Complement of a set

Fix a universal set $U$ and every subset $A$ instantly carves $U$ into two pieces: the elements inside $A$ , and everything left over. That leftover piece is the complement, written $A'$ , and it is one of the most-used ideas in the whole "Sets" chapter, because once $U$ is agreed upon, talking about "everything except $A$ " becomes a single symbol instead of a clumsy sentence. This article — the closing topic of Unit I's "Sets" chapter — builds the complement from the set difference you already know, collects the handful of properties every question expects you to quote on sight, and proves De Morgan's laws, which tell you exactly what happens to a complement when it meets a union or an intersection. The same reasoning resurfaces later whenever set language describes combinations of events, so the goal here is fluency, not just recognition.

The core insight

The complement $A' = U - A$ splits $U$ cleanly into $A$ and $A'$ , and De Morgan's laws say a complement distributed over a union or intersection flips that operation to the other one.

The core idea

Given a universal set $U$ and $A \subseteq U$ , the complement of $A$ is defined as

A' = U - A = \{x \in U : x \notin A\}

where: $U$ is the fixed universal set for the problem, $A$ is the subset whose complement is wanted, and $A'$ (also written $A^c$ ) collects every element of $U$ that fails to belong to $A$ . It is nothing more than the familiar set difference $U - A$ , given its own name because it is used so constantly once $U$ has been fixed for a question. A short list of properties follows directly from this definition, and each is easy to see on a Venn diagram where $A'$ is the region inside $U$ but outside $A$ : $A \cup A' = U$ , since every element of $U$ is either in $A$ or outside it. $A \cap A' = \varnothing$ , since no element can simultaneously be inside $A$ and outside it. $(A')' = A$ — taking the complement twice returns the original set, because "everything outside everything outside $A$ " is just $A$ again. $U' = \varnothing$ and $\varnothing' = U$ , the two extreme cases. $n(A) + n(A') = n(U)$ for finite sets, which turns complement into a quick tool for counting problems.

Figure

A rectangle for the universal set with a circle for set A inside it and the region outside the circle shaded to mark the complement of A

De Morgan's laws connect complement to union and intersection:

(A \cup B)' = A' \cap B'

(A \cap B)' = A' \cup B'

where: $A$ and $B$ are subsets of the same universal set $U$ , and both statements say that when a complement is pushed inward past a union or intersection sign, that operation flips to the other one. In words, the complement of a union is the intersection of the complements, and the complement of an intersection is the union of the complements.

Derivation of De Morgan's law $(A \cup B)' = A' \cap B'$

Derivation ADe Morgan's law for the complement of a union

The goal of this derivation is to show, for any two subsets $A$ and $B$ of a universal set $U$ , that the set of elements lying outside $A \cup B$ is exactly the same set as the elements lying outside $A$ and outside $B$ at once, so that the single equation $(A \cup B)' = A' \cap B'$ can be trusted for every choice of $A$ , $B$ and $U$ , not merely checked on one example.

Step 1. Take any element $x$ that belongs to $(A \cup B)'$ . By the definition of complement, $x \in U$ and $x \notin A \cup B$ .

Step 2. Unpack $x \notin A \cup B$ . An element fails to be in a union only if it fails to be in both pieces separately, so $x \notin A$ and $x \notin B$ .

Step 3. Since $x \in U$ with $x \notin A$ , the definition of complement gives $x \in A'$ ; likewise $x \in U$ with $x \notin B$ gives $x \in B'$ . Having both at once means $x \in A' \cap B'$ . This shows every element of $(A \cup B)'$ also lies in $A' \cap B'$ , so $(A \cup B)' \subseteq A' \cap B'$ .

Step 4. Now run the argument the other way. Take any $x \in A' \cap B'$ , so $x \in A'$ and $x \in B'$ , which means $x \notin A$ and $x \notin B$ . If $x$ were in $A \cup B$ it would have to be in $A$ or in $B$ , contradicting both facts, so $x \notin A \cup B$ , giving $x \in (A \cup B)'$ . This shows $A' \cap B' \subseteq (A \cup B)'$ .

Step 5. With both inclusions established, the two sets contain exactly the same elements, so $(A \cup B)' = A' \cap B'$ holds for every $A$ , $B \subseteq U$ , completing the proof. The mirror argument, swapping every "and" for "or" (and union for intersection) at each step, proves $(A \cap B)' = A' \cup B'$ in exactly the same way.

Worked example

Let $U = \{1,2,\dots,12\}$ , $A = \{1,2,3,4,5,6\}$ and $B = \{4,5,6,7,8,9\}$ . Verify De Morgan's law $(A \cup B)' = A' \cap B'$ for this pair of sets.

Step 1. Compute $A \cup B$ by combining every element that appears in either set: $A \cup B = \{1,2,3,4,5,6,7,8,9\}$ .

Step 2. Compute $(A \cup B)'$ by removing these from $U = \{1,\dots,12\}$ : $(A \cup B)' = \{10, 11, 12\}$ .

Step 3. Compute $A'$ and $B'$ separately. $A' = U - A = \{7,8,9,10,11,12\}$ , and $B' = U - B = \{1,2,3,10,11,12\}$ .

Step 4. Compute $A' \cap B'$ by keeping only the elements common to both lists: $10, 11, 12$ appear in each, so $A' \cap B' = \{10,11,12\}$ .

Step 5. Compare the two results: $(A \cup B)' = \{10,11,12\}$ and $A' \cap B' = \{10,11,12\}$ match exactly, confirming $(A \cup B)' = A' \cap B'$ for this example, exactly as the derivation guarantees. Answer: $A' \cap B' = \{10, 11, 12\}$ , equal to $(A \cup B)'$ .

Competitive stretch

Using the same $U$ , $A$ and $B$ , check the second law $(A \cap B)' = A' \cup B'$ . First, $A \cap B = \{4,5,6\}$ (the elements common to both), so $(A \cap B)' = U - \{4,5,6\} = \{1,2,3,7,8,9,10,11,12\}$ . Next, $A' \cup B' = \{7,8,9,10,11,12\} \cup \{1,2,3,10,11,12\} = \{1,2,3,7,8,9,10,11,12\}$ — the same nine elements. Both sides equal $\{1,2,3,7,8,9,10,11,12\}$ , confirming the intersection form of De Morgan's law too.

Common traps

Computing $A'$ without a clearly stated universal set — the complement is meaningless until $U$ is fixed, and the same $A$ has different complements relative to different $U$ .
Mixing up the two De Morgan's laws — the operation always flips (union becomes intersection, and vice versa) once the complement is pushed through.
Forgetting $(A')' = A$ and treating a double complement as some new set instead of the original one.
Thinking complement is a different operation from set difference — it is exactly $U - A$ , just given a shorter name once $U$ is fixed.
Quoting De Morgan's law from memory in only one direction without checking which operation, union or intersection, the actual problem is distributing a complement over.

Ask the tutor: Want to verify De Morgan's laws on a set of your own, or check a complement calculation relative to a given universal set?

Why do we even need a universal set to define a complement?

Because "everything not in $A$ " is meaningless unless you say everything within what. Without bounds, "not in $A$ " could mean every number, every word, every object in the universe — an absurd and useless set. Fixing $U$ first turns the question into a well-defined one: the complement is only ever taken relative to the $U$ stated in the problem, and the same $A$ can have very different complements under two different choices of $U$ .

Is complement really the same thing as set difference?

Yes, exactly. $A' = U - A$ by definition — there is no extra rule hiding inside "complement." The reason it gets its own name and symbol is convenience: once $U$ is fixed for an entire problem, writing $A'$ is faster than repeatedly writing $U - A$ , and it makes identities like $A \cup A' = U$ read more cleanly. If you ever forget what $A'$ means, just mentally replace it with $U - A$ and proceed as usual.

How do I remember which way De Morgan's law flips the operation?

Think of the complement as "outside." Being outside a union ( $A \cup B$ , the region covered by either circle) means being outside both circles at once — that is an intersection of the two "outside" regions, giving $(A \cup B)' = A' \cap B'$ . Being outside an intersection ( $A \cap B$ , the small overlap) means being outside at least one of the circles — that is a union of "outside" regions, giving $(A \cap B)' = A' \cup B'$ . Picturing the Venn diagram, rather than memorising symbols, makes the flip automatic.

Complement of a set

The core idea

Derivation of De Morgan's law $(A \cup B)' = A' \cap B'$

Worked example

Competitive stretch

Common traps

Before this

Where this leads

Complement of a set

The core idea

Derivation of De Morgan's law (A∪B)′=A′∩B′(A \cup B)' = A' \cap B'(A∪B)′=A′∩B′

Worked example

Competitive stretch

Common traps

Before this

Where this leads

Derivation of De Morgan's law $(A \cup B)' = A' \cap B'$