Union of Sets

Combining Sets by Membership

In Universal Sets, a universal set was used to specify the objects under consideration. Within such a context, we often need to describe a set formed from two sets already given. The union is one basic way to do this: it gathers together the elements of either set.

Let $A$ and $B$ be sets. Their union, written $A\cup B$, is the set of all objects that belong to $A$, belong to $B$, or belong to both. In set-builder notation, $$ A\cup B=\{x:x\in A\text{ or }x\in B\}. $$ The symbol $\cup$ names the operation of taking a union. As with other sets, the result is determined by its elements, not by the order in which those elements are listed.

The word “or” here is inclusive. An object that belongs to both $A$ and $B$ still belongs to $A\cup B$. The definition does not require an object to belong to exactly one of the two sets. In logical notation, the membership rule is $$ x\in A\cup B \quad\Longleftrightarrow\quad (x\in A)\lor(x\in B). $$ Here $\lor$ means that at least one of the statements is true, including the possibility that both are true.

Union means “in at least one.” An element is in $A\cup B$ when it is in $A$, in $B$, or in both. If it appears in both sets, it is still just one element of the resulting set.

Evaluating a Union

For finite sets, list the elements from both sets and remove repetitions. Repetition in a list does not create a new set element: by the basic set conventions established in Sets and Elements, a set records which objects it contains, not how many times an object is written. Thus, the union combines membership rather than adding the sizes of the two sets.

Worked Example: Two Finite Sets With an Overlap

Let $$ A=\{2,4,6,8\},\qquad B=\{1,2,3,4\}. $$ The elements of $A$ are $2,4,6,8$, and the elements of $B$ are $1,2,3,4$. Collecting every element that occurs in at least one set gives $1,2,3,4,6,8$. The values $2$ and $4$ belong to both sets, but they are listed only once. Therefore $$ A\cup B=\{1,2,3,4,6,8\}. $$ For instance, $2\in A\cup B$ because $2\in A$ and $2\in B$; $6\in A\cup B$ because $6\in A$; and $3\in A\cup B$ because $3\in B$. The value $5$ belongs to neither set, so $5\notin A\cup B$.

It would be incorrect to add the displayed list lengths and claim that the union has eight elements. The repeated elements do not become separate elements of the union. In this example, the union has six distinct elements.

The membership rule also gives a systematic way to evaluate unions described by conditions. One can test an object against both membership statements. It belongs to the union if at least one test succeeds.

Worked Example: A Union Defined by Conditions

Use the universal set $U=\{-4,-2,-1,0,1,3,4\}$, and define $$ A=\{x\in U:x^2=1\},\qquad B=\{x\in U:x\geq3\}. $$ For $A$, the elements of $U$ whose squares equal $1$ are $-1$ and $1$, so $A=\{-1,1\}$. For $B$, the elements of $U$ that are at least $3$ are $3$ and $4$, so $B=\{3,4\}$. Consequently, $$ A\cup B=\{-1,1,3,4\}. $$ To verify the membership rule on a few candidates, $-1\in A$, so $-1\in A\cup B$; $4\in B$, so $4\in A\cup B$; and $0$ belongs to neither $A$ nor $B$, so $0\notin A\cup B$. The domain $U$ determines the candidates, as explained in Universal Sets.

The Membership Characterization of Union

The defining rule for union is useful not only for calculating examples but also for proving statements about sets. In a set proof, take an arbitrary object $x$ and translate its membership in a union into the two alternatives in the definition. The resulting “or” is often the central step.

Theorem (Membership in a Union). Let $A$, $B$, and $C$ be sets. Then $$ C\subseteq A\cup B \quad\Longleftrightarrow\quad (C\subseteq A\text{ or }C\subseteq B) $$ is not true in general. Instead, the elementwise characterization is that for every object $x$, $$ x\in A\cup B \quad\Longleftrightarrow\quad (x\in A\text{ or }x\in B). $$

Proof. By the definition of union, an object $x$ belongs to $A\cup B$ exactly when it belongs to $A$ or belongs to $B$. This is precisely the stated equivalence.

The distinction in the statement matters. An element-by-element “or” does not usually turn into a single choice that applies to every element of a set. For example, if $A=\{1\}$, $B=\{2\}$, and $C=\{1,2\}$, then $C\subseteq A\cup B$, but $C\not\subseteq A$ and $C\not\subseteq B$. The valid subset criterion is $$ C\subseteq A\cup B \quad\Longleftrightarrow\quad \forall x\in C,\ (x\in A\text{ or }x\in B). $$ This follows by applying the definition of subset inclusion and then the membership characterization of union.

Worked Example: Checking a Subset of a Union

Let $$ A=\{a,c\},\qquad B=\{b,c\},\qquad C=\{a,b\}. $$ First, $A\cup B=\{a,b,c\}$. Both elements of $C$ belong to this union: $a\in A$, and $b\in B$. Hence every element of $C$ belongs to $A\cup B$, which proves $C\subseteq A\cup B$.

However, $C$ is not a subset of $A$, because $b\in C$ and $b\notin A$. It is also not a subset of $B$, because $a\in C$ and $a\notin B$. Thus the subset $C$ can be contained in the union even though it is contained in neither set individually. The elements of $C$ can satisfy the union membership condition through different alternatives.

The Union Is the Smallest Set Containing Both Sets

The union has a useful characterization in terms of subset inclusion: it is contained in every set that contains both $A$ and $B$. First, each of $A$ and $B$ is a subset of their union. Then, if some set $D$ contains both $A$ and $B$, every element of $A\cup B$ belongs to $D$. This describes $A\cup B$ as the smallest set, with respect to subset inclusion, that contains both sets.

Theorem (Least Set Containing Two Sets). Let $A$, $B$, and $D$ be sets. Then $$ A\cup B\subseteq D \quad\Longleftrightarrow\quad (A\subseteq D\text{ and }B\subseteq D). $$

Proof. First suppose $A\cup B\subseteq D$. If $x\in A$, then $x\in A\cup B$ by the definition of union. The assumed inclusion gives $x\in D$. Since this holds for every $x\in A$, $A\subseteq D$. If $y\in B$, then $y\in A\cup B$, so $y\in D$. Thus $B\subseteq D$.

Conversely, suppose $A\subseteq D$ and $B\subseteq D$. Let $x$ be an arbitrary element of $A\cup B$. By the membership characterization, $x\in A$ or $x\in B$. If $x\in A$, then $x\in D$ because $A\subseteq D$. If $x\in B$, then $x\in D$ because $B\subseteq D$. In either case $x\in D$. Since every element of $A\cup B$ belongs to $D$, the definition of subset inclusion gives $A\cup B\subseteq D$.

This theorem gives a practical proof technique. To show that a union is contained in $D$, it is enough to show separately that each of its two component sets is contained in $D$. In the other direction, if the union is already known to lie in $D$, then both sets lie in $D$ as well.

To prove $A\cup B\subseteq D$: take $x\in A\cup B$, so $x\in A$ or $x\in B$.

Use the appropriate inclusion: if $x\in A$, use $A\subseteq D$; if $x\in B$, use $B\subseteq D$.

Finish by the definition of subset: in either case $x\in D$, so $A\cup B\subseteq D$.

Worked Example: Containment in a Specified Universe

Let $U=\{0,1,2,3,4,5,6\}$, $A=\{1,3,5\}$, and $B=\{2,4\}$. Each element of $A$ belongs to $U$, so $A\subseteq U$. Each element of $B$ also belongs to $U$, so $B\subseteq U$. The Least Set Containing Two Sets theorem now gives $A\cup B\subseteq U$.

Evaluating the union explicitly gives $$ A\cup B=\{1,2,3,4,5\}. $$ This confirms directly that every element of the union is in $U$. Notice that $0$ and $6$ are in the universal set but not in $A\cup B$. Being an available element of $U$ does not by itself make an object a member of the union.

Basic Identities and Their Proofs

Union also has simple identities that can be established directly from membership. These identities help simplify expressions and clarify how order, repetition, and the empty set affect the result. The following proof of commutativity uses the Equality by Double Inclusion theorem from Set Equality.

Theorem (Commutativity of Union). For any sets $A$ and $B$, $$ A\cup B=B\cup A. $$

Proof. Let $x$ be any object. By the membership characterization, $$ x\in A\cup B \quad\Longleftrightarrow\quad (x\in A\text{ or }x\in B). $$ The statement “$x\in A$ or $x\in B$” has the same truth value as “$x\in B$ or $x\in A$.” Applying the membership characterization again gives $$ (x\in A\text{ or }x\in B) \quad\Longleftrightarrow\quad x\in B\cup A. $$ Therefore, $x\in A\cup B$ if and only if $x\in B\cup A$. Since this holds for every object $x$, the sets have exactly the same elements. By equality by double inclusion, $A\cup B=B\cup A$.

Another identity concerns the empty set. It contributes no elements to a union. In contrast, taking the union of a set with itself does not add any new elements. These facts are useful checks when evaluating or rewriting a union.

Proposition (Identity and Idempotence for Union). For every set $A$, $$ A\cup\varnothing=A \qquad\text{and}\qquad A\cup A=A. $$

Proof. For any object $x$, the empty-set characterization gives $x\notin\varnothing$. Thus $$ x\in A\cup\varnothing \quad\Longleftrightarrow\quad (x\in A\text{ or }x\in\varnothing) \quad\Longleftrightarrow\quad x\in A. $$ The two sets have the same elements, so $A\cup\varnothing=A$. Also, $$ x\in A\cup A \quad\Longleftrightarrow\quad (x\in A\text{ or }x\in A) \quad\Longleftrightarrow\quad x\in A. $$ Again the two sets have the same elements, giving $A\cup A=A$.

Expression	Membership test	Result
$x\in A\cup B$	$x\in A$ or $x\in B$	At least one membership statement must hold.
$A\cup B$	Collect elements from both sets	Every element is included once, even if it lies in both.
$A\cup\varnothing$	$x\in A$ or $x\in\varnothing$	Equals $A$.
$A\cup A$	$x\in A$ or $x\in A$	Equals $A$.
$A\cup B\subseteq D$	Both $A\subseteq D$ and $B\subseteq D$	The union lies in $D$ exactly when both sets do.

Common Errors to Avoid

The most common mistake is to interpret “or” as exclusive. Exclusive “or” would include an object only when it belongs to exactly one of the sets. The union does not impose that restriction. If $x$ belongs to both $A$ and $B$, then at least one of the membership statements is true—in fact both are—and $x\in A\cup B$.

A second mistake is to treat a union as though it were a list formed by writing the elements of $A$, followed by the elements of $B$, and keeping every repeated entry. A union is a set, so only membership matters. If an object occurs in both lists, it appears only once as an element of the union.

Finally, be careful when moving between element statements and subset statements. The rule for one object is $x\in A$ or $x\in B$. For a set $C$ to be a subset of $A\cup B$, every $x\in C$ must satisfy this alternative. It does not follow that all of $C$ must be contained in $A$, or all of $C$ must be contained in $B$; different elements of $C$ may belong to different sets.

Keep the logical level clear. For an individual element, membership in the union requires an “or.” For a subset of the union, every element must satisfy that “or.” Do not replace this element-by-element condition with the stronger claim that the entire subset lies in one component set.

Use the membership rule for union in each question. Where a domain is given, test only the candidates from that domain.

Check Your Understanding

Let $A=\{0,2,4\}$ and $B=\{2,3,5\}$. List $A\cup B$, making clear how the repeated element is handled.
Let $U=\{-2,-1,0,1,2,3\}$, $A=\{x\in U:x<0\}$, and $B=\{x\in U:x^2=4\}$. Find $A$, $B$, and $A\cup B$.
If $x\in A$ and $x\in B$, must $x\in A\cup B$? State which part of the definition answers the question.
Let $A=\{r\}$, $B=\{s\}$, and $C=\{r,s\}$. Is $C\subseteq A\cup B$? Is $C\subseteq A$ or $C\subseteq B$? Explain the difference.
Suppose $A\subseteq D$ and $B\subseteq D$. What theorem gives $A\cup B\subseteq D$, and what is the key membership step in its proof?
Use the union identities to evaluate $\{4,7\}\cup\varnothing$ and $\{4,7\}\cup\{4,7\}$.

Expression	Membership test	Result
\(x\in A\cup B\)	\(x\in A\) or \(x\in B\)	At least one membership statement must hold.
\(A\cup B\)	Collect elements from both sets	Every element is included once, even if it lies in both.
\(A\cup\varnothing\)	\(x\in A\) or \(x\in\varnothing\)	Equals \(A\).
\(A\cup A\)	\(x\in A\) or \(x\in A\)	Equals \(A\).
\(A\cup B\subseteq D\)	Both \(A\subseteq D\) and \(B\subseteq D\)	The union lies in \(D\) exactly when both sets do.

What You'll Learn