Complement of a Set and Its Properties

Complement of a set. Once a universal set $U$ has been fixed, the complement

of a set $A$ (with $A \subseteq U$), written $A'$ (or $A^c$, or $U - A$), is the set of

all elements of $U$ that do not belong to $A$:

Worked illustration. Let $U = \{1, 2, \dots, 12\}$ and $A = \{2, 4, 6, 8, 10, 12\}$

(the even numbers up to 12). Then $A'$ consists of everything in $U$ that is not in

$A$, namely the odd numbers: $A' = \{1, 3, 5, 7, 9, 11\}$.

Properties of the complement. For any set $A \subseteq U$:

1. Complement laws: $A \cup A' = U$ and $A \cap A' = \varnothing$. (Every element of $U$ is either in $A$ or in $A'$, never both — so together they fill up all of $U$ with no overlap.)
2. Law of double complementation: $(A')' = A$. Taking the complement of the complement returns the original set.
3. $\varnothing' = U$ and $U' = \varnothing$.
4. De Morgan's laws:

Proof of $(A')' = A$. By definition, $(A')' = U - A' = \{x \in U : x \notin A'\}$.

But $x \notin A'$ means $x$ is not in "everything outside $A$," which (since

$x \in U$) can only mean $x \in A$. Hence $(A')' = \{x \in U : x \in A\} = A$.

$\blacksquare$

Proof of De Morgan's law $(A \cup B)' = A' \cap B'$. Let $x$ be any element of

$U$.

Since this chain of "if and only if" holds for every $x \in U$, the two sets

$(A \cup B)'$ and $A' \cap B'$ have exactly the same elements, so they are equal.

$\blacksquare$ The second law, $(A \cap B)' = A' \cup B'$, is proved the same way, …