Keeping Only Shared Elements
In Union of Sets, the union \(A\cup B\) was defined by an inclusive “or”: an object belongs to the union if it belongs to at least one of the two sets. This tutorial introduces the complementary membership requirement. The intersection keeps an object only when it belongs to both sets.
Let \(A\) and \(B\) be sets. Their intersection, written \(A\cap B\), is the set of all objects that belong to \(A\) and belong to \(B\). In set-builder notation, $$ A\cap B=\{x:x\in A\text{ and }x\in B\}. $$ The symbol \(\cap\) names the intersection operation. Its defining membership rule is $$ x\in A\cap B \quad\Longleftrightarrow\quad (x\in A)\land(x\in B). $$ The connective \(\land\) means that both membership statements must hold.
For a finite list of elements, one way to find an intersection is to check each candidate and keep it exactly when it appears in both sets. An object that appears in only one set is excluded. If an object belongs to both sets, it is a single element of the intersection; as with other sets, writing it more than once does not create additional elements.
Evaluating Intersections
For finite sets, compare their elements and retain the shared ones. It is useful to check both membership conditions explicitly, especially when the sets are given by descriptions rather than lists. If the sets are defined within a universal set \(U\), as in Universal Sets, only elements of \(U\) are candidates.
Worked Example: Shared Elements of Finite Sets
Let $$ A=\{1,3,5,7,9\},\qquad B=\{0,3,6,7,9\}. $$ The elements \(3\), \(7\), and \(9\) occur in both sets. The elements \(1\) and \(5\) belong only to \(A\), while \(0\) and \(6\) belong only to \(B\). Therefore $$ A\cap B=\{3,7,9\}. $$ For instance, \(7\in A\cap B\) because \(7\in A\) and \(7\in B\). In contrast, \(5\notin A\cap B\): although \(5\in A\), we have \(5\notin B\), so the two required membership statements do not both hold.
The order of the written lists does not change the membership test. We are not looking for elements in the same position in each list; we are looking for objects that occur as elements of both sets.
With condition-defined sets, both conditions must be satisfied by the same candidate. This is an important point: one candidate cannot satisfy the first condition while another satisfies the second. The intersection contains an object only if that object meets each condition.
Worked Example: Applying Two Conditions on a Domain
Take \(U=\{-3,-2,-1,0,1,2,3,4\}\), and define $$ A=\{x\in U:x^2\leq4\},\qquad B=\{x\in U:x\geq0\}. $$ Among the elements of \(U\), the values satisfying \(x^2\leq4\) are \(-2,-1,0,1,2\). Thus \(A=\{-2,-1,0,1,2\}\). The values in \(U\) that are at least \(0\) are \(0,1,2,3,4\), so \(B=\{0,1,2,3,4\}\). The elements common to these sets are \(0,1,2\), giving $$ A\cap B=\{0,1,2\}. $$ The same answer follows by combining the conditions: \(x\) must lie in \(U\), satisfy \(x^2\leq4\), and satisfy \(x\geq0\). For example, \(-1\) satisfies the square condition but fails \(x\geq0\), so it is not in the intersection. The value \(3\) satisfies \(x\geq0\) but has \(3^2=9>4\), so it is not in the intersection either.
The condition-combining description is a direct application of the result in Set Builder Notation: successive conditions define the set of candidates satisfying both. In particular, if \(A=\{x\in U:P(x)\}\) and \(B=\{x\in U:Q(x)\}\), then $$ A\cap B=\{x\in U:P(x)\text{ and }Q(x)\}. $$ This equivalence is often a convenient way to translate a set operation into a single condition.
Membership and Subset Inclusion
The membership characterization of intersection leads to a useful criterion for proving that one set is contained in an intersection. Because an element of \(A\cap B\) must be in both \(A\) and \(B\), a set \(C\) is contained in the intersection exactly when it is contained in each component set.
Theorem (Subset of an Intersection). Let \(A\), \(B\), and \(C\) be sets. Then $$ C\subseteq A\cap B \quad\Longleftrightarrow\quad (C\subseteq A\text{ and }C\subseteq B). $$
Proof. First suppose \(C\subseteq A\cap B\). Let \(x\in C\) be arbitrary. The assumed inclusion gives \(x\in A\cap B\). By the membership characterization of intersection, \(x\in A\) and \(x\in B\). Since this holds for every \(x\in C\), the definition of subset inclusion gives \(C\subseteq A\) and \(C\subseteq B\).
Conversely, suppose \(C\subseteq A\) and \(C\subseteq B\). Let \(x\in C\) be arbitrary. The two assumed inclusions imply \(x\in A\) and \(x\in B\), respectively. Therefore \(x\in A\cap B\) by the membership characterization. Every element of \(C\) consequently belongs to \(A\cap B\), so \(C\subseteq A\cap B\). This proves both directions.
The result differs from the union criterion in a useful way. For a union, the Least Set Containing Two Sets theorem says that \(A\cup B\subseteq D\) exactly when both \(A\subseteq D\) and \(B\subseteq D\). For an intersection, the theorem just proved characterizes a set \(C\) contained in \(A\cap B\): \(C\) must be contained in both \(A\) and \(B\). These statements reflect the “or” and “and” in their respective membership rules.
Worked Example: Establishing Containment in an Intersection
Let $$ A=\{2,4,6,8\},\qquad B=\{1,2,4,7,8\},\qquad C=\{2,8\}. $$ Every element of \(C\) belongs to \(A\), so \(C\subseteq A\). Every element of \(C\) also belongs to \(B\), so \(C\subseteq B\). By the Subset of an Intersection theorem, $$ C\subseteq A\cap B. $$ Indeed, direct evaluation gives \(A\cap B=\{2,4,8\}\), which contains both elements of \(C\).
By contrast, let \(D=\{2,6\}\). We have \(D\subseteq A\), but \(D\not\subseteq B\), because \(6\in D\) and \(6\notin B\). Accordingly, \(D\not\subseteq A\cap B\). A single element failing either required component inclusion is enough to show that the whole subset claim fails.
The Intersection Is the Largest Set Contained in Both
There is a useful way to describe the role of \(A\cap B\) using subset inclusion. Every element of \(A\cap B\) is in \(A\), and every element is in \(B\), so the intersection is contained in each set. Moreover, any set contained in both \(A\) and \(B\) must be contained in \(A\cap B\). Thus the intersection is the largest set, with respect to subset inclusion, that is contained in both.
Theorem (Greatest Set Contained in Two Sets). Let \(A\), \(B\), and \(C\) be sets. Then $$ C\subseteq A\cap B \quad\Longleftrightarrow\quad (C\subseteq A\text{ and }C\subseteq B). $$ In particular, \(A\cap B\subseteq A\) and \(A\cap B\subseteq B\), and every set contained in both \(A\) and \(B\) is contained in \(A\cap B\).
The displayed equivalence is the Subset of an Intersection theorem proved above; its two immediate consequences show the “greatest” characterization. To verify those consequences, use the equivalence first with \(C=A\cap B\). Since \(A\cap B\subseteq A\cap B\) by reflexivity of subset inclusion, the theorem gives \(A\cap B\subseteq A\) and \(A\cap B\subseteq B\). For any set \(C\) contained in both \(A\) and \(B\), the reverse implication of the theorem gives \(C\subseteq A\cap B\).
This is a characterization, not a claim that the intersection must be larger in number of elements than either set. “Largest” here refers to the subset relation: the intersection contains every common subset. It may be empty, or it may equal one of the sets.
Basic Intersection Identities
Membership also proves identities involving the order of the sets, repetition, and the empty set. As with the union identities, equality of sets can be established by showing that an arbitrary object belongs to one side exactly when it belongs to the other. This is equality by double inclusion from Set Equality.
Theorem (Commutativity of Intersection). For any sets \(A\) and \(B\), $$ A\cap B=B\cap A. $$
Proof. Let \(x\) be any object. By the membership characterization, $$ x\in A\cap B \quad\Longleftrightarrow\quad (x\in A\text{ and }x\in B). $$ The conjunction “\(x\in A\) and \(x\in B\)” has the same truth value as “\(x\in B\) and \(x\in A\).” Applying the membership characterization in the other order gives $$ (x\in B\text{ and }x\in A) \quad\Longleftrightarrow\quad x\in B\cap A. $$ Thus \(x\in A\cap B\) if and only if \(x\in B\cap A\). Since this holds for every object \(x\), the sets have the same elements. By equality by double inclusion, \(A\cap B=B\cap A\).
The empty set and a set combined with itself give two more identities. The empty set has no elements, so no object can satisfy the requirement of belonging to both \(A\) and \(\varnothing\). Requiring membership in \(A\) twice does not change the membership condition.
Proposition (Empty Set and Idempotence for Intersection). For every set \(A\), $$ A\cap\varnothing=\varnothing \qquad\text{and}\qquad A\cap A=A. $$
Proof. Let \(x\) be any object. By the membership characterization, $$ x\in A\cap\varnothing \quad\Longleftrightarrow\quad (x\in A\text{ and }x\in\varnothing). $$ By the Characterization of the Empty Set, no object belongs to \(\varnothing\). Therefore the conjunction on the right is false for every \(x\), so no object belongs to \(A\cap\varnothing\). The Characterization of the Empty Set gives \(A\cap\varnothing=\varnothing\).
For the second identity, for any object \(x\), $$ x\in A\cap A \quad\Longleftrightarrow\quad (x\in A\text{ and }x\in A) \quad\Longleftrightarrow\quad x\in A. $$ Thus \(A\cap A\) and \(A\) have exactly the same elements, and consequently \(A\cap A=A\).
The two basic operations treat the empty set differently. The union identity from Union of Sets is \(A\cup\varnothing=A\), because union requires membership in at least one set. The intersection identity is \(A\cap\varnothing=\varnothing\), because intersection requires membership in both sets, which is impossible when one set has no elements.
| Expression | Membership test | Result or interpretation |
|---|---|---|
| \(x\in A\cap B\) | \(x\in A\) and \(x\in B\) | Both membership statements must hold. |
| \(A\cap B\) | Keep elements shared by the two sets | Each common element is included once. |
| \(A\cap\varnothing\) | \(x\in A\) and \(x\in\varnothing\) | Equals \(\varnothing\). |
| \(A\cap A\) | \(x\in A\) and \(x\in A\) | Equals \(A\). |
| \(C\subseteq A\cap B\) | Both \(C\subseteq A\) and \(C\subseteq B\) | A subset of the intersection lies in each set. |
Common Errors to Avoid
The most common error is to confuse the “and” for intersection with the “or” for union. If an object belongs to \(A\) but not \(B\), it belongs to \(A\cup B\) but not to \(A\cap B\). The words describe different membership tests, so they cannot be interchanged.
Another error occurs with condition-defined sets when the conditions are applied to different candidates. Suppose \(A=\{x\in U:P(x)\}\) and \(B=\{x\in U:Q(x)\}\). Membership in \(A\cap B\) requires a single candidate \(x\) for which both \(P(x)\) and \(Q(x)\) hold. It is not enough that some element of \(U\) satisfies \(P\) and a possibly different element satisfies \(Q\).
Finally, distinguish an element statement from a subset statement. The statement \(x\in A\cap B\) requires two memberships for the same object. The statement \(C\subseteq A\cap B\) requires that every element of \(C\) have both memberships. Equivalently, the entire set \(C\) must be contained in each of \(A\) and \(B\).
Use the membership characterization of intersection in each question. When a universal set is specified, consider only its elements.
Check Your Understanding
- Let \(A=\{1,2,5,8\}\) and \(B=\{0,2,4,8\}\). Find \(A\cap B\), and identify which elements of \(A\) are excluded from the intersection.
- Let \(U=\{-3,-2,-1,0,1,2,3\}\), \(A=\{x\in U:x^2=1\}\), and \(B=\{x\in U:x<2\}\). Find \(A\), \(B\), and \(A\cap B\).
- If \(x\in A\) but \(x\notin B\), must \(x\in A\cap B\)? Explain using the membership definition.
- Suppose \(C\subseteq A\) and \(C\subseteq B\). Which theorem gives \(C\subseteq A\cap B\)?
- What are \(A\cap\varnothing\) and \(A\cap A\)? Explain why the empty-set identity differs from \(A\cup\varnothing=A\).
- If \(A\cap B=A\), what can be concluded about the relation between \(A\) and \(B\)? Justify your answer using the subset characterization.