Two concrete sets are disjoint if they have no common elements.

Given an abstract set $V$, two subsets $A$ and $B$ of $V$ **meet** if their intersection is inhabited; $A$ and $B$ are **disjoint** if they do not meet, in other words if their intersection is empty.

In material set theory, we may speak of take $A$ and $B$ to be simply sets, rather than subsets of some ambient set? $V$. Equivalently, one may take $V$ to be the class of all sets by default. (In this context, it’s important that whether $A$ and $B$ meet or are disjoint is independent of the ambient set or class.)

In constructive mathematics, the default meaning of ‘disjoint’ is as above, but sometimes one wants a definition relative to some inequality relation $\ne$ on $V$. Then $A$ and $B$ are **$\ne$-disjoint** if, whenever $x \in A$ and $y \in B$, $x \ne y$. (Ordinary disjointness is relative to the denial inequality.)

The concrete sets $A$ and $B$ are disjoint iff they have an internal disjoint union, in other words if their inclusions into their union $A \cup B$ form a coproduct diagram in the category of sets. (Etymologically, of course, this is backwards.)

Many authors are unfamiliar with disjoint unions. When the disjoint union oid two abstract sets $A$ and $B$ is needed, they will typically lapse into material set theory (even when the work is otherwise perfectly structural), and make some comment such as ‘without loss of generality, assume that $A$ and $B$ are disjoint’ or (especially when $A = B$) ‘take two isomorphic copies of $A$ and $B$’, then call the disjoint union simply a ‘union’. (This works by the previous paragraph.)

In any category with an object $V$, two subobjects $A$ and $B$ are **disjoint** if their pullback is initial in $C$. Then disjoint subsets are precisely disjoint subobjects in Set.

To internalize the characterization in terms of internal disjoint unions is harder. If $A$ and $B$ have a join in the poset of subobjects $Sub(V)$, then we may ask whether this forms a coproduct diagram in $C$. This should be equivalent if $C$ has disjoint coproducts.

Last revised on September 1, 2017 at 18:48:00. See the history of this page for a list of all contributions to it.