ( a ) must be one-to-one. A contradiction. W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 3 / 14

Remark (Sierpi´ nski) If S has the above property then | S | � ℵ 1 . Proof. Fix A ∈ [ S ] ℵ 1 . For each x ∈ A let F x = { f n ( x ): n ∈ ω } . The set � x ∈ A F x has cardinality � ℵ 1 . Suppose p ∈ S is such that p / ∈ F x for x ∈ A . For each a ∈ A there is n ( a ) ∈ ω such that a = f n ( a ) ( p ) . The map a �→ n ( a ) must be one-to-one. A contradiction. W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 3 / 14

Remark (Sierpi´ nski) If S has the above property then | S | � ℵ 1 . Proof. Fix A ∈ [ S ] ℵ 1 . For each x ∈ A let F x = { f n ( x ): n ∈ ω } . The set � x ∈ A F x has cardinality � ℵ 1 . Suppose p ∈ S is such that p / ∈ F x for x ∈ A . For each a ∈ A there is n ( a ) ∈ ω such that a = f n ( a ) ( p ) . The map a �→ n ( a ) must be one-to-one. A contradiction. W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 3 / 14

Remark (Sierpi´ nski) If S has the above property then | S | � ℵ 1 . Proof. Fix A ∈ [ S ] ℵ 1 . For each x ∈ A let F x = { f n ( x ): n ∈ ω } . The set � x ∈ A F x has cardinality � ℵ 1 . Suppose p ∈ S is such that p / ∈ F x for x ∈ A . For each a ∈ A there is n ( a ) ∈ ω such that a = f n ( a ) ( p ) . The map a �→ n ( a ) must be one-to-one. A contradiction. W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 3 / 14

Remark (Sierpi´ nski) If S has the above property then | S | � ℵ 1 . Proof. Fix A ∈ [ S ] ℵ 1 . For each x ∈ A let F x = { f n ( x ): n ∈ ω } . The set � x ∈ A F x has cardinality � ℵ 1 . Suppose p ∈ S is such that p / ∈ F x for x ∈ A . For each a ∈ A there is n ( a ) ∈ ω such that a = f n ( a ) ( p ) . The map a �→ n ( a ) must be one-to-one. A contradiction. W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 3 / 14

Question Is it possible that the square of some uncountable subset of R is covered by countably many continuous real functions and their inverses? In other words: Question Does there exist a family { f n : R → R } n ∈ ω consisting of continuous functions such that � ( f n ∪ f − 1 S × S ⊆ n ) n ∈ ω for some S ∈ [ R ] ℵ 1 ? W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 4 / 14

Question Is it possible that the square of some uncountable subset of R is covered by countably many continuous real functions and their inverses? In other words: Question Does there exist a family { f n : R → R } n ∈ ω consisting of continuous functions such that � ( f n ∪ f − 1 S × S ⊆ n ) n ∈ ω for some S ∈ [ R ] ℵ 1 ? W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 4 / 14

Question Is it possible that the square of some uncountable subset of R is covered by countably many continuous real functions and their inverses? In other words: Question Does there exist a family { f n : R → R } n ∈ ω consisting of continuous functions such that � ( f n ∪ f − 1 S × S ⊆ n ) n ∈ ω for some S ∈ [ R ] ℵ 1 ? W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 4 / 14

Question How about covering by (continuous) non-decreasing functions? n ∈ ω ( f n ∪ f − 1 Suppose S × S ⊆ � n ) , where each f n : S → S is a non-decreasing function. Then both f n and f − 1 are chains in S × S . n Thus, if | S | > ℵ 0 then S is a Countryman type! W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 5 / 14

Question How about covering by (continuous) non-decreasing functions? n ∈ ω ( f n ∪ f − 1 Suppose S × S ⊆ � n ) , where each f n : S → S is a non-decreasing function. Then both f n and f − 1 are chains in S × S . n Thus, if | S | > ℵ 0 then S is a Countryman type! W.Kubi´ s (http://www.pu.kielce.pl/ ∼ wkubis/) Covering ℵ 1 × ℵ 1 by ℵ 0 functions Be ¸dlewo, 11 September 2007 5 / 14