Difference Set¶

Definition¶

Given two sets $A$ and $B\text{,}$ we define the difference set $A\setminus B$ or $A-B$ to be the set $A$ with any elements of $B$ removed. In other words, we have $x\in A\setminus B\;\iff\; (x\in A)\land (x\notin B)\text{.}$

Interpretation¶

The difference $A\setminus B$ keeps the elements of $A$ and removes those that also lie in $B$. The set $B$ acts only as a filter: its elements outside $A$ play no role, and $B$ need not be a subset of $A$.

The shaded crescent is $A\setminus B$$B$, the part of disc $A$ outside disc $B$. The lens $A\cap B$ is removed, and the right-hand crescent $B\setminus A$ is never part of the result. The two crescents are different sets in general, which is the asymmetry in the Key Results, and $A\setminus B=\varnothing$ if and only if $A\subseteq B$$B$. Inside a fixed universal set, $A\setminus B=A\cap B^{\complement}$ with the complement of $B$: "in $A$ and not in $B$".

A common slip is to count elements as $|A\setminus B|=|A|-|B|$$|B|$. For finite sets the correct count is $|A\setminus B|=|A|-|A\cap B|$$B|$: with $A=\{1,2,3\}$ and $B=\{2,4\}$ it is $3-1=2$$2$, not $3-2=1$$1$. The two agree only when $B\subseteq A$$A$.

Key Results¶

• Set difference is not commutative: in general $A\setminus B\neq B\setminus A$ (for example $\{1\}\setminus\{2\}=\{1\}$ but $\{2\}\setminus\{1\}=\{2\}$$\{2\}$).
• The difference of a set with itself is empty.

Examples¶

• If $A=\{1,\,2,\,3\}$ and $B=\{2,\,4\}\text{,}$ then $A\setminus B=\{1,\,3\}\text{.}$
• $A\setminus\varnothing=A\text{.}$

Prerequisites¶

• Set

Related Concepts¶

• Intersection of Sets
• Union of Sets
• Disjoint Sets
• Subtraction and Difference

Sources¶

• Math I: Calculus · Lecture 1

Aliases: set difference, difference of sets, relative complement, set minus, set-theoretic difference Symbols: element of, intersection, subset

Last updated: 20 October 2026 03:02