Union and Intersection

Suppose you are told that a number belongs to the set of even integers, or to the set of multiples of $3$ , or possibly both. How should mathematics record "or" and "and" between sets as precisely as it records them between statements? This question is not idle bookkeeping. Nearly every construction in analysis, algebra, and computer science combines collections of objects. It does this by asking what is in either of two sets, or what is in both. Union and intersection are the two most basic operations for building new sets from old ones. They turn the loose vocabulary of "or" and "and" into operations with exact, checkable defining conditions. The importance of getting this right is that these two operations generate an entire algebra of sets. Together with Set Equality and Extensionality, they mirror the algebra of logical connectives almost exactly. Once union and intersection are pinned down by precise membership conditions, statements like "union distributes over intersection" become theorems. They are provable from the definitions of $\in$ , $\lor$ , and $\land$ , rather than appeals to pictures. The big picture is that set theory builds all of its machinery from a small number of primitive operations. Each is defined by a membership condition: an if-and-only-if statement. It tells you exactly when an object belongs to the newly constructed set. Union and intersection are the first and most fundamental examples of this pattern. The technique used to prove facts about them — element-chasing, guided by extensionality — is the technique used throughout the rest of set theory. This article gives the formal definitions of $A \cup B$ and $A \cap B$ . It proves the core algebraic laws (commutativity, associativity, distributivity, absorption, and idempotence) with full element-chasing proofs. It works two concrete asymmetric examples, and shows exactly what fails when a hypothesis is dropped. It closes by placing union and intersection in the wider landscape. That landscape includes Set Difference and Complement, Proving Set Identities, and Indexed Families of Sets.

Key takeaways

$A \cup B$ collects everything in $A$ or $B$ (inclusive or); $A \cap B$ collects everything in both.
Both are defined by a single biconditional membership condition, then equality of sets is checked via Set Equality and Extensionality.
Union and intersection are commutative, associative, idempotent, and each distributes over the other.
$A \cap B \subseteq A \subseteq A \cup B$ always holds, for any sets $A, B$ .
$A \cap B = \varnothing$ characterizes disjointness; $A \cup B = A$ exactly when $B \subseteq A$ .
Key idea: union and intersection translate "or" and "and" from logic into operations on sets, and every set-algebra identity about them is really a logic identity in disguise.

Where this lives today

Union and intersection are the backbone of measure theory and probability, where events are subsets of a sample space and $P(A\cup B) = P(A)+P(B)-P(A\cap B)$ (inclusion–exclusion) governs the probability of " $A$ or $B$ ." In topology, arbitrary unions of open sets and finite intersections of open sets are declared open by definition, making union and intersection the literal generators of a topology. In computer science, database query languages implement `UNION` and `INTERSECT` as direct set operations on result tables, and Boolean circuit design uses OR-gates and AND-gates as the physical realization of $\cup$ and $\cap$ at the level of truth values. In formal logic and proof theory, the correspondence $\cup \leftrightarrow \lor$ , $\cap \leftrightarrow \land$ used throughout the proofs above is an instance of the broader algebraic structure called a Boolean algebra, which unifies set algebra, propositional logic, and switching circuits under one axiomatic roof. In cryptography and coding theory, intersections of codewords or key sets determine collision and security bounds, while unions describe the total space of possible messages or attacks under consideration.

Ask the tutor: How do the distributive laws for union and intersection generalize when one of the sets is replaced by an infinite indexed family?

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Intuition

Picture $A$ and $B$ as two overlapping regions in a plane — the classical Venn diagram. The union is the entire shaded area covered by at least one of the two regions: you are asking "is $x$ in the left circle, the right circle, or the lens where they overlap?" The intersection is only the lens, the strip common to both: you are asking the much stronger question "is $x$ in the left circle and simultaneously in the right circle?" Because "and" is a stronger demand than "or," intersection always produces a smaller (or equal) set than union, and both are squeezed between the empty set and the full union of everything under discussion. A second useful picture is arithmetic with sets of numbers. If $A = \{1,2,3,4\}$ and $B = \{3,4,5,6\}$ , then $A \cup B$ merges the lists (discarding repeats) to get $\{1,2,3,4,5,6\}$ , while $A \cap B$ keeps only what survives in both lists, namely $\{3,4\}$ . This "merge vs. filter" intuition — union merges, intersection filters — is exactly the logical content of $\lor$ versus $\land$ , transported wholesale into the language of sets.

Figure

Venn diagram: A ∪ B (all shaded) versus A ∩ B (only the overlap)

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X = Y

Worked examples

Example 1 (finite sets of mixed type, asymmetric sizes). Let

A = \{2, 4, 6, 8, 10\}, \qquad B = \{3, 6, 9, 12\}.

These sets have different sizes ( $|A|=5$ , $|B|=4$ ) and only partial overlap, which is exactly the kind of case that exposes errors a symmetric example would hide. Union. By definition $x \in A\cup B$ iff $x\in A$ or $x \in B$ . Listing every element of $A$ , then adjoining every element of $B$ not already listed:

A \cup B = \{2,3,4,6,8,9,10,12\}.

Check: $|A\cup B| = 8 = |A| + |B| - |A\cap B| = 5 + 4 - 1$ , consistent with the inclusion–exclusion count computed below. Intersection. By definition $x \in A \cap B$ iff $x$ is in both lists. Scanning $A = \{2,4,6,8,10\}$ against $B=\{3,6,9,12\}$ , the only common value is $6$ . Hence

A \cap B = \{6\}.

Check via Theorem, part (7). We must have $A\cap B \subseteq A \subseteq A \cup B$ . Indeed $\{6\}\subseteq\{2,4,6,8,10\}$ (true, $6$ is listed), and $\{2,4,6,8,10\} \subseteq \{2,3,4,6,8,9,10,12\}$ (true, every element of $A$ appears in the union list). Both inclusions hold, confirming the computation is consistent with monotonicity. Check via part (8). Since $A \neq B$ and neither is a subset of the other ( $3 \in B\setminus A$ and $2 \in A \setminus B$ ), the theorem predicts $A \cup B \neq B$ and $A \cap B \neq A$ — and indeed $A\cup B = \{2,3,4,6,8,9,10,12\} \neq \{3,6,9,12\}=B$ , and $A\cap B=\{6\}\neq\{2,4,6,8,10\}=A$ , matching the prediction exactly.

Example 2 (sets defined by conditions, not by listing). Let $A$ be the set of integers divisible by $4$ , and let $B$ be the set of integers divisible by $6$ :

A = \{n \in \mathbb{Z} : 4 \mid n\}, \qquad B = \{n \in \mathbb{Z} : 6 \mid n\}.

These are infinite sets, so we cannot list elements; we must reason directly from the membership condition. Union. By definition, $n \in A \cup B$ iff $4\mid n$ or $6 \mid n$ . So, for instance, $8 \in A\cup B$ (since $4\mid 8$ ), $18 \in A \cup B$ (since $6 \mid 18$ ), and $12 \in A\cup B$ (since both hold), but $10 \notin A \cup B$ (neither $4$ nor $6$ divides $10$ ). There is no simpler single-modulus description of $A\cup B$ ; it genuinely is the union, not reducible to "divisible by $d$ " for any one $d$ . Intersection. By definition, $n \in A \cap B$ iff $4\mid n$ and $6\mid n$ , i.e. $n$ is a common multiple of $4$ and $6$ . A standard number-theoretic fact (provable by unique factorization) is that the common multiples of $4=2^2$ and $6=2\cdot 3$ are exactly the multiples of $\operatorname{lcm}(4,6) = 12$ . So

A \cap B = \{n \in \mathbb{Z} : 12 \mid n\}.

Concretely $12, 24, -12, 0 \in A\cap B$ , while $6 \notin A \cap B$ (since $4 \nmid 6$ , even though $6\mid 6$ ) — a useful reminder that membership in $B$ alone is not enough; both conditions in the $\land$ must hold. Check via Theorem, part (7). Every multiple of $12$ is a multiple of $4$ , so $A \cap B \subseteq A$ ; and every multiple of $4$ satisfies "divisible by $4$ or by $6$ ," so $A \subseteq A\cup B$ . Both containments hold as predicted, and here neither is an equality (e.g. $4 \in A$ but $4 \notin A\cap B$ , and $6\in A\cup B$ but $6 \notin A$ ), showing the inclusions in part (7) are generally strict.

x \in A \cup B \iff (x\in A)\lor(x\in B), \qquad x \in A\cap B \iff (x\in A)\land(x\in B).

Scope, edge cases & common traps

Situation	What to watch for
$A = B$	$A\cup B = A\cap B = A$ ; union and intersection collapse to the common value (idempotence).
$A$ , $B$ disjoint	$A \cap B = \varnothing$ ; union has size $\lvert A\rvert + \lvert B\rvert$ when finite.
One set empty	$A \cup \varnothing = A$ and $A\cap\varnothing = \varnothing$ always — never assume $A\cap\varnothing = A$ .
$A \subseteq B$	$A\cup B = B$ and $A\cap B = A$ (Theorem part 8); a common source of "obvious" but unproved claims.
Infinite sets given by conditions	Cannot list elements; must reason from the membership biconditional directly, as in Example 2.
Confusing $\lor$ with exclusive or	$A\cup B$ uses inclusive or; elements in both $A$ and $B$ are still counted once in $A\cup B$ .

The mathematician behind it

Origins · 1 minGeorg Cantor and how this was discovered

Georg Cantor

1845–1918 · Halle, German Empire

Georg Cantor created set theory as an independent mathematical discipline in the 1870s–1880s, seeking a rigorous foundation for treating collections of infinite objects — such as the set of all real numbers or all points on a line — as legitimate mathematical entities subject to precise operations, including union and intersection of arbitrarily infinite families. His work provoked fierce controversy (Kronecker and others rejected completed infinities outright) but ultimately became the bedrock on which nearly all of modern mathematics, including the formal treatment given here, is built.

History

- 1874 — Cantor publishes his first paper proving the real numbers are uncountable, launching set theory as a subject.

- 1895–1897 — Cantor systematically develops operations on sets, including union and intersection of infinite families, in his Beiträge.

- 1901 — Russell's paradox exposes the need for careful axiomatic restriction on which "sets" can exist.

- 1908 — Zermelo axiomatizes set theory, including the Axiom of Union, giving union and intersection a rigorous existential footing.

- 1963 — Paul Cohen's independence results confirm set theory's axioms leave certain questions (e.g. the continuum hypothesis) genuinely open, cementing set theory's mature axiomatic status.

Is A ∪ B always at least as large as A?

Yes, whenever these are finite sets: $A \subseteq A \cup B$ by Theorem part (7), and a subset of a finite set has cardinality no greater than the whole set. In fact for finite sets, $|A \cup B| = |A| + |B| - |A \cap B|$ (inclusion–exclusion), which is at least $|A|$ since $|B| - |A\cap B| \ge 0$ (because $A\cap B \subseteq B$ , so $|A \cap B| \le |B|$ ). For infinite sets "size" needs cardinal arithmetic, but the inclusion $A \subseteq A \cup B$ itself holds unconditionally, regardless of finiteness.

Does A ∩ B always contain something, i.e. can it be empty?

Yes, it can absolutely be empty, and often is. Whenever $A$ and $B$ share no common element — for example $A = \{1,2\}$ and $B=\{3,4\}$ — the definition gives $x \in A\cap B \iff (x\in A)\land(x\in B)$ , and no $x$ satisfies both, so $A\cap B = \varnothing$ . This is precisely the definition of $A$ and $B$ being disjoint. There is no hypothesis in the definitions of union or intersection that forces overlap; disjointness is a special, extra condition on the pair $(A,B)$ , not a general guarantee.

How is union different from the logical "or" it is built from?

Union is a set-forming operation, while "or" ( $\lor$ ) is a connective between two truth-valued statements. They are linked, not identical: $A \cup B$ is defined precisely so that the statement " $x \in A\cup B$ " is logically equivalent to the statement " $x\in A$ or $x \in B$ ." Every algebraic law about $\cup$ (commutativity, associativity, distributivity) is literally a restatement, via this biconditional, of the corresponding tautology about $\lor$ . The set operation packages the logical connective into an object you can then intersect, take subsets of, or apply further operations to.

Why do we need extensionality to prove things like commutativity of union?

The definitions of $A\cup B$ and $A\cap B$ only tell you the membership condition — when an object belongs to the set. To conclude that two differently described sets, such as $A\cup B$ and $B\cup A$ , are the same set, you need a principle converting "same membership condition for every $x$ " into "same set." That principle is exactly Set Equality and Extensionality: two sets are equal if and only if they have exactly the same elements. Every proof in this article's Theorem block ends by invoking extensionality after establishing the biconditional on memberships.

What happens with unions and intersections of more than two sets, or infinitely many sets?

Everything here generalizes, but requires new notation and a new definition, covered separately under Indexed Families of Sets. Briefly, for a family $\{A_i\}_{i\in I}$ , one defines $\bigcup_{i\in I} A_i$ by $x \in \bigcup_i A_i \iff \exists i \in I,\, x \in A_i$ , and $\bigcap_{i \in I} A_i$ by $x \in \bigcap_i A_i \iff \forall i \in I,\, x \in A_i$ , mirroring the quantifiers $\exists$ and $\forall$ the same way binary union and intersection mirror $\lor$ and $\land$ . Associativity of the binary operations is exactly what makes finite unions and intersections well-defined regardless of parenthesization before this generalization is even needed.

Is there an operation capturing "in A but not B"? Is that related to union and intersection?

Yes — that is set difference, $A \setminus B = \{x : x \in A \land x \notin B\}$ , treated under Set Difference and Complement. It is built from the same logical toolkit ( $\land$ together with negation $\lnot$ ), and interacts with union and intersection through identities such as De Morgan's laws, $A \setminus (B\cup C) = (A\setminus B)\cap(A\setminus C)$ . Understanding union and intersection first, with their biconditional-and-truth-table proof method, is exactly the preparation needed to prove those difference/complement identities rigorously rather than just diagrammatically.

Can A ∪ B ever equal A ∩ B without A = B?

No. If $A\cup B = A \cap B$ , take any $x \in A$ . Then $x \in A \cup B$ (since $A \subseteq A\cup B$ always), so by the assumed equality $x \in A\cap B$ , which means $x \in B$ . This shows $A \subseteq B$ . By a symmetric argument starting from $x \in B$ , one gets $B \subseteq A$ . Two mutual inclusions give $A = B$ by extensionality. So $A\cup B = A\cap B$ forces $A = B$ ; it is not a coincidence but a provable equivalence, and conversely if $A=B$ then trivially $A\cup B = A = A\cap B$ .