Textbook
The union of sets $A$ and $B$ contains the elements belonging to at least one of them:
$$A\cup B=\{x\mid x\in A\lor x\in B\}.$$
This logical “or” permits both conditions to hold. A shared element is included only once. The definition gives $A\cup B=B\cup A$, $A\cup A=A$, and $A\cup\varnothing=A$: the membership conditions on each side coincide.
For an indexed family $(A_i)_{i\in I}$, the rule is unchanged: an element must belong to at least one set indexed by $I$. The symbol $\exists$ means “there exists”:
$$\bigcup_{i\in I}A_i=\{x\mid\exists i\in I:\ x\in A_i\}.$$
If the index set $I$ is empty, no qualifying index exists, so the union of the empty family is $\varnothing$.
Each input set is included in the union. If a set $D$ contains both $A$ and $B$, it contains their union: every union element comes from at least one input and therefore belongs to $D$. Thus union is the smallest set, by inclusion, containing both inputs. Successively adding sets never removes elements already admitted. With several unions, parentheses do not change the condition “in at least one,” so grouping does not affect the result.
Related concepts: Set membership.
In plain language
Union resembles a guest list assembled from several groups: one invitation is enough for inclusion. A guest invited by two groups still arrives once; the operation does not issue a spare body. Each new group adds only names not already recorded. The result keeps the guest list, but not invitation counts or the identity of each sender.
Example
The illustrative sets are $A=\{1,2,3\}$, $B=\{3,4\}$, and $C=\{2,3,5\}$. First unite two sets, then all three.
Set union
Initial set A
Every element of A already belongs to at least one set.
Step 1 of 3: Initial set A
Read all steps
Initial set A
Every element of A already belongs to at least one set.
- Result
- 1included2included3included
Addition of B
The number 3 is already present; only 4 is new.
- B
- 34current
- Result
- 1included2included3included4current
Addition of C
The numbers 2 and 3 are already present; 5 is added. Membership in any one set is sufficient.
- C
- 235current
- Result
- 1included2included3included4included5current
After adding $B$, three remains: “or” permits simultaneous membership. Adding $C$ likewise keeps two and three without extra copies, while five expands the result to five distinct elements. The result satisfies both directions: every final number has an input set as its source, and no input number is missing. Simply adding the input sizes would produce eight entries, but that counts appearances rather than distinct elements of the union.