Intersection of Sets¶

Definition¶

The intersection $A\cap B$ of two sets $A$ and $B$ is defined to be the set

\begin{aligned} &A\cap B=\{x\in A\colon x\in B\},\quad \text{or equivalently}\\ &x\in A\cap B\;\iff\; x\in A \text{ and }x\in B \iff \; (x\in A) \land (x\in B). \end{aligned}

Thus, $A\cap B$ consists of all the elements which belong to both $A$ and $B\text{.}$

Interpretation¶

The intersection of sets captures the elements common to all sets involved. An element of $A\cap B$ must pass both membership tests at once, so intersection is the set version of the logical "and".

Two overlapping discs A and B with their common lens shaded as the intersection.

The shaded lens is where the two discs overlap; its points lie in $A$ and in $B$ . Points in only one of the discs are left out. The lens lies inside each disc, so $A\cap B\subseteq A$ and $A\cap B\subseteq B$ $B$ ; if the discs were moved apart, the lens would vanish and the sets would be disjoint.

The form $\{x\in A\colon x\in B\}$ reads as a filter: start from $A$ and keep the elements that also lie in $B$ . Although this notation treats $A$ and $B$ differently, the result does not depend on the order. If $A\subseteq B$ $B$ , the filter removes nothing: $A\cap B=A$ if and only if $A\subseteq B$ $B$ . By associativity, $A_1\cap\cdots\cap A_n$ needs no brackets.

In probability, $A\cap B$ returns as the event that both $A$ and $B$ happen (see union and intersection of events).

Key Results¶

An element belongs to the intersection if and only if it belongs to each set.
Intersection is commutative (i.e., $A\cap B=B\cap A$ $A$ ) and associative (i.e., $A\cap(B\cap C)=(A\cap B)\cap C$ $C$ ).
The intersection with the empty set is empty.

Examples¶

$\{1,\,2,\,3\}\cap\{2,\,3,\,4\}=\{2,\,3\}\text{.}$
$A\cap\varnothing=\varnothing\text{.}$

Prerequisites¶

Set

Sources¶

Math I: Calculus · Lecture 1

Aliases: set intersection Symbols: element of, intersection, subset

Last updated: 22 October 2026 19:04