Complements of Sets

A Set's Complement Depends on Its Universe

Set difference keeps elements of a first set while excluding elements of a second. A complement is a particular use of that idea: it records what is left out of a set when we look only inside a specified universal set. The universe is not a decorative extra. It determines which objects are available to be included in the complement.

Let $U$ be a set, and let $A\subseteq U$. The complement of $A$ relative to $U$ is $$ A^c=U\setminus A. $$ Equivalently, its membership rule is $$ x\in A^c \quad\Longleftrightarrow\quad (x\in U\text{ and }x\notin A). $$ The superscript $c$ is read as “complement.” When the universe is understood from context, it is common to write $A^c$ without repeating $U$. If there is any possibility of confusion, write $U\setminus A$ or specify the universe explicitly.

The condition $A\subseteq U$ ensures that every element of $A$ is being considered within the chosen universe. Objects outside $U$ are not candidates for $A^c$, whether or not they belong to $A$. Complements are therefore relative: there is no single complement of $A$ independent of a universe.

Always identify the universe. The complement $A^c$ means “in $U$, but not in $A$.” It does not mean every object that is not in $A$, without restriction.

Worked Example: Changing the Universe Changes the Complement

Let $A=\{2,4\}$. First choose $U_1=\{1,2,3,4,5\}$. Since the elements of $U_1$ not in $A$ are $1,3,5$, $$ A^{c_{U_1}}=U_1\setminus A=\{1,3,5\}. $$ Now choose the larger universe $U_2=\{1,2,3,4,5,6,7\}$. The same set $A$ has complement $$ A^{c_{U_2}}=U_2\setminus A=\{1,3,5,6,7\}. $$ The extra elements $6$ and $7$ appear because they belong to $U_2$ and do not belong to $A$. They could not appear in the complement relative to $U_1$, because they are not in that universe. Here the subscripts make the two choices of universe explicit.

Finding a Complement

For a finite universe, list its elements and retain exactly those that fail to belong to the set being complemented. This gives a reliable two-part check: a proposed complement must contain only elements of $U$, and it must contain no elements of $A$. It must also include every element of $U$ that is not in $A$.

Worked Example: A Complement in a Finite Universe

Let $$ U=\{a,b,c,d,e,f,g\},\qquad A=\{b,d,f\}. $$ The candidates are the elements of $U$. Of these, $a,c,e,g$ are not in $A$, while $b,d,f$ are in $A$. Therefore $$ A^c=U\setminus A=\{a,c,e,g\}. $$ For example, $c\in U$ and $c\notin A$, so $c\in A^c$. The element $h$, even though $h\notin A$, is not in $A^c$, because $h\notin U$. Thus being outside $A$ alone is insufficient.

When a set is defined by a condition on $U$, its complement is obtained by keeping the same universe and negating the membership condition. This follows directly from the complement membership rule. In particular, candidates outside the stated domain are not added when the condition is negated.

Worked Example: Complement of a Condition-Defined Set

Let $U=\{-4,-3,-2,-1,0,1,2,3,4\}$, and define $$ A=\{x\in U:x^2\leq4\}. $$ The members of $U$ satisfying $x^2\leq4$ are $-2,-1,0,1,2$, so $$ A=\{-2,-1,0,1,2\}. $$ The complement consists of the elements of $U$ that fail $x^2\leq4$. For real $x$, this failure is $x^2>4$, and the members of the stated finite universe satisfying it are $-4,-3,3,4$. Hence $$ A^c=\{x\in U:x^2>4\}=\{-4,-3,3,4\}. $$ The domain restriction $x\in U$ remains part of the description. For instance, $x=5$ satisfies $x^2>4$, but it does not belong to this complement because $5\notin U$.

Two Basic Complement Properties

Every element of $U$ either belongs to $A$ or does not. Accordingly, $A$ and $A^c$ together contain the whole universe, and they have no element in common. This makes them a two-part division of $U$. The two parts are determined by membership in $A$, not by an arbitrary choice.

Theorem (Complement Identities). Let $U$ be a set and $A\subseteq U$. Then $$ A\cup A^c=U,\qquad A\cap A^c=\varnothing. $$

Proof. We first prove the union equality. Let $x$ be any object. If $x\in A\cup A^c$, then $x\in A$ or $x\in A^c$. In the first case, $A\subseteq U$ gives $x\in U$. In the second case, the complement membership rule gives $x\in U$. Thus $A\cup A^c\subseteq U$.

Conversely, let $x\in U$. If $x\in A$, then $x\in A\cup A^c$. If $x\notin A$, then $x\in U$ and $x\notin A$, so the complement membership rule gives $x\in A^c$, and again $x\in A\cup A^c$. Therefore $U\subseteq A\cup A^c$. Equality by double inclusion gives $A\cup A^c=U$.

For the intersection, suppose an object $x$ belonged to $A\cap A^c$. Membership in the intersection would give both $x\in A$ and $x\in A^c$. But $x\in A^c$ implies $x\notin A$, a contradiction. Thus no object belongs to $A\cap A^c$. By the Characterization of the Empty Set, $A\cap A^c=\varnothing$. This proves both identities.

Worked Example: Checking the Two Complement Identities

Let $U=\{p,q,r,s,t\}$ and $A=\{p,r\}$. Then $A^c=\{q,s,t\}$. Taking the union gives $$ A\cup A^c=\{p,r\}\cup\{q,s,t\}=\{p,q,r,s,t\}=U. $$ No element occurs in both sets, so $$ A\cap A^c=\{p,r\}\cap\{q,s,t\}=\varnothing. $$ The union check confirms that no element of the universe was omitted; the intersection check confirms that no element was placed in both parts.

The empty set and the universe have especially simple complements. Relative to $U$, the empty set has complement $U$, because every element of $U$ is outside $\varnothing$. The complement of $U$ is empty, because an element of $U$ cannot also fail to belong to $U$: $$ \varnothing^c=U,\qquad U^c=\varnothing. $$ These statements use the same fixed universe $U$; the first is the complement of $\varnothing$ relative to $U$.

Taking a Complement Twice

Once the universe is fixed, the complement operation can be applied again. The complement of $A^c$ consists of the elements in $U$ that are not in $A^c$. Since $A^c$ contains exactly the elements of $U$ outside $A$, the elements remaining are precisely those in $A$.

Proposition (Double Complement). If $A\subseteq U$, then $$ (A^c)^c=A, $$ where both complements are taken relative to $U$.

Proof. First, $A^c\subseteq U$ by the complement membership rule, so its complement relative to $U$ is defined. Let $x$ be any object. Then $$ x\in(A^c)^c \quad\Longleftrightarrow\quad (x\in U\text{ and }x\notin A^c). $$ If $x\in U$, the statement $x\in A^c$ is equivalent to $x\notin A$. Consequently, for $x\in U$, the statement $x\notin A^c$ is equivalent to $x\in A$. Also, $A\subseteq U$, so $x\in A$ already implies $x\in U$. It follows that $$ x\in(A^c)^c\quad\Longleftrightarrow\quad x\in A. $$ This equivalence holds for every object $x$. Equality by double inclusion gives $(A^c)^c=A$.

The fixed-universe condition matters here too. Applying the complement operation twice means using the same $U$ both times. If the universe changed between operations, the notation would refer to different operations and the proposition would not apply as stated.

Complements Reverse Subset Inclusion

Suppose $A\subseteq B\subseteq U$. Anything outside $B$ must also be outside $A$: if an object belonged to $A$, the inclusion would force it to belong to $B$. Therefore $B^c\subseteq A^c$. The direction reverses: the complement of the larger set is contained in the complement of the smaller set.

Theorem (Complement Reverses Inclusion). Let $A,B\subseteq U$. Then $$ A\subseteq B \quad\Longleftrightarrow\quad B^c\subseteq A^c. $$

Proof. First suppose $A\subseteq B$. To prove $B^c\subseteq A^c$, let $x\in B^c$ be arbitrary. Then $x\in U$ and $x\notin B$. We claim $x\notin A$. If $x\in A$, the inclusion $A\subseteq B$ would imply $x\in B$, contradicting $x\notin B$. Thus $x\notin A$. Together with $x\in U$, this gives $x\in A^c$. Since every $x\in B^c$ belongs to $A^c$, we have $B^c\subseteq A^c$.

Conversely, suppose $B^c\subseteq A^c$. We prove $A\subseteq B$. Let $x\in A$ be arbitrary. Since $A\subseteq U$, we have $x\in U$. If $x\notin B$, then $x\in U$ and $x\notin B$, so $x\in B^c$. The assumed inclusion would then give $x\in A^c$, which means $x\notin A$. This contradicts $x\in A$. Therefore $x\in B$. Since this holds for every $x\in A$, $A\subseteq B$. Both implications are proved.

Worked Example: Using Complement Inclusion

Let $U=\{1,2,3,4,5,6,7\}$, $A=\{2,5\}$, and $B=\{2,3,5,7\}$. We have $A\subseteq B$, since each element of $A$ also belongs to $B$. Their complements relative to $U$ are $$ A^c=\{1,3,4,6,7\},\qquad B^c=\{1,4,6\}. $$ Every element of $B^c$ belongs to $A^c$, so $B^c\subseteq A^c$, as the theorem predicts. Notice that the inclusion goes in the reverse direction. In fact, $A^c$ has more elements here because $A$ is the smaller of the two sets.

The theorem also works in reverse as a test: if two sets are known to lie in the same universe and $B^c\subseteq A^c$, then $A\subseteq B$. Without a common universe, the complements in this test would not be specified consistently.

Name the universe: identify $U$ and verify that the set being complemented is a subset of $U$.

Keep only universe elements: candidates for $A^c$ must belong to $U$.

Exclude the original set: retain exactly the candidates $x$ for which $x\notin A$.

Check both directions: every listed element should be in $U\setminus A$, and every element of $U\setminus A$ should be listed.

Common Pitfalls

The most common error is to treat “not in $A$” as a complete description of the complement. It is only half of the membership test. An object must be both in the universe and outside $A$. For example, if $U=\{1,2,3\}$ and $A=\{2\}$, then $4\notin A$, but $4\notin A^c$ because $4\notin U$.

A second error is to forget which universe is being used. A set may have different complements relative to different universes, as the examples show. When a problem defines $A$ as a subset of a particular $U$, use that same $U$ when taking its complement.

Finally, do not confuse the complement with the set difference in the wrong order. Relative to $U$, the complement is $U\setminus A$, not $A\setminus U$. Since $A\subseteq U$, the latter contains no elements: it asks for elements in $A$ that are not in $U$, which cannot occur. The order of set difference determines the candidates, and here the universe supplies them.

Expression	Membership test	Interpretation
$A^c$	$x\in U$ and $x\notin A$	Elements of the specified universe outside $A$.
$A\cup A^c$	Elements in $A$ or outside $A$, within $U$	Equals $U$.
$A\cap A^c$	Would require $x\in A$ and $x\notin A$	Equals $\varnothing$.
$(A^c)^c$	In $U$ and not in $A^c$	Equals $A$, with the same universe.
$A\subseteq B$	$B^c\subseteq A^c$	Inclusion reverses under complementation.

Complement is a relative operation. Use the same universe throughout a complement argument, and verify both parts of membership: the object must be in $U$ and not in the set being complemented.

For each question, keep the universe explicit. A complement includes exactly the members of the universe that are not members of the original set.

Check Your Understanding

Let $U=\{0,1,2,3,4,5,6\}$ and $A=\{1,3,5\}$. Find $A^c$ relative to $U$.
If $A=\{a,c\}$, compare its complements relative to $U_1=\{a,b,c,d\}$ and $U_2=\{a,b,c,d,e\}$. What accounts for any difference?
Let $U=\{-3,-2,-1,0,1,2,3\}$ and $A=\{x\in U:x\geq0\}$. Describe $A^c$ using set-builder notation and list its elements.
For $A\subseteq U$, what are $A\cup A^c$ and $A\cap A^c$?
If $A,B\subseteq U$ and $A\subseteq B$, which inclusion relates $A^c$ and $B^c$?
Why is an object outside $U$ not a member of $A^c$, even if it is not a member of $A$?

Expression	Membership test	Interpretation
\(A^c\)	\(x\in U\) and \(x\notin A\)	Elements of the specified universe outside \(A\).
\(A\cup A^c\)	Elements in \(A\) or outside \(A\), within \(U\)	Equals \(U\).
\(A\cap A^c\)	Would require \(x\in A\) and \(x\notin A\)	Equals \(\varnothing\).
\((A^c)^c\)	In \(U\) and not in \(A^c\)	Equals \(A\), with the same universe.
\(A\subseteq B\)	\(B^c\subseteq A^c\)	Inclusion reverses under complementation.

What You'll Learn