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University of South Alabama Monday, August 27, 2012. MA 542 Fall Semester 2012 Courtesy a Student Topology Homework 1 Problem. Show that the power set,P(S), together with two binary operations ∪, ∩, and a unary operation (the compliment) is a Boolean algebra. 0 Proof. We must show that the two binary operations commute. Let P(S) denote the power set of some arbitrary set S. Let A ∈ P(S), B ∈ P(S) and C ∈ P(S) be arbitrary sets. Let α ∈ (A ∪ B) ⇔ (α ∈ A) ∨ (α ∈ B) ⇔ (α ∈ B) ∨ (α ∈ A) ⇔ α ∈ (B ∪ A) thus A ∪ B = B ∪ A. Let α ∈ (A ∩ B) ⇔ (α ∈ A) ∧ (α ∈ B) ⇔ (α ∈ B) ∧ (α ∈ A) ⇔ α ∈ (B ∩ A) thus A ∩ B = B ∩ A. Next we will show that there exist elements ∅, S with A ∪ ∅ = A, A ∩ S = A for all A ∈ P(S). For any set S, ∅ ∈ P(S) and S ∈ P(S) by definition. Let α ∈ (A ∪ ∅) ⇒ (α ∈ A) ∨ (α ∈ ∅) but α ∈ / ∅ thus α ∈ A hence (A ∪ ∅) ⊂ A. Let α ∈ A ⇒ α ∈ (A ∪ ∅) thus A ⊂ (A ∪ ∅) and we conclude A ∪ ∅ = A for all A ∈ P(S). A ⊂ S ⇔ ∀α : (α ∈ A) ⇒ (α ∈ S) thus A ∩ S = A for all A ∈ P(S). Now we must show that the distributive laws hold, i.e., A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) and A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). Let α ∈ A ∩ (B ∪ C) ⇔ (α ∈ A) ∧ (α ∈ (B ∪ C)) ⇔ (α ∈ A) ∧ ((α ∈ B) ∨ (α ∈ C)) ⇔ ((α ∈ A) ∧ (α ∈ B)) ∨ ((α ∈ A) ∧ (α ∈ C)) ⇔ (α ∈ (A ∩ B)) ∨ (α ∈ (A ∩ C)) ⇔ α ∈ ((A ∩ B) ∪ (A ∩ C)) Let α ∈ A ∪ (B ∩ C) ⇔ (α ∈ A) ∨ (α ∈ (B ∩ C)) ⇔ (α ∈ A) ∨ ((α ∈ B) ∧ (α ∈ C)) ⇔ ((α ∈ A) ∨ (α ∈ B)) ∧ ((α ∈ A) ∨ (α ∈ C)) ⇔ (α ∈ (A ∪ B)) ∧ (α ∈ (A ∪ C)) ⇔ α ∈ ((A ∪ B) ∩ (A ∪ C)) Hence the distributive laws A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) and A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) hold. Finally we will show A ∩ A = ∅ and A ∪ A = S for each A ∈ P(S). Assmue A ∩ A 6= ∅, then there exist some α ∈ (A ∩ A). Let α ∈ (A ∩ A) ⇔ (α ∈ A) ∧ (α ∈ A) ⇔ (α ∈ A) ∧ (α ∈ / A) but this is a contradiction. Therefore A ∩ A = ∅ for each A ∈ P(S). Lastly observe that A ∪ A = A ∪ (S \ A) = S follows from the definition of the compliment of A with respect to S. Theorem 6.4. If f : X → Y , then for the induced map f : P(X) → P(Y ): S S 1. f ( Aα ) = f (Aα ). α α T T 2. f ( Aα ) ⊂ f (Aα ). α α Proof. [ 1. Let y ∈ f ( Aα ) ⇔ ∃x ∈ α [ Aα : f (x) = y α ⇔ ∃α : x ∈ Aα s.t. f (x) = y ⇔ ∃α : f (x) ∈ f (Aα ) [ [ [ ⇔ f (x) ∈ f (Aα ) thus f ( Aα ) = f (Aα ) α \ 2. Let y ∈ f ( Aα ) ⇒ α ∃x ∈ \ α α Aα : f (x) = y α ⇒ ∀α : x ∈ Aα ⇒ ∀α : f (x) ∈ f (Aα ) \ \ \ f (Aα ) f (Aα ) thus f ( Aα ) ⊂ ⇒ f (x) ∈ α α α T T To see that f ( Aα ) 6= f (Aα ) let f : R → R be the constant map x 7→ 2. Then, with A = [0, 1], B = [2, 3], α α we get ∅ = f (A ∩ B) 6= f (A) ∩ f (B) = {2}. Theorem 6.5. If f : X → Y , then: 1. For each A ⊂ X, f −1 [f (A)] ⊃ A. 2. For each A ⊂ X and B ⊂ Y , f [f −1 (B) ∩ A] = B ∩ f (A); in particular, f [f −1 (B)] = B ∩ f (X). Proof. 1. Let x ∈ A ⇒ f (x) ∈ f (A) ⇒ x ∈ f −1 (f (A)) thus A ⊂ f −1 [f (A)] To see that f −1 [f (A)] 6= A let f : R → R be the map x 7→ x2 . Then with A = {2}, we get f (A) = 4 ⇒ f −1 (f (A)) = f −1 (4) = {±2}. 2. Let y ∈ f [f −1 (B) ∩ A] ⇔ ⇔ ∃x ∈ (f −1 (B) ∩ A) : f (x) = y x ∈ f −1 (B) ∧ (x ∈ A) ⇔ (α ∈ B) ∧ (α ∈ f (A)) ⇔ α ∈ (B ∩ f (A)) thus f [f −1 (B) ∩ A] = B ∩ f (A) Furthermore if we let A = X, then we get f [f −1 (B)] = B ∩ f (X).