# Disjoint union

In mathematics, the disjoint union (or discriminated union) of a family ${\displaystyle (A_{i}\colon i\in I)}$ of sets is a set ${\displaystyle A=\bigsqcup _{i\in I}A_{i},}$ with an injective function of each ${\displaystyle A_{i}}$ into A, such that the union of the images of these injections form a partition of A (that is, each element of A belongs to exactly one of these images).

The disjoint union of a family of pairwise disjoint sets is their set union.

In terms of category theory, the disjoint union is the coproduct of the category of sets.

The disjoint union is thus defined up to a bijection.

A standard way for building the disjoint union is to define A as the set of ordered pairs (x, i) such that ${\displaystyle x\in A_{i},}$ and the injective functions ${\displaystyle A_{i}\to A}$ by ${\displaystyle x\mapsto (x,i).}$

## Example

Consider the sets ${\displaystyle A_{0}=\{5,6,7\}}$ and ${\displaystyle A_{1}=\{5,6\}}$. We can index the set elements according to set origin by forming the associated sets

{\displaystyle {\begin{aligned}A_{0}^{*}&=\{(5,0),(6,0),(7,0)\}\\A_{1}^{*}&=\{(5,1),(6,1)\},\end{aligned}}}

where the second element in each pair matches the subscript of the origin set (e.g., the ${\displaystyle 0}$ in ${\displaystyle (5,0)}$ matches the subscript in ${\displaystyle A_{0}}$, etc.). The disjoint union ${\displaystyle A_{0}\sqcup A_{1}}$ can then be calculated as follows:

${\displaystyle A_{0}\sqcup A_{1}=A_{0}^{*}\cup A_{1}^{*}=\{(5,0),(6,0),(7,0),(5,1),(6,1)\}.}$

## Set theory definition

Formally, let {Ai : iI} be a family of sets indexed by I. The disjoint union of this family is the set

${\displaystyle \bigsqcup _{i\in I}A_{i}=\bigcup _{i\in I}\{(x,i):x\in A_{i}\}.}$

The elements of the disjoint union are ordered pairs (x, i). Here i serves as an auxiliary index that indicates which Ai the element x came from.

Each of the sets Ai is canonically isomorphic to the set

${\displaystyle A_{i}^{*}=\{(x,i):x\in A_{i}\}.}$

Through this isomorphism, one may consider that Ai is canonically embedded in the disjoint union. For ij, the sets Ai* and Aj* are disjoint even if the sets Ai and Aj are not.

In the extreme case where each of the Ai is equal to some fixed set A for each iI, the disjoint union is the Cartesian product of A and I:

${\displaystyle \bigsqcup _{i\in I}A_{i}=A\times I.}$

One may occasionally see the notation

${\displaystyle \sum _{i\in I}A_{i}}$

for the disjoint union of a family of sets, or the notation A + B for the disjoint union of two sets. This notation is meant to be suggestive of the fact that the cardinality of the disjoint union is the sum of the cardinalities of the terms in the family. Compare this to the notation for the Cartesian product of a family of sets.

Disjoint unions are also sometimes written ${\displaystyle \biguplus _{i\in I}A_{i}}$ or ${\displaystyle \ \cdot \!\!\!\!\!\bigcup _{i\in I}A_{i}}$.

In the language of category theory, the disjoint union is the coproduct in the category of sets. It therefore satisfies the associated universal property. This also means that the disjoint union is the categorical dual of the Cartesian product construction. See coproduct for more details.

For many purposes, the particular choice of auxiliary index is unimportant, and in a simplifying abuse of notation, the indexed family can be treated simply as a collection of sets. In this case ${\displaystyle A_{i}^{*}}$ is referred to as a copy of ${\displaystyle A_{i}}$ and the notation ${\displaystyle {\underset {A\in C}{\,\,\bigcup \nolimits ^{*}\!}}A}$ is sometimes used.

## Category theory point of view

In category theory the disjoint union is defined as a coproduct in the category of sets.

As such, the disjoint union is defined up to an isomorphism, and the above definition is just one realization of the coproduct, among others. When the sets are pairwise disjoint, the usual union is another realization of the coproduct. This justifies the second definition in the lead.

This categorical aspect of the disjoint union explains why ${\displaystyle \coprod }$ is frequently used, instead of ${\displaystyle \bigsqcup }$, to denote coproduct.