Textbook
The intersection of $A$ and $B$ consists of elements belonging to both sets:
$$A\cap B=\{x\mid x\in A\land x\in B\}.$$
Logical “and” requires both conditions. Hence $A\cap B\subseteq A$ and $A\cap B\subseteq B$. Order does not matter; $A\cap A=A$, and $A\cap\varnothing=\varnothing$.
Given a universe $U$ and a family of subsets $A_i\subseteq U$ indexed by $i\in I$. The symbol $\forall$ means “for every”:
$$\bigcap_{i\in I}A_i=\{x\in U\mid\forall i\in I:\ x\in A_i\}.$$
If even one set excludes $x$, the element is rejected. When $I$ is empty, no index violates the condition, so this relative intersection equals $U$. The universe must therefore be stated explicitly for an empty family.
The intersection is the largest set, by inclusion, contained in both $A$ and $B$. Indeed, if $D\subseteq A$ and $D\subseteq B$, every element of $D$ satisfies both conditions and belongs to $A\cap B$. In particular, when $A\subseteq B$, the intersection equals $A$. Adding another set to a nonempty family of requirements can only preserve or reduce the result: passing all checks requires passing every earlier check as well.
Related concepts: Set membership.
In plain language
Intersection resembles finding an evening free in every friend’s calendar. A free Tuesday in one calendar and a free Wednesday in another cannot rescue the meeting: the same evening must work throughout. Each additional calendar adds a requirement, so the number of suitable dates cannot increase. One surprise meeting can empty the result; a changed schedule requires a fresh comparison.
Example
The illustrative sets are $A=\{1,2,3\}$, $B=\{3,4\}$, and $C=\{2,3,5\}$. Elements failing each successive requirement are removed.
Set intersection
Candidates from A
The elements of the first set are the initial candidates.
Step 1 of 3: Candidates from A
Read all steps
Candidates from A
The elements of the first set are the initial candidates.
- Candidates
- 1current2current3current
Membership in B
Only 3 among the candidates belongs to B. The number 4 cannot enter: it was not in A.
- A
- 1excluded2excluded3included
- B
- 3included4
Membership in C
The number 3 is also in C, so it satisfies all three requirements.
- C
- 23included5
- Result
- 3included
After the second step, only three remains, so the third set checks that surviving candidate. Two occurs in $A$ and $C$ but not in $B$; it cannot reappear during the last step. One later success does not erase an earlier failure. The final three belongs to all three original sets, and every rejected candidate failed at least one requirement. This verifies both admission and exclusion rather than trusting the final highlight alone.