Set complements

Sometimes we need "not $A$" - all elements that aren't in a set. To define this precisely, we first need a universal set $U$ containing all elements under consideration.

To find $A^c$, we list every element from $U$ that doesn't appear in $A$. If $A$ contains the even numbers, then $A^c$ contains the odd numbers - every element that fails to satisfy the original condition.

On a Venn diagram, the complement $A^c$ is the region outside circle $A$ but still inside the rectangle $U$. Use the button below to toggle the shading.

The complement captures everything that's "left over" when we exclude $A$ from the universal set.

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

Find $A^c$.

Solution

The complement $A^c$ contains all elements in $U$ that are not in $A$. We check each element:

• $1$: not in $A$, so $1 \in A^c$
• $2$: in $A$, so $2 \notin A^c$
• $3$: not in $A$, so $3 \in A^c$
• $4$: in $A$, so $4 \notin A^c$
• $5$: not in $A$, so $5 \in A^c$
• $6$: in $A$, so $6 \notin A^c$

The complement contains the odd numbers - all elements not in the original set of even numbers.

Complements have three fundamental properties that follow directly from their definition.

Property 1: Double complement

Taking the complement twice returns the original set:

If $A$ contains the even numbers and $A^c$ contains the odd numbers, then $(A^c)^c$ is "not odd" - which brings us back to even.

Property 2: Disjoint

A set and its complement share no elements:

No element can be both in $A$ and not in $A$ simultaneously. Sets with empty intersection are disjoint.

Property 3: Exhaustive coverage

A set and its complement together cover the entire universal set:

Every element is either in $A$ or not in $A$ - there's no third option. Together, $A$ and $A^c$ account for everything.

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

Verify all three complement properties for this set.

Solution

First, we find the complement:

Property 1: $(A^c)^c = A$

The complement of $A^c = \{2, 4, 6\}$ contains all elements not in $\{2, 4, 6\}$:

Property 2: $A \cap A^c = \emptyset$

The intersection contains elements in both $A$ and $A^c$:

• $A = \{1, 3, 5\}$
• $A^c = \{2, 4, 6\}$

These sets share no elements:

Property 3: $A \cup A^c = U$

The union contains all elements in either $A$ or $A^c$:

When we take the complement of a union or intersection, something interesting happens. De Morgan's laws tell us exactly how complements interact with these operations.

The first law says: to be outside "$A$ or $B$", an element must be outside $A$ and outside $B$. If it were in either one, it would be in the union.

The second law says: to be outside "$A$ and $B$", an element just needs to be outside $A$ or outside $B$. Being missing from either set is enough to be missing from their intersection.

Use the interactive widget below to see why these laws work.

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

1. Find $(A \cup B)^c$ directly
2. Find $A^c \cap B^c$ and verify they match

Solution

Part 1: Finding $(A \cup B)^c$ directly

First, compute the union:

Now take the complement - elements in $U$ but not in $A \cup B$:

Part 2: Finding $A^c \cap B^c$

First, find each complement:

Now take the intersection - elements in both $A^c$ and $B^c$:

Both methods give $\{5\}$, confirming De Morgan's first law: $(A \cup B)^c = A^c \cap B^c$.

The element $5$ is the only one that's in neither $A$ nor $B$.