Union collects everything that belongs to either of two sets. Intersection keeps only the elements shared by both. These operations turn membership conditions into new sets.

For two study clubs, the union contains students in at least one club. The intersection contains students who belong to both clubs.

For any object x, membership in the union means membership in A or in B, including the case where both hold. Membership in the intersection requires membership in both sets.

$x\in A\cup B$ means $x\in A$ or $x\in B$ (or both).

• $x\in A\cap B$ means $x\in A$ and $x\in B$.
• The sets are disjoint when .

Three places and three answers: $p$ is in $A$ alone, $r$ in $B$ alone, and $q$ in both. The union is all three together; the intersection is $q$’s place by itself.

The intervals $A_n=(0,\tfrac1n)$ are nested and none is empty, yet : any is missed by every with , and itself is in none of them. An infinite intersection can lose everything that each finite one kept.

The number 5 is shared and stays in the result. The result also contains 1 and 9.

Only 5 passes both membership checks.

The number 4 differs from 7, so no element belongs to both sets.

If $P=\{0,1\}$, $Q=\{1,2\}$ and $R=\{3\}$, then .

The inner union contains 1, 2, and 3. Of these, only 1 also belongs to P.

When there are more than two sets, they are named by an index and the operations run over the whole family.

• $x\in A\cup B$ is an inclusive or; $x\in A\cap B$ is an and.
• A family $\{A_i\}_{i\in I}$ has for "some index" and for "every index".

Great work! You now know how to:

• Compute unions and intersections of listed sets
• Read and use the notation for a family indexed by a set
• Tell "for some index" from "for every index"
• Give a family of non-empty sets with empty intersection