The difference $a \setminus b$, the symmetric difference $a \triangle b$, and the complement $X \setminus a$ relative to a set $X$

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

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

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

$a \triangle b := (a \setminus b) \cup (b \setminus a),$

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 $x \subsetneq y$, and the separation notation $\{\, z \in x : \varphi(z) \,\}$), 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, $z \in a \triangle b$ holds exactly when $z$ belongs to $a$ or to $b$ but not to both.