Let $C$ be a class of sets. The intersection of $C$ is
\[\bigcap C = \left\{x : (\forall X\in C:x\in X)\right\}.\]This is a set: $\bigcap C\subseteq X$ for any $X\in C$.
We write
\[\begin{align*} A\cap B &= \bigcap\{A,B\}, \\ A\cap B\cap C &= (A\cap B)\cap C, \\ A\cap B\cap C\cap D &= (A\cap B\cap C)\cap D \\ \end{align*}\]and so on.
- \[\bigcap\varnothing = \varnothing.\]
For any $X$,
\[X\cap\varnothing = \varnothing.\]
For any $X$,
\[X\cap X = X.\]
For any $X$ and $Y$,
\[X\cap Y = Y\cap X.\]
For any $X$, $Y$ and $Z$,
\[(X\cap Y)\cap Z = X\cap(Y\cap Z).\]
$(\V,\cap)$ is an abelian semigroup.
Proof.By commutativity and associativity of $\cap$.