The Timpson Lyceum · Discrete Math (2025)

Set Operations and Combinatorial Proofs - Lecture 6 (Discrete Math)

Overview

This lecture moves from set relations (which return true or false) to set operations (which take sets and return new sets), then uses those operations to count. We define intersection, union, difference, symmetric difference, and the Cartesian (cross) product, and we collect the core algebraic laws — commutativity, associativity, identity with the empty set, distributivity, and De Morgan's laws — all of which follow directly from the corresponding rules of propositional logic. We then introduce the inclusion–exclusion principle for the size of a union and prove it by counting, applying it to a divisibility problem. The lecture closes with two combinatorial (counting) proofs of summation identities, illustrating the general technique of showing that both sides of an equation count the same thing.

Key Concepts

Relations vs. operations. A relation on sets (e.g. $x \in A$ , $A \subseteq B$ , $A = B$ ) is a predicate that returns true or false. A set operation (intersection, union, difference, …) takes sets and returns a new set, just as $+$ and $\times$ take numbers and return numbers.

Subset review.

Subset: $A \subseteq B$ means $\forall x\,(x \in A \Rightarrow x \in B)$ — everything in $A$ is also in $B$ .
Equality: $A = B$ iff $A \subseteq B$ and $B \subseteq A$ (the same elements).
Proper subset: $A \subsetneq B$ means $A \subseteq B$ and $A \neq B$ . A set is a subset of itself but never a proper subset of itself. (Think $\subseteq$ as $\le$ and $\subsetneq$ as $<$ .)

The set operations. For all $x$ :

Intersection: $A \cap B = \{\,x : x \in A \text{ and } x \in B\,\}$ — elements in both.
Union: $A \cup B = \{\,x : x \in A \text{ or } x \in B\,\}$ — elements in either (inclusive "or": being in both still counts).
Difference: $A - B = \{\,x : x \in A \text{ and } x \notin B\,\}$ — in $A$ but not in $B$ .
Symmetric difference: $A \,\triangle\, B = (A - B) \cup (B - A)$ — in exactly one of the two sets (everything except the overlap).

Commutativity. $A \cap B = B \cap A$ and $A \cup B = B \cup A$ (because "and"/"or" are commutative). But $A - B \neq B - A$ in general — difference and the cross product are the two operations here that are not commutative.

Concrete computation. For $A = \{1,2,3,4\}$ , $B = \{1,4,7,11\}$ :

$A \cup B = \{1,2,3,4,7,11\}$
$A \cap B = \{1,4\}$
$A - B = \{2,3\}$ , while $B - A = \{7,11\}$
$A \,\triangle\, B = \{2,3,7,11\}$

Algebra of sets (Theorem 12.3). For sets $A, B, C$ :

Commutative: $A \cup B = B \cup A$ , $A \cap B = B \cap A$ .
Associative: $A \cup (B \cup C) = (A \cup B) \cup C$ , $A \cap (B \cap C) = (A \cap B) \cap C$ .
Identity / absorbing element: $A \cup \varnothing = A$ (like adding $0$ ), $A \cap \varnothing = \varnothing$ .
Distributive: $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ and $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ .

Each law mirrors a propositional-logic identity: union/intersection are defined by "or"/"and," so any property of "or"/"and" transfers directly to the corresponding set operation (verifiable by a truth table).

Inclusion–exclusion (two sets). For finite sets, $|A| + |B| = |A \cup B| + |A \cap B|$ , equivalently $|A \cup B| = |A| + |B| - |A \cap B|.$ Adding $|A|$ and $|B|$ counts the overlap twice, so subtract $|A \cap B|$ once. Forgetting to subtract the intersection is the classic error.

Disjoint and pairwise disjoint.

$A$ and $B$ are disjoint iff $A \cap B = \varnothing$ (no shared elements).
Sets $A_1, A_2, \dots, A_n$ are pairwise disjoint iff $A_i \cap A_j = \varnothing$ for all $i, j$ with $1 \le i < j \le n$ — every pair is disjoint.
Corollary: if $A, B$ are disjoint, then $|A \cup B| = |A| + |B|$ (the intersection term vanishes).

Cartesian (cross) product. $A \times B = \{\,(a,b) : a \in A,\ b \in B\,\}$ — ordered pairs with first coordinate from $A$ , second from $B$ . Size: $|A \times B| = |A| \cdot |B|$ . Like difference, it is not commutative: $A \times B \neq B \times A$ in general (the pairs have their coordinates swapped).

De Morgan's laws for sets. $A - (B \cup C) = (A - B) \cap (A - C)$ and $A - (B \cap C) = (A - B) \cup (A - C)$ . Distributing a "minus" over a union flips it to an intersection (and vice versa) — the set-theory image of the logical $\lnot(\alpha \lor \beta) = \lnot\alpha \land \lnot\beta$ .

Combinatorial proof technique. To prove an identity, find a collection of objects that both sides count; if they count the same set of objects, they are equal. Useful facts: $2^n$ counts all subsets of an $n$ -element set, and $n!$ counts the arrangements (orderings) of $n$ distinct objects.

Worked Example

Inclusion–exclusion applied to divisibility. How many integers in $\{1, 2, \dots, 1000\}$ are divisible by $2$ or $5$ ?

Let $A$ = multiples of $2$ and $B$ = multiples of $5$ in that range. Numbers divisible by both $2$ and $5$ are exactly the multiples of $10$ . Then

\begin{aligned} |A \cup B| &= |A| + |B| - |A \cap B| \\ &= 500 + 200 - 100 = 600. \end{aligned}

So 600 integers from $1$ to $1000$ are divisible by $2$ or $5$ . (Naively answering $500 + 200 = 700$ double-counts the $100$ multiples of $10$ .)

Combinatorial proof that $\;2^0 + 2^1 + \cdots + 2^{\,n-1} = 2^n - 1$ .

Let $A = \{1, 2, \dots, n\}$ , so $|A| = n$ . Since $A$ has $2^n$ subsets, it has $2^n - 1$ non-empty subsets (all subsets except $\varnothing$ ) — that is the right-hand side.

Now count the non-empty subsets a second way, by their maximum element $k$ . A subset whose largest element is $k$ consists of $k$ together with any subset of $\{1, 2, \dots, k-1\}$ , and there are $2^{\,k-1}$ such choices. Summing over every possible maximum $k = 1, 2, \dots, n$ :

\begin{aligned} \#\{\text{non-empty subsets of } A\} &= \sum_{k=1}^{n} 2^{\,k-1} \\ &= 2^0 + 2^1 + 2^2 + \cdots + 2^{\,n-1}. \end{aligned}

Both expressions count the non-empty subsets of $A$ , so they are equal: $2^0 + 2^1 + \cdots + 2^{\,n-1} = 2^n - 1$ . (Check $n=4$ : $1+2+4+8 = 15 = 2^4 - 1$ .)

Summary

Set relations return true/false; set operations ( $\cap$ , $\cup$ , $-$ , $\triangle$ , $\times$ ) take sets and return sets.
Union and intersection are commutative and associative; difference and the cross product are not commutative ( $A - B \neq B - A$ , $A \times B \neq B \times A$ ).
The empty set acts as an identity for union ( $A \cup \varnothing = A$ ) and absorbs under intersection ( $A \cap \varnothing = \varnothing$ ); distributive and De Morgan laws hold, each inherited from propositional logic.
Inclusion–exclusion: $|A \cup B| = |A| + |B| - |A \cap B|$ ; for disjoint sets it simplifies to $|A \cup B| = |A| + |B|$ .
Sizes multiply for the cross product: $|A \times B| = |A| \cdot |B|$ .
A combinatorial (counting) proof establishes an identity by exhibiting one collection of objects that both sides count — e.g. counting non-empty subsets by maximum element gives $\sum_{k=1}^{n} 2^{k-1} = 2^n - 1$ , and counting restricted arrangements gives $\sum_{k=1}^{n} k\cdot k! = (n+1)! - 1$ .