April 4, 2025
A partition of a set S is a collection of disjoint subsets with union equal to S. The elements of the partition are called atoms. The coarsest partition containing only the atom S represents no information. The finest partition of singletons \{s\}, s\in S, represents complete information. Knowing which atom of a partition s\in S belongs to represents partial information. Finite algebras of set are in one-to-one correspondence with partitions. An algebra of sets on S is a collection of subsets of S closed under union and complement. We assume every algebra contains the empty set. By De Morgan’s laws it is also closed under intersection.
If \mathcal{A} is an algebra of sets on S then A\in\mathcal{A} is an atom if B\subseteq A and B\in\mathcal{A} imply B = \emptyset or B = A.
George Boole showed how to reduce De Morgan’s laws for sets to an algebraic calculation. For A\subseteq S define the indicator function 1_A\colon S\to\boldsymbol{R} by 1_A(s) = 1 if s\in A and 1_A(s) = 0 if {s\in S\setminus A = \{s\in S\mid s\not\in A\}}. Recall the union A\cup B of sets A and B is the set if elements belonging to either A or B and and the intersection A\cap B of sets A and B is the set of elements belonging to both A and B.
Exercise. Show 1_{S\setminus A} = 1_S - 1_A, 1_{A\cup B} = 1_A + 1_B - 1_{A\cap B}, and 1_{A\cap B} = 1_A1_B.
Exercise. (De Morgan) Show S\setminus (A\cup B) = (S\setminus A)\cap(S\setminus B) and S\setminus (A\cap B) = (S\setminus A)\cup(S\setminus B).
Hint: 1_{(S\setminus A)\cap(S\setminus B)} = 1_{S\setminus A}1_{S\setminus B} and 1_{(S\setminus A)\cup(S\setminus B)} = 1_{S\setminus A} + 1_{S\setminus B} - 1_{S\setminus A}1_{S\setminus B}.
For example, let S = [0,1) be the real numbers greater or equal to 0 and strictly less than 1. Every s\in S can be written in a base-2 expansion s = \sum_{n>0} s_n/2^n where s_n is either 0 or 1.
The partition \{[0,1/2),[1/2,1)\} represents knowing the first base-2 digit of s. If s\in[0,1/2) then s_1 = 0 and if s\in[1/2,1) then s_1 = 1.
The partition \{[j/2^n, (j+1)/2^n)\mid 0\le j\le 2^n\} represents knowing the first n base-2 digits of s.