The difference $a\setminus b$, the symmetric difference $a\mathrm{△}b$, and the complement $X\setminus a$ relative to a set $X$

Let $a$ and $b$ be sets. The difference of $a$ and $b$ is

$a\setminus b:=\{z\in a:z\mathrm{\notin }b\},$

a set by The Axiom Schema of Separation: for each formula $\phi$, $\mathrm{\forall }\bar{p}\mathrm{\forall }x\mathrm{\exists }y\mathrm{\forall }z(z\in y\leftrightarrow (z\in x\wedge \phi (z,\bar{p})))$ and the only one with those elements by The Axiom of Extensionality: $\mathrm{\forall }x\mathrm{\forall }y(\mathrm{\forall }z(z\in x\leftrightarrow z\in y)\to x=y)$; so $z\in a\setminus b$ holds exactly when $z\in a$ and $z\mathrm{\notin }b$. The symmetric difference is

$a\mathrm{△}b:=(a\setminus b)\cup (b\setminus a),$

formed with the binary union of The union $\bigcup x$ of a set, and the binary union $a\cup b:=\bigcup \{a,b\}$.

When $a\subseteq X$ (Subset $x\subseteq y$, proper subset $x⊊y$, and the separation notation $\{z\in x:\phi (z)\}$), the difference $X\setminus a$ is called the complement of $a$ relative to $X$. The qualification is not optional: there is no set of all sets (There is no set $U$ with $y\in U$ for every set $y$), so an absolute complement, the class of all sets not in $a$, is never a set.

• Difference needs no hypothesis. $a\setminus b$ is defined for all sets $a$ and $b$, with no assumption that $b\subseteq a$ or that the two meet. When they are disjoint, $a\setminus b=a$.

• Symmetric difference collects what lies in exactly one of the two sets. Unfolding the definition, $z\in a\mathrm{△}b$ holds exactly when $z$ belongs to $a$ or to $b$ but not to both.