Within the subject of algebra, there is a structure called algebra. In order to meet our needs, we need to strongly modify this concept to obtain Boolean algebras. Definition 1.1 (Boolean algebras): A Boolean algebra is a set A {\displaystyle A} together with two binary operations ∨ {\displaystyle \vee } and ∧ {\displaystyle \wedge } , an unary operation ¬ {\displaystyle \neg } and 0 , 1 ∈ A {\displaystyle 0,1\in A} such that the following axioms hold for all a , b , c ∈ A {\displaystyle a,b,c\in A} : Associativity of ∨ {\displaystyle \vee } and ∧ {\displaystyle \wedge } : a ◻ ( b ◻ c ) = ( a ◻ b ) ◻ c {\displaystyle a\Box (b\Box c)=(a\Box b)\Box c} , ◻ ∈ { ∨ , ∧ } {\displaystyle \Box \in \{\vee ,\wedge \}} Commutativity of ∨ {\displaystyle \vee } and ∧ {\displaystyle \wedge } : a ◻ b = b ◻ a {\displaystyle a\Box b=b\Box a} , ◻ ∈ { ∨ , ∧ } {\displaystyle \Box \in \{\vee ,\wedge \}} Absorbtion laws: a ◻ ( a ▽ b ) = a {\displaystyle a\Box (a\triangledown b)=a} , { ◻ , ▽ } = { ∨ , ∧ } {\displaystyle \{\Box ,\triangledown \}=\{\vee ,\wedge \}} Distributivity laws: a ◻ ( b ▽ c ) = ( a ◻ b ) ▽ ( a ◻ c ) {\displaystyle a\Box (b\triangledown c)=(a\Box b)\triangledown (a\Box c)} , { ◻ , ▽ } = { ∨ , ∧ } {\displaystyle \{\Box ,\triangledown \}=\{\vee ,\wedge \}} Neutral elements: a ∨ 0 = a {\displaystyle a\vee 0=a} , a ∧ 1 = a {\displaystyle a\wedge 1=a} Complementation laws: a ∨ ¬ a = 1 {\displaystyle a\vee \neg a=1} , a ∧ ¬ a = 0 {\displaystyle a\wedge \neg a=0} Fundamental example 1.2 (logic): If we take A = { ⊥ , ⊤ } {\displaystyle A=\{\bot ,\top \}} and ∨ , ∧ , ¬ {\displaystyle \vee ,\wedge ,\neg } to be the usual operations from logic, we obtain a Boolean algebra. Fundamental example and theorem 1.3: Let S {\displaystyle S} be an arbitrary set, and let A ⊂ 2 S {\displaystyle {\mathcal {A}}\subset 2^{S}} such that S , ∅ ∈ A {\displaystyle S,\emptyset \in {\mathcal {A}}} A , B ∈ S ⇒ A ∪ B ∈ S , A ∩ B ∈ S {\displaystyle A,B\in S\Rightarrow A\cup B\in S,A\cap B\in S} A ∈ S ⇒ A  ̄ ∈ S {\displaystyle A\in S\Rightarrow {\overline {A}}\in S} , where A  ̄ {\displaystyle {\overline {A}}} denotes the complement of A {\displaystyle A} . We set 1 := S {\displaystyle 1:=S} , 0 := ∅ {\displaystyle 0:=\emptyset } , ∨ := ∪ {\displaystyle \vee :=\cup } , ∧ := ∩ {\displaystyle \wedge :=\cap } , and ¬ A := A  ̄ {\displaystyle \neg A:={\overline {A}}} for all A ⊆ S {\displaystyle A\subseteq S} . Then ( A , 0 , 1 , ∨ , ∧ , ¬ ) {\displaystyle ({\mathcal {A}},0,1,\vee ,\wedge ,\neg )} is a Boolean algebra, called an algebra of subsets of S {\displaystyle S} . Proof: Closedness under the operations follows from 1. - 3. We have to verify 1. - 6. from definition 1.1. 1. A ∪ ( B ∪ C ) = { x ∈ S : x ∈ A ∨ ( x ∈ B ∪ C ) } = { x ∈ S : x ∈ A ∨ ( x ∈ B ∨ x ∈ C ) } = { x ∈ S : ( x ∈ A ∨ x ∈ B ) ∨ x ∈ C ) } = { x ∈ S : ( x ∈ A ∪ B ) ∨ x ∈ C } = ( A ∪ B ) ∪ C {\displaystyle {\begin{aligned}A\cup (B\cup C)&=\{x\in S:x\in A\vee (x\in B\cup C)\}\\&=\{x\in S:x\in A\vee (x\in B\vee x\in C)\}\\&=\{x\in S:(x\in A\vee x\in B)\vee x\in C)\}\\&=\{x\in S:(x\in A\cup B)\vee x\in C\}\\&=(A\cup B)\cup C\end{aligned}}} A ∩ ( B ∩ C ) = { x ∈ S : x ∈ A ∧ ( x ∈ B ∩ C ) } = { x ∈ S : x ∈ A ∧ ( x ∈ B ∧ x ∈ C ) } = { x ∈ S : ( x ∈ A ∧ x ∈ B ) ∧ x ∈ C ) } = { x ∈ S : ( x ∈ A ∩ B ) ∧ x ∈ C } = ( A ∩ B ) ∩ C {\displaystyle {\begin{aligned}A\cap (B\cap C)&=\{x\in S:x\in A\wedge (x\in B\cap C)\}\\&=\{x\in S:x\in A\wedge (x\in B\wedge x\in C)\}\\&=\{x\in S:(x\in A\wedge x\in B)\wedge x\in C)\}\\&=\{x\in S:(x\in A\cap B)\wedge x\in C\}\\&=(A\cap B)\cap C\end{aligned}}} 2. A ∪ B = { x ∈ S : x ∈ A ∨ x ∈ B } = { x ∈ S : x ∈ B ∨ x ∈ A } = B ∪ A {\displaystyle A\cup B=\{x\in S:x\in A\vee x\in B\}=\{x\in S:x\in B\vee x\in A\}=B\cup A} A ∩ B = { x ∈ S : x ∈ A ∧ x ∈ B } = { x ∈ S : x ∈ B ∧ x ∈ A } = B ∩ A {\displaystyle A\cap B=\{x\in S:x\in A\wedge x\in B\}=\{x\in S:x\in B\wedge x\in A\}=B\cap A} 3. A ∪ ( A ∩ B ) = { x ∈ S : x ∈ A ∨ x ∈ ( A ∩ B ) } = { x ∈ S : x ∈ A ∨ ( x ∈ A ∧ x ∈ B ) } = { x ∈ S : ( x ∈ A ∨ x ∈ A ) ∧ ( x ∈ A ∨ x ∈ B ) } = { x ∈ S : x ∈ A ∧ ( x ∈ A ∨ x ∈ B ) } = { x ∈ S : x ∈ A } = A {\displaystyle {\begin{aligned}A\cup (A\cap B)&=\{x\in S:x\in A\vee x\in (A\cap B)\}\\&=\{x\in S:x\in A\vee (x\in A\wedge x\in B)\}\\&=\{x\in S:(x\in A\vee x\in A)\wedge (x\in A\vee x\in B)\}\\&=\{x\in S:x\in A\wedge (x\in A\vee x\in B)\}\\&=\{x\in S:x\in A\}=A\end{aligned}}} A ∩ ( A ∪ B ) = { x ∈ S : x ∈ A ∧ x ∈ ( A ∪ B ) } = { x ∈ S : x ∈ A ∧ ( x ∈ A ∨ x ∈ B ) } = { x ∈ S : ( x ∈ A ∨ x ∈ A ) ∧ ( x ∈ A ∨ x ∈ B ) } = { x ∈ S : x ∈ A ∨ ( x ∈ A ∧ x ∈ B ) } = { x ∈ S : x ∈ A } = A {\displaystyle {\begin{aligned}A\cap (A\cup B)&=\{x\in S:x\in A\wedge x\in (A\cup B)\}\\&=\{x\in S:x\in A\wedge (x\in A\vee x\in B)\}\\&=\{x\in S:(x\in A\vee x\in A)\wedge (x\in A\vee x\in B)\}\\&=\{x\in S:x\in A\vee (x\in A\wedge x\in B)\}\\&=\{x\in S:x\in A\}=A\end{aligned}}} 4. A ∪ ( B ∩ C ) = { x ∈ S | x ∈ A ∨ x ∈ ( B ∩ C ) } = { x ∈ S | x ∈ A ∨ ( x ∈ B ∧ x ∈ C ) } = { x ∈ S | ( x ∈ A ∨ x ∈ B ) ∧ ( x ∈ A ∨ x ∈ C ) } = { x ∈ S | ( x ∈ A ∩ B ) ∧ ( x ∈ A ∩ C ) } = ( A ∩ B ) ∪ ( A ∩ C ) {\displaystyle {\begin{aligned}A\cup (B\cap C)&=\{x\in S|x\in A\vee x\in (B\cap C)\}\\&=\{x\in S|x\in A\vee (x\in B\wedge x\in C)\}\\&=\{x\in S|(x\in A\vee x\in B)\wedge (x\in A\vee x\in C)\}\\&=\{x\in S|(x\in A\cap B)\wedge (x\in A\cap C)\}\\&=(A\cap B)\cup (A\cap C)\end{aligned}}} A ∩ ( B ∪ C ) = { x ∈ S | x ∈ A ∧ x ∈ ( B ∪ C ) } = { x ∈ S | x ∈ A ∧ ( x ∈ B ∨ x ∈ C ) } = { x ∈ S | ( x ∈ A ∧ x ∈ B ) ∨ ( x ∈ A ∧ x ∈ C ) } = { x ∈ S | ( x ∈ A ∪ B ) ∨ ( x ∈ A ∪ C ) } = ( A ∪ B ) ∩ ( A ∪ C ) {\displaystyle {\begin{aligned}A\cap (B\cup C)&=\{x\in S|x\in A\wedge x\in (B\cup C)\}\\&=\{x\in S|x\in A\wedge (x\in B\vee x\in C)\}\\&=\{x\in S|(x\in A\wedge x\in B)\vee (x\in A\wedge x\in C)\}\\&=\{x\in S|(x\in A\cup B)\vee (x\in A\cup C)\}\\&=(A\cup B)\cap (A\cup C)\end{aligned}}} 5. A ∩ S = { x ∈ S | x ∈ A ∧ x ∈ S } = { x ∈ S | x ∈ A ∧ ⊤ } = { x ∈ S | x ∈ A } = A {\displaystyle A\cap S=\{x\in S|x\in A\wedge x\in S\}=\{x\in S|x\in A\wedge \top \}=\{x\in S|x\in A\}=A} A ∪ ∅ = { x ∈ S | x ∈ A ∨ x ∈ ∅ } = { x ∈ S | x ∈ A ∨ ⊥ } = { x ∈ S | x ∈ A } = A {\displaystyle A\cup \emptyset =\{x\in S|x\in A\vee x\in \emptyset \}=\{x\in S|x\in A\vee \bot \}=\{x\in S|x\in A\}=A} 6. A ∪ A  ̄ = { x ∈ S | x ∈ A ∨ ( x ∈ A  ̄ ) } = { x ∈ S | x ∈ A ∨ ( x ∉ A ) } = { x ∈ S | ⊤ } = S {\displaystyle {\begin{aligned}A\cup {\overline {A}}&=\{x\in S|x\in A\vee (x\in {\overline {A}})\}\\&=\{x\in S|x\in A\vee (x\notin A)\}\\&=\{x\in S|\top \}\\&=S\end{aligned}}} A ∩ A  ̄ = { x ∈ S | x ∈ A ∧ ( x ∈ A  ̄ ) } = { x ∈ S | x ∈ A ∧ ( x ∉ A ) } = { x ∈ S | ⊥ } = S {\displaystyle {\begin{aligned}A\cap {\overline {A}}&=\{x\in S|x\in A\wedge (x\in {\overline {A}})\}\\&=\{x\in S|x\in A\wedge (x\notin A)\}\\&=\{x\in S|\bot \}\\&=S\end{aligned}}} We thus see that the laws of a Boolean algebra are "elevated" from the Boolean algebra of logic to the Boolean algebra of sets. Exercise 1.1.1: Let A {\displaystyle A} be a Boolean algebra and a ∈ A {\displaystyle a\in A} . Prove that a ∧ a = a {\displaystyle a\wedge a=a} and a ∨ a = a {\displaystyle a\vee a=a} . During the remainder of the book, we shall adhere to the following notation conventions (due to Felix Hausdorff). If the sets A 1 , ... , A n {\displaystyle A_{1},\ldots ,A_{n}} are pairwise disjoint, we shall write ∑ j = 1 n A j {\displaystyle \sum _{j=1}^{n}A_{j}} for ⋃ j = 1 n A j {\displaystyle \bigcup _{j=1}^{n}A_{j}} ; with this notation we already indicate that the A j {\displaystyle A_{j}} are pairwise disjoint. That is, if we encounter an expression such as ∑ j = 1 n A j {\displaystyle \sum _{j=1}^{n}A_{j}} and the A j {\displaystyle A_{j}} are sets, the A j {\displaystyle A_{j}} are assumed to be pairwise disjoint. If A , B {\displaystyle A,B} are sets and A ⊆ B {\displaystyle A\subseteq B} , we replace B ∖ A {\displaystyle B\setminus A} by B − A {\displaystyle B-A} . This means: In any occasion where you find the notation B − A {\displaystyle B-A} within this book, it means B ∖ A {\displaystyle B\setminus A} and A ⊆ B {\displaystyle A\subseteq B} (note that in this way a set obtains a unique "additive inverse").
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