Definition:Set Union

Definition
Let $S$ and $T$ be any two sets.

The union (or logical sum, or sum) of $S$ and $T$ is written $S \cup T$.

It means the set which consists of all the elements which are contained in either (or both) of $S$ and $T$:
 * $x \in S \cup T \iff x \in S \lor x \in T$

or, slightly more formally:
 * $A = S \cup T \iff \forall z: \left({z \in A \iff z \in S \lor z \in T}\right)$

We can write:
 * $S \cup T = \left\{{x: x \in S \lor x \in T}\right\}$

For example, let $S = \left \{{1,2,3}\right\}$ and $T = \left \{{2,3,4}\right\}$. Then $S \cup T = \left \{{1,2,3,4}\right\}$.

It can be seen that $\cup$ is an operator.

Disjoint Union
A commonly used special case of union is disjoint union which is simply a union of two disjoint sets. The possible notations are $\sqcup$ or $\dot{\cup}$.

By definition
 * $R = S \sqcup T \iff (R = S \cup T) \land (S \cap T = \emptyset)$.

Generalized Notation
Let $S = S_1 \cup S_2 \cup \ldots \cup S_n$. Then:


 * $\displaystyle \bigcup_{i \in \N^*_n} S_i = \left\{{x: \exists i \in \N^*_n: x \in S_i}\right\}$

If it is clear from the context that $i \in \N^*_n$, we can also write $\displaystyle \bigcup_{\N^*_n} S_i$.

An alternative notation for the same concept is $\displaystyle \bigcup_{i=1}^n S_i$.

If $\mathbb S$ is a set of sets, then the union of $\mathbb S$ is:
 * $\displaystyle \bigcup \mathbb S = \left\{{x: \exists X \in \mathbb S: x \in X}\right\}$

That is, the set of all elements of all elements of $\mathbb S$.

Thus:
 * $\displaystyle S \cup T = \bigcup \left\{{S, T}\right\}$

Illustration by Venn Diagram
The red area in the following Venn diagram illustrates $S \cup T$:


 * VennDiagramSetUnion.png

Axiomatic Set Theory
The concept of set union is axiomatised in the Axiom of Unions in Zermelo-Fraenkel set theory:
 * $\forall A: \exists x: \forall y: \left({y \in x \iff \exists z: \left({z \in A \land y \in z}\right)}\right)$

Also see

 * Set Intersection, a related operation.


 * Union of One Set, where it is shown that $\displaystyle \mathbb S = \left\{{S}\right\} \implies \bigcup \mathbb S = S$
 * Union of Empty Set, where it is shown that $\displaystyle \mathbb S = \left\{{\varnothing}\right\} \implies \bigcup \mathbb S = \varnothing$