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===Disjoint union=== {{main|Disjoint union}} The ''disjoint union'' of two or more sets is similar to the union, but, if two sets have elements in common, these elements are considered as distinct in the disjoint union. This is obtained by labelling the elements by the indexes of the set they are coming from. The disjoint union of two sets {{tmath|A}} and {{tmath|B}} is commonly denoted {{tmath|A\sqcup B}} and is thus defined as <math display=block>A\sqcup B=\{(a,i)\mid (i=1 \land a\in A)\lor (i=2 \land a\in B\}.</math> If {{tmath|1=A=B}} is a set with {{tmath|n}} elements, then {{tmath|1=A\cup A =A}} has {{tmath|n}} elements, while {{tmath|1=A\sqcup A}} has {{tmath|2n}} elements. The disjoint union of two sets is a particular case of the disjoint union of an indexed family of sets, which is defined as <math display=block>\bigsqcup_{i \in \mathcal I}=\{(a,i)\mid i\in \mathcal I \land a\in A_i\}.</math> The disjoint union is the [[coproduct]] in the [[category (mathematics)|category]] of sets. Therefore the notation <math display=block>\coprod_{i \in \mathcal I}=\{(a,i)\mid i\in \mathcal I \land a\in A_i\}</math> is commonly used. ==== Internal disjoint union ==== Given an indexed family of sets {{tmath|(A_i)_{i\in \mathcal I} }}, there is a [[canonical map|natural map]] <math display=block>\begin{align} \bigsqcup_{i\in \mathcal I} A_i&\to \bigcup_{i\in \mathcal I} A_i\\ (a,i)&\mapsto a , \end{align}</math> which consists in "forgetting" the indices. This maps is always surjective; it is bijective if and only if the {{tmath|A_i}} are [[pairwise disjoint]], that is, all intersections of two sets of the family are empty. In this case, <math display=inline>\bigcup_{i\in \mathcal I} A_i</math> and <math display=inline>\bigsqcup_{i\in \mathcal I} A_i</math> are commonly identified, and one says that their union is the ''disjoint union'' of the members of the family. If a set is the disjoint union of a family of subsets, one says also that the family is a [[partition of a set|partition]] of the set.
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