Union, Intersection and Complement

Set operations combine membership conditions. Union asks whether an object belongs to either collection. Intersection asks whether it belongs to both. Complement asks whether it lies outside a collection while remaining inside the stated universe.

The operations are easier to understand when the objects stay fixed. Throughout the main examples, use $U=\{1,2,\ldots,12\}$ , let A contain the even members of U, and let B contain the members divisible by 3.

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Union uses inclusive or

The union $A\cup B$ contains every element in A or B, including elements in both. Here

A=\{2,4,6,8,10,12\},\qquad B=\{3,6,9,12\},

A\cup B=\{2,3,4,6,8,9,10,12\}.

The numbers 6 and 12 are listed once. A set records membership, not how many qualifying conditions an element satisfies. “A or B” is therefore different from “exactly one of A and B”. The latter excludes the overlap.

Intersection uses and

The intersection $A\cap B$ consists of objects that satisfy both conditions:

A\cap B=\{6,12\}.

The union contains eight elements, but the intersection contains two. Comparing the definitions, rather than memorising the shapes of the symbols, helps keep them straight. An intersection is always a subset of each of its input sets.

Two sets are disjoint when their intersection is empty. Disjoint does not mean unrelated in every sense: the odd and even members of U are disjoint, but together they partition U.

Calculate each disjoint region before combining regions

Membership, not the drawn circle area, determines the sets.

Diagram explanation and exact data

A only contains 2,4,8,10; both contains 6,12; B only contains 3,9; neither contains 1,5,7,11. Union combines the first three regions; symmetric difference omits the overlap.

Calculate each disjoint region before combining regions: exact reference data
Region	Members	Count
A only	2, 4, 8, 10	4
Both	6, 12	2
B only	3, 9	2
Neither	1, 5, 7, 11	4

Complement depends on the universe

Write $A^c=U\setminus A$ for the complement of A relative to U. In this example,

A^c=\{1,3,5,7,9,11\}.

These are the members of U that are not even. Numbers such as 13 are not added: they are outside U. If U were all integers, the complement would instead include every odd integer, positive or negative.

Both $A\cup A^c=U$ and $A\cap A^c=\varnothing$ follow directly from the membership definition. When A is a subset of finite U, $|A^c|=|U|-|A|$ . Do not apply that subtraction as ordinary finite arithmetic when U is infinite.

Difference and symmetric difference

The difference $A\setminus B$ means “in A but not in B”. It is

A\setminus B=\{2,4,8,10\}.

Reversing the sets gives $B\setminus A=\{3,9\}$ . Difference is therefore not commutative. It can be rewritten as $A\cap B^c$ , provided the complement uses a universe containing both sets.

The symmetric difference, $A\triangle B$ , means exactly one membership:

A\triangle B=(A\setminus B)\cup(B\setminus A)=\{2,3,4,8,9,10\}.

The two pieces on the right are disjoint. This operation is useful when asking which labels changed between two versions of a list: elements retained in both versions are not part of the change.

Grouping determines the condition

Add $C=\{1,4,9\}$ , the perfect-square members of U. Then

A\cap(B\cup C)=\{4,6,12\},

because each result is even and either divisible by 3 or a square. By contrast,

(A\cap B)\cup C=\{1,4,6,9,12\}.

The second expression includes every square in C, whether it is even or not. Parentheses specify which operation happens first. They cannot be dropped merely because all the same set letters appear.

Set identities come from membership reasoning

Union and intersection are commutative and associative. They also distribute over one another. For example,

A\cap(B\cup C)=(A\cap B)\cup(A\cap C).

To justify this identity, take an arbitrary object x. Membership on the left means x belongs to A and belongs to B or C. That is exactly the condition that x belongs to both A and B, or to both A and C. Hence it is also the membership condition on the right.

This argument works for every x, rather than merely the particular finite example. Checking that both sides give $\{4,6,12\}$ is a useful verification, but it is the arbitrary-member argument that establishes the general identity.

De Morgan’s laws change the connective

Not belonging to a union means belonging to neither set. Not belonging to an intersection means failing at least one membership requirement. With complements taken in the same U,

(A\cup B)^c=A^c\cap B^c,

(A\cap B)^c=A^c\cup B^c.

For the running example, $(A\cup B)^c=\{1,5,7,11\}$ . In contrast, $(A\cap B)^c$ is every member of U except 6 and 12. “Not both” is broader than “neither”. These identities connect directly to negating “and” and “or” in Logic and Mathematical Statements.

A complement of a difference can be unpacked in the same way. An object fails “in A and not in B” exactly when it is outside A or inside B:

(A\setminus B)^c=A^c\cup B.

Count a union without counting overlap twice

For finite sets,

|A\cup B|=|A|+|B|-|A\cap B|.

Adding $6+4$ counts 6 and 12 twice. Subtracting the two-element overlap once gives $6+4-2=8$ . The formula does not require equal-sized sets or disjoint sets. If the sets are disjoint, the intersection term is zero.

Counts alone can also reveal impossible data. Two subsets of a twelve-element universe cannot have a union containing fourteen distinct elements. Before solving for an unknown region, confirm that every resulting count is a non-negative integer and the regions total the universe.

Practice questions

Use $U=\{1,2,3,4,5,6\}$ , $A=\{1,2,4\}$ and $B=\{2,3,4\}$ .

Find $A\cup B$ and $A\cap B$ .
Find $A\setminus B$ and $A^c$ .
Find $(A\cup B)^c$ and verify it equals $A^c\cap B^c$ .

Check your answers

The union is $\{1,2,3,4\}$ ; the intersection is $\{2,4\}$ . Union keeps either membership, while intersection requires both.
$A\setminus B=\{1\}$ and $A^c=\{3,5,6\}$ . The complement includes only members of U.
$(A\cup B)^c=\{5,6\}$ . Since $B^c=\{1,5,6\}$ , intersecting it with $A^c$ also gives $\{5,6\}$ .