Ramsey regularity, MAD families, and their relatives David - PowerPoint PPT Presentation

Ramsey regularity, MAD families, and their relatives David Schrittesser (KGRC) Joint work with Asger Törnquist Arctic Set Theory 4 January 22, 2019 David Schrittesser (KGRC) Regularity and MAD families Arctic 4 1 / 12

Why there are no analytic MAD families Theorem 1 (Mathias) There are no analytic MAD families. David Schrittesser (KGRC) Regularity and MAD families Arctic 4 2 / 12

Why there are no analytic MAD families Theorem 1 (Mathias) There are no analytic MAD families. I will sketch a proof based on invariance and the fact that analytic sets are completely Ramsey. David Schrittesser (KGRC) Regularity and MAD families Arctic 4 2 / 12

Why there are no analytic MAD families Theorem 1 (Mathias) There are no analytic MAD families. I will sketch a proof based on invariance and the fact that analytic sets are completely Ramsey. Sketch of proof. Suppose that T is tree on 2 × ω such that A = p [ T ] is an a.d. family. We show A is not maximal. David Schrittesser (KGRC) Regularity and MAD families Arctic 4 2 / 12

An Invariant Tree For X ∈ [ ω ] ω define T X = { s ∈ T | ( ∃ A ∈ p [ T s ]) A ∩ X is infinite } David Schrittesser (KGRC) Regularity and MAD families Arctic 4 3 / 12

An Invariant Tree For X ∈ [ ω ] ω define T X = { s ∈ T | ( ∃ A ∈ p [ T s ]) A ∩ X is infinite } X ∆ Y ∈ Fin ⇒ T X = T Y , i.e, the tree is invariant , 1 David Schrittesser (KGRC) Regularity and MAD families Arctic 4 3 / 12

An Invariant Tree For X ∈ [ ω ] ω define T X = { s ∈ T | ( ∃ A ∈ p [ T s ]) A ∩ X is infinite } X ∆ Y ∈ Fin ⇒ T X = T Y , i.e, the tree is invariant , 1 T X is a sub-tree of T , 2 David Schrittesser (KGRC) Regularity and MAD families Arctic 4 3 / 12

An Invariant Tree For X ∈ [ ω ] ω define T X = { s ∈ T | ( ∃ A ∈ p [ T s ]) A ∩ X is infinite } X ∆ Y ∈ Fin ⇒ T X = T Y , i.e, the tree is invariant , 1 T X is a sub-tree of T , 2 s ∈ T X ⇐ s ] � = ∅ (that is, T X is pruned), ⇒ [ T X 3 David Schrittesser (KGRC) Regularity and MAD families Arctic 4 3 / 12

An Invariant Tree For X ∈ [ ω ] ω define T X = { s ∈ T | ( ∃ A ∈ p [ T s ]) A ∩ X is infinite } X ∆ Y ∈ Fin ⇒ T X = T Y , i.e, the tree is invariant , 1 T X is a sub-tree of T , 2 s ∈ T X ⇐ s ] � = ∅ (that is, T X is pruned), ⇒ [ T X 3 ∈ T X ⇐ ∅ / ⇒ ( ∀ A ∈ A ) A ∩ X ∈ Fin ⇐ ⇒ X is a counterexample 4 to maximaliy of A . David Schrittesser (KGRC) Regularity and MAD families Arctic 4 3 / 12

The Main Lemma We need the following crucial lemma. David Schrittesser (KGRC) Regularity and MAD families Arctic 4 4 / 12

The Main Lemma We need the following crucial lemma. Main Lemma Suppose s, t ∈ T X , lh ( s ) = lh ( t ) but p ( s ) � = p ( t ) . David Schrittesser (KGRC) Regularity and MAD families Arctic 4 4 / 12

The Main Lemma We need the following crucial lemma. Main Lemma Suppose s, t ∈ T X , lh ( s ) = lh ( t ) but p ( s ) � = p ( t ) . Then there are s ′ ∈ T X s and t ′ ∈ T X such that t �� � �� � p [ T X p [ T X ⊆ p ( s ′ ) ∩ p ( t ′ ) . s ′ ] ∩ t ′ ] Proof. Otherwise, we could construct s = s 0 ⊏ s 1 ⊏ . . . and t = t 0 ⊏ t 1 ⊏ . . . from T such that � � � � � � p s n ∩ p t n ∈ Fin / n ∈ ω n ∈ ω which contradicts that A is an a.d. family. David Schrittesser (KGRC) Regularity and MAD families Arctic 4 4 / 12

The tilde-operator Fix a sequence � A = � A 0 , A 1 , A 2 , . . . � of distinct elements from A = p [ T ] . David Schrittesser (KGRC) Regularity and MAD families Arctic 4 5 / 12

The tilde-operator Fix a sequence � A = � A 0 , A 1 , A 2 , . . . � of distinct elements from A = p [ T ] . A l ( m ) be the m th element from A l (in its increasing enumeration). Let ˆ David Schrittesser (KGRC) Regularity and MAD families Arctic 4 5 / 12

The tilde-operator Fix a sequence � A = � A 0 , A 1 , A 2 , . . . � of distinct elements from A = p [ T ] . A l ( m ) be the m th element from A l (in its increasing enumeration). Let ˆ Define a map ∼ : [ ω ] ω → [ ω ] ω , B �→ ˜ B by B = { ˆ ˜ A l ( m ) | l ∈ B, m = min B \ ( l + 1) } . David Schrittesser (KGRC) Regularity and MAD families Arctic 4 5 / 12

Properties of the tilde-operator Given any A ∈ A , 1 ( ∀ B ∈ [ ω ] ω )( ∃ B ′ ∈ [ B ] ω ) ˜ B ′ ∩ A ∈ Fin . David Schrittesser (KGRC) Regularity and MAD families Arctic 4 6 / 12

Properties of the tilde-operator Given any A ∈ A , 1 ( ∀ B ∈ [ ω ] ω )( ∃ B ′ ∈ [ B ] ω ) ˜ B ′ ∩ A ∈ Fin . Given any X ∈ [ ω ] ω , 2 ( ∀ B ∈ [ ω ] ω )( ∃ B ′ ∈ [ B ] ω ) ˜ B ′ ⊆ X or ˜ B ′ ⊆ ω \ X, David Schrittesser (KGRC) Regularity and MAD families Arctic 4 6 / 12

Properties of the tilde-operator Given any A ∈ A , 1 ( ∀ B ∈ [ ω ] ω )( ∃ B ′ ∈ [ B ] ω ) ˜ B ′ ∩ A ∈ Fin . Given any X ∈ [ ω ] ω , 2 ( ∀ B ∈ [ ω ] ω )( ∃ B ′ ∈ [ B ] ω ) ˜ B ′ ⊆ X or ˜ B ′ ⊆ ω \ X, Proof of Item 2: Ramsey’s Theorem for pairs, or directly using the pigeon hole principle. David Schrittesser (KGRC) Regularity and MAD families Arctic 4 6 / 12

The Argument There is B 0 ∈ [ ω ] ω and T ∗ such that 1 B = T ∗ ˜ ( ∀ B ∈ [ B 0 ] ω ) T Proof: Using that analytic sets are Ramsey, make X �→ T ˜ X continuous; by invariance, this map must be constant. David Schrittesser (KGRC) Regularity and MAD families Arctic 4 7 / 12

The Argument There is B 0 ∈ [ ω ] ω and T ∗ such that 1 B = T ∗ ˜ ( ∀ B ∈ [ B 0 ] ω ) T Proof: Using that analytic sets are Ramsey, make X �→ T ˜ X continuous; by invariance, this map must be constant. We show p [ T ∗ ] ≤ 1 . 2 David Schrittesser (KGRC) Regularity and MAD families Arctic 4 7 / 12

The Argument There is B 0 ∈ [ ω ] ω and T ∗ such that 1 B = T ∗ ˜ ( ∀ B ∈ [ B 0 ] ω ) T Proof: Using that analytic sets are Ramsey, make X �→ T ˜ X continuous; by invariance, this map must be constant. We show p [ T ∗ ] ≤ 1 . 2 Proof: Use the Main Lemma and properties of the tilde operator! David Schrittesser (KGRC) Regularity and MAD families Arctic 4 7 / 12

The Argument There is B 0 ∈ [ ω ] ω and T ∗ such that 1 B = T ∗ ˜ ( ∀ B ∈ [ B 0 ] ω ) T Proof: Using that analytic sets are Ramsey, make X �→ T ˜ X continuous; by invariance, this map must be constant. We show p [ T ∗ ] ≤ 1 . 2 Proof: Use the Main Lemma and properties of the tilde operator! In fact T ∗ = ∅ . 3 David Schrittesser (KGRC) Regularity and MAD families Arctic 4 7 / 12

The Argument There is B 0 ∈ [ ω ] ω and T ∗ such that 1 B = T ∗ ˜ ( ∀ B ∈ [ B 0 ] ω ) T Proof: Using that analytic sets are Ramsey, make X �→ T ˜ X continuous; by invariance, this map must be constant. We show p [ T ∗ ] ≤ 1 . 2 Proof: Use the Main Lemma and properties of the tilde operator! In fact T ∗ = ∅ . 3 ∈ T ∗ = T ˜ B 0 it follows that ˜ Since ∅ / B 0 is a counterexample to 4 maximality of A . David Schrittesser (KGRC) Regularity and MAD families Arctic 4 7 / 12

‘No MAD families’ from regularity The previous argument can be generalized to show the following: Theorem 2 Suppose the following hold: Dependent Choice (DC), 1 Every relation can be uniformized on a Ramsey positive set, 2 Every subset of [ ω ] ω is completely Ramsey. 3 Then there are no MAD families. David Schrittesser (KGRC) Regularity and MAD families Arctic 4 8 / 12

‘No MAD families’ from regularity The previous argument can be generalized to show the following: Theorem 2 Suppose the following hold: Dependent Choice (DC), 1 Every relation can be uniformized on a Ramsey positive set, 2 Every subset of [ ω ] ω is completely Ramsey. 3 Then there are no MAD families. These hypothesis are true, e.g., in Solvay’s model or under AD in L ( R ) . David Schrittesser (KGRC) Regularity and MAD families Arctic 4 8 / 12

‘No MAD families’ from regularity The previous argument can be generalized to show the following: Theorem 2 Suppose the following hold: Dependent Choice (DC), 1 Every relation can be uniformized on a Ramsey positive set, 2 Every subset of [ ω ] ω is completely Ramsey. 3 Then there are no MAD families. These hypothesis are true, e.g., in Solvay’s model or under AD in L ( R ) . There is a projective version of Theorem 2 whose its hypotheses hold after collapsing an inaccessible, or under PD + DC. David Schrittesser (KGRC) Regularity and MAD families Arctic 4 8 / 12

Sketch of a proof of Theorem 2 Towards a contradiction, suppose uniformization and the Ramsey property and let A be a MAD family. David Schrittesser (KGRC) Regularity and MAD families Arctic 4 9 / 12

Sketch of a proof of Theorem 2 Towards a contradiction, suppose uniformization and the Ramsey property and let A be a MAD family. Define a relation R ⊆ ([ ω ] ω ) 2 as follows: ˜ ( X, A ) ∈ R ⇐ ⇒ X ∩ A / ∈ Fin ∧ A ∈ A David Schrittesser (KGRC) Regularity and MAD families Arctic 4 9 / 12