Definition:Set Union

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

The (set) union 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, in this form, $\cup$ is a binary operation which acts on sets.

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

Historical Note
The symbol $\cup$, informally known as cup, was first used by Hermann Grassmann in Die Ausdehnungslehre from 1844. However, he was using it as a general operation symbol, not specialized for union.

It was Giuseppe Peano who took this symbol and used it for union, in his 1888 work Calcolo geometrico secondo l'Ausdehnungslehre di H. Grassmann.

Peano also created the large symbol $\bigcup$ for general union of more than two sets. This appeared in his Formulario Mathematico (5th edtion, 1908).

Also known as
The union of sets is also known as the logical sum, or just sum, but these terms are usually considered old-fashioned nowadays.

Some authors use the notation $S + T$ for $S \cup T$, but this is non-standard and can be confusing, so its use is not recommended.

Also, $S + T$ is sometimes used for disjoint union.

Also see

 * Definition:Set Intersection, a related operation.


 * Definition:Disjoint Union (Set Theory)


 * Union of Singleton, 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 $\displaystyle \mathbb S = \varnothing \implies \bigcup \mathbb S = \varnothing$