Probability Theory/The algebra of sets

Boolean algebras
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.

Fundamental example 1.2 (logic):

If we take $$A = \{\bot, \top\}$$ and $$\vee, \wedge, \neg$$ to be the usual operations from logic, we obtain a Boolean algebra.

Proof: Closedness under the operations follows from 1. - 3. We have to verify 1. - 6. from definition 1.1.

1.


 * $$\begin{align}

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{align}$$


 * $$\begin{align}

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{align}$$

2.


 * $$A \cup B = \{x \in S : x \in A \vee x \in B\} = \{x \in S : x \in B \vee x \in A\} = B \cup A$$


 * $$A \cap B = \{x \in S : x \in A \wedge x \in B\} = \{x \in S : x \in B \wedge x \in A\} = B \cap A$$

3.


 * $$\begin{align}

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{align}$$


 * $$\begin{align}

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{align}$$

4.


 * $$\begin{align}

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{align}$$


 * $$\begin{align}

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{align}$$

5.


 * $$A \cap S = \{x \in S | x \in A \wedge x \in S\} = \{x \in S| x \in A \wedge \top \}= \{x \in S|x \in A\} = A$$


 * $$A \cup \emptyset = \{x \in S | x \in A \vee x \in \emptyset\} = \{x \in S| x \in A \vee \bot \}= \{x \in S|x \in A\} = A$$

6.


 * $$\begin{align}

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{align}$$


 * $$\begin{align}

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{align}$$

We thus see that the laws of a Boolean algebra are "elevated" from the Boolean algebra of logic to the Boolean algebra of sets.

Exercises

 * Exercise 1.1.1: Let $$A$$ be a Boolean algebra and $$a \in A$$. Prove that $$a \wedge a = a$$ and $$a \vee a = a$$.

Notation
During the remainder of the book, we shall adhere to the following notation conventions (due to Felix Hausdorff).
 * 1) If the sets $$A_1, \ldots, A_n$$ are pairwise disjoint, we shall write $$\sum_{j=1}^n A_j$$ for $$\bigcup_{j=1}^n A_j$$; with this notation we already indicate that the $$A_j$$ are pairwise disjoint. That is, if we encounter an expression such as $$\sum_{j=1}^n A_j$$ and the $$A_j$$ are sets, the $$A_j$$ are assumed to be pairwise disjoint.
 * 2) If $$A, B$$ are sets and $$A \subseteq B$$, we replace $$B \setminus A$$ by $$B - A$$. This means: In any occasion where you find the notation $$B - A$$ within this book, it means $$B \setminus A$$ and $$A \subseteq B$$ (note that in this way a set obtains a unique "additive inverse").