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On Properties of Families of Sets Lecture 2 On properties of families of sets Lecture 2 Lajos Soukup Alfréd Rényi Institute of Mathematics Hungarian Academy of Sciences http://www.renyi.hu/∼soukup 7th Young Set Theory Workshop Recapitulation Recapitulation Definition: A family A⊂P(X) has property B iff χ(A)= 2, where the chromatic number of A is defined as follows: χ(A)=min{λ | ∃f : X → λ ∀A ∈ A |f [A]|≥ 2}. Recapitulation Definition: A family A⊂P(X) has property B iff χ(A)= 2, where the chromatic number of A is defined as follows: χ(A)=min{λ | ∃f : X → λ ∀A ∈ A |f [A]|≥ 2}. Theorem (E. W. Miller, 1937) ω There is an almost disjoint A⊂ ω with χ(A)= ω. Recapitulation Definition: A family A⊂P(X) has property B iff χ(A)= 2, where the chromatic number of A is defined as follows: χ(A)=min{λ | ∃f : X → λ ∀A ∈ A |f [A]|≥ 2}. Theorem (E. W. Miller, 1937) ω There is an almost disjoint A⊂ ω with χ(A)= ω. Theorem (Gy. Elekes, Gy Hoffman, 1973) ω For all infinite cardinal κ there is an almost disjoint A⊂ X with χ(A) ≥ κ. Recapitulation Definition: A family A⊂P(X) has property B iff χ(A)= 2, where the chromatic number of A is defined as follows: χ(A)=min{λ | ∃f : X → λ ∀A ∈ A |f [A]|≥ 2}. Theorem (E. W. Miller, 1937) ω There is an almost disjoint A⊂ ω with χ(A)= ω. Theorem (Gy. Elekes, Gy Hoffman, 1973) ω For all infinite cardinal κ there is an almost disjoint A⊂ X with χ(A) ≥ κ. Definition: A is n-almost-disjoint iff |A ∩ A′| < n for all A 6= A′ ∈ A Recapitulation Definition: A family A⊂P(X) has property B iff χ(A)= 2, where the chromatic number of A is defined as follows: χ(A)=min{λ | ∃f : X → λ ∀A ∈ A |f [A]|≥ 2}. Theorem (E. W. Miller, 1937) ω There is an almost disjoint A⊂ ω with χ(A)= ω. Theorem (Gy. Elekes, Gy Hoffman, 1973) ω For all infinite cardinal κ there is an almost disjoint A⊂ X with χ(A) ≥ κ. Definition: A is n-almost-disjoint iff |A ∩ A′| < n for all A 6= A′ ∈ A Theorem (E. W. Miller, 1937) An n-almost disjoint family of infinite sets has property B. Covering of the plain. Covering of the plain. Theorem (Sierpinski) CH holds iff R2 is the union of countably many functions and their inverses. Covering of the plain. Theorem (Sierpinski) CH holds iff R2 is the union of countably many functions and their inverses. −1 • f = R(90◦)(−f ) Covering of the plain. Theorem (Sierpinski) CH holds iff R2 is the union of countably many functions and their inverses. −1 • f = R(90◦)(−f ) 2 2 • R(α) : R → R the rotation by α degree around the origin. Covering of the plain. Theorem (Sierpinski) CH holds iff R2 is the union of countably many functions and their inverses. −1 • f = R(90◦)(−f ) 2 2 • R(α) : R → R the rotation by α degree around the origin. • If CH holds, then R2 is the union of countably many rotations of functions. Covering of the plain. Theorem (Sierpinski) CH holds iff R2 is the union of countably many functions and their inverses. −1 • f = R(90◦)(−f ) 2 2 • R(α) : R → R the rotation by α degree around the origin. • If CH holds, then R2 is the union of countably many rotations of functions. • Sierpinski, 1951: Is the converse true? Covering of the plain. Theorem (Sierpinski) CH holds iff R2 is the union of countably many functions and their inverses. −1 • f = R(90◦)(−f ) 2 2 • R(α) : R → R the rotation by α degree around the origin. • If CH holds, then R2 is the union of countably many rotations of functions. • Sierpinski, 1951: Is the converse true? Theorem (Davies, 1963) R2 is the union of countably many rotations of functions. Covering of the plain. Theorem (Sierpinski) CH holds iff R2 is the union of countably many functions and their inverses. −1 • f = R(90◦)(−f ) 2 2 • R(α) : R → R the rotation by α degree around the origin. • If CH holds, then R2 is the union of countably many rotations of functions. • Sierpinski, 1951: Is the converse true? Theorem (Davies, 1963) R2 is the union of countably many rotations of functions. If α0, α1,... are pairwise different angles between 0 and π, then there are function f0, f1 ... such that 2 R = Sn∈ω R(αn)(fn). 2 Thm. There are function f0, f1 ... such that R = Sn∈ω R(αn )(fn). 2 Thm. There are function f0, f1 ... such that R = Sn∈ω R(αn )(fn). x1 = R(α1)(x) x0 = R(α0)(x) 2 Thm. There are function f0, f1 ... such that R = Sn∈ω R(αn )(fn). e(P, x0) x1 = R(α1)(x) e(P, x1) P x0 = R(α0)(x) 2 Thm. There are function f0, f1 ... such that R = Sn∈ω R(αn )(fn). e(P, x0) x1 = R(α1)(x) e(P, x1) P x0 = R(α0)(x) 2 • Write E(P)= {e(P, xn): n ∈ ω}. Let E= {E(P): P ∈ R }. 2 Thm. There are function f0, f1 ... such that R = Sn∈ω R(αn )(fn). e(P, x0) x1 = R(α1)(x) e(P, x1) P Rα0 (f0) Rα1 (f1) x0 = R(α0)(x) 2 • Write E(P)= {e(P, xn): n ∈ ω}. Let E= {E(P): P ∈ R }. 2 • Assume R = Sn∈ω R(αn)(fn). 2 Thm. There are function f0, f1 ... such that R = Sn∈ω R(αn )(fn). e(P, x0))= e(P) x1 = R(α1)(x) e(P, x1) P Rα0 (f0) Rα1 (f1) x0 = R(α0)(x) 2 • Write E(P)= {e(P, xn): n ∈ ω}. Let E= {E(P): P ∈ R }. 2 • Assume R = Sn∈ω R(αn)(fn). • if P ∈ R(αn)(fn), let e(P)= e(P, xn) 2 Thm. There are function f0, f1 ... such that R = Sn∈ω R(αn )(fn). e(P, x0))= e(P) x1 = R(α1)(x) e(P, x1) P Rα0 (f0) Rα1 (f1) x0 = R(α0)(x) 2 • Write E(P)= {e(P, xn): n ∈ ω}. Let E= {E(P): P ∈ R }. 2 • Assume R = Sn∈ω R(αn)(fn). • if P ∈ R(αn)(fn), let e(P)= e(P, xn) • P 6= Q implies e(P) 6= e(Q). 2 Thm. There are function f0, f1 ... such that R = Sn∈ω R(αn )(fn). e(P, x0))= e(P) x1 = R(α1)(x) e(P, x1) P Q Rα0 (f0) Rα1 (f1) x0 = R(α0)(x) 2 • Write E(P)= {e(P, xn): n ∈ ω}. Let E= {E(P): P ∈ R }. 2 • Assume R = Sn∈ω R(αn)(fn). • if P ∈ R(αn)(fn), let e(P)= e(P, xn) • P 6= Q implies e(P) 6= e(Q). 2 Thm. There are function f0, f1 ... such that R = Sn∈ω R(αn )(fn). e(P, x0))= e(P) x1 = R(α1)(x) e(P, x1) P e(Q, x0)= e(Q) Q Rα0 (f0) Rα1 (f1) x0 = R(α0)(x) 2 • Write E(P)= {e(P, xn): n ∈ ω}. Let E= {E(P): P ∈ R }. 2 • Assume R = Sn∈ω R(αn)(fn). • if P ∈ R(αn)(fn), let e(P)= e(P, xn) • P 6= Q implies e(P) 6= e(Q). 2 Thm. There are function f0, f1 ... such that R = Sn∈ω R(αn )(fn). e(P, x0))= e(P) x1 = R(α1)(x) e(P, x1) P e(Q, x0)= e(Q) Q Rα0 (f0) Rα1 (f1) e(Q, x1)= e(Q) x0 = R(α0)(x) 2 • Write E(P)= {e(P, xn): n ∈ ω}. Let E= {E(P): P ∈ R }. 2 • Assume R = Sn∈ω R(αn)(fn). • if P ∈ R(αn)(fn), let e(P)= e(P, xn) • P 6= Q implies e(P) 6= e(Q). 2 Thm. There are function f0, f1 ... such that R = Sn∈ω R(αn )(fn). e(P, x0))= e(P) x1 = R(α1)(x) e(P, x1) P e(Q, x0)= e(Q) Q Rα0 (f0) Rα1 (f1) e(Q, x1)= e(Q) x0 = R(α0)(x) 2 • Write E(P)= {e(P, xn): n ∈ ω}. Let E= {E(P): P ∈ R }. 2 • Assume R = Sn∈ω R(αn)(fn). • if P ∈ R(αn)(fn), let e(P)= e(P, xn) • P 6= Q implies e(P) 6= e(Q). • E has a transversal, i.e. an injective choice function. e(P, x0) x1 = R(α1)(x) e(P, x1) P x0 = R(α0)(x) 2 • Write E(P)= {e(P, xn): n ∈ ω} Let E= {E(P): P ∈ R }. e(P, x0))= e(P) x1 = R(α1)(x) e(P, x1) P x0 = R(α0)(x) 2 • Write E(P)= {e(P, xn): n ∈ ω} Let E= {E(P): P ∈ R }. • Assume E has a transversal (an injective choice function) e. e(P, x0))= e(P) x1 = R(α1)(x) e(P, x1) P e(Q, x0)= e(Q) Q x0 = R(α0)(x) 2 • Write E(P)= {e(P, xn): n ∈ ω} Let E= {E(P): P ∈ R }. • Assume E has a transversal (an injective choice function) e. • Let Fn={P : e(P)= e(P, xn)}. e(P, x0))= e(P) x1 = R(α1)(x) e(P, x1) P e(Q, x0)= e(Q) Q Rα0 (f0) x0 = R(α0)(x) 2 • Write E(P)= {e(P, xn): n ∈ ω} Let E= {E(P): P ∈ R }.
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