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 $A=\{\mathrm{\perp },\mathrm{\top }\}$ and $\vee ,\wedge ,\mathrm{\neg }$ 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.

$\begin{array}{cc}A\cup (B\cup C)& =\{x\in S:x\in A\vee (x\in B\cup C)\}\\ & =\{x\in S:x\in A\vee (x\in B\vee x\in C)\}\\ & =\{x\in S:(x\in A\vee x\in B)\vee x\in C)\}\\ & =\{x\in S:(x\in A\cup B)\vee x\in C\}\\ & =(A\cup B)\cup C\end{array}$

$\begin{array}{cc}A\cap (B\cap C)& =\{x\in S:x\in A\wedge (x\in B\cap C)\}\\ & =\{x\in S:x\in A\wedge (x\in B\wedge x\in C)\}\\ & =\{x\in S:(x\in A\wedge x\in B)\wedge x\in C)\}\\ & =\{x\in S:(x\in A\cap B)\wedge x\in C\}\\ & =(A\cap B)\cap C\end{array}$

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.

$\begin{array}{cc}A\cup (A\cap B)& =\{x\in S:x\in A\vee x\in (A\cap B)\}\\ & =\{x\in S:x\in A\vee (x\in A\wedge x\in B)\}\\ & =\{x\in S:(x\in A\vee x\in A)\wedge (x\in A\vee x\in B)\}\\ & =\{x\in S:x\in A\wedge (x\in A\vee x\in B)\}\\ & =\{x\in S:x\in A\}=A\end{array}$

$\begin{array}{cc}A\cap (A\cup B)& =\{x\in S:x\in A\wedge x\in (A\cup B)\}\\ & =\{x\in S:x\in A\wedge (x\in A\vee x\in B)\}\\ & =\{x\in S:(x\in A\vee x\in A)\wedge (x\in A\vee x\in B)\}\\ & =\{x\in S:x\in A\vee (x\in A\wedge x\in B)\}\\ & =\{x\in S:x\in A\}=A\end{array}$

4.

$\begin{array}{cc}A\cup (B\cap C)& =\{x\in S|x\in A\vee x\in (B\cap C)\}\\ & =\{x\in S|x\in A\vee (x\in B\wedge x\in C)\}\\ & =\{x\in S|(x\in A\vee x\in B)\wedge (x\in A\vee x\in C)\}\\ & =\{x\in S|(x\in A\cap B)\wedge (x\in A\cap C)\}\\ & =(A\cap B)\cup (A\cap C)\end{array}$

$\begin{array}{cc}A\cap (B\cup C)& =\{x\in S|x\in A\wedge x\in (B\cup C)\}\\ & =\{x\in S|x\in A\wedge (x\in B\vee x\in C)\}\\ & =\{x\in S|(x\in A\wedge x\in B)\vee (x\in A\wedge x\in C)\}\\ & =\{x\in S|(x\in A\cup B)\vee (x\in A\cup C)\}\\ & =(A\cup B)\cap (A\cup C)\end{array}$

5.

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

$A\cup \mathrm{\varnothing }=\{x\in S|x\in A\vee x\in \mathrm{\varnothing }\}=\{x\in S|x\in A\vee \mathrm{\perp }\}=\{x\in S|x\in A\}=A$

6.

$\begin{array}{cc}A\cup \bar{A}& =\{x\in S|x\in A\vee (x\in \bar{A})\}\\ & =\{x\in S|x\in A\vee (x\notin A)\}\\ & =\{x\in S|\mathrm{\top }\}\\ & =S\end{array}$

$\begin{array}{cc}A\cap \bar{A}& =\{x\in S|x\in A\wedge (x\in \bar{A})\}\\ & =\{x\in S|x\in A\wedge (x\notin A)\}\\ & =\{x\in S|\mathrm{\perp }\}\\ & =S\end{array}$

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 $A_1,\dots ,A_n$ 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").