Does a set exist containing the elements shared between two sets? How might we construct such a set?

Let $A$ and $B$ denote sets. Consider the set $\Set{x \in A}{x \in B}$. This set exists by the principle of specification (see Set Specification). Moreover $(y \in \Set{x \in A}{x \in B}) \iff (y \in A \land y \in B)$. In other words, $\Set{x \in A}{x \in B}$ contains all the elements of $A$ that are also elements of $B$.

We can also consider $\Set{x \in B}{x \in A}$,
in which we have swapped the positions of $A$
and $B$.
Similarly, the set exists by the principle of
specification (see Set Specification) and again $y \in \Set{x \in B}{x \in A}
\iff (y \in B \land y \in B)$.
Of course, $y \in A \land y \in B$ means the
same as^{1}
$y \in B \land y \in A$ and so by the
principle of extension (see Set Equality)

\[ \Set{x \in A}{x \in B} = \Set{x \in B}{x \in A}. \]

We call this set the pair intersection of the set denoted by $A$ with the set denoted by $B$.We denote the intersection fo the set denoted by $A$ with the set denoted by $B$ by $A \cap B$. We read this notation aloud as “A intersect B”.

All the following results are immediate.^{2}

$A \cap \varnothing = \varnothing$

$A \cap B = B \cap A$

$(A \cap B) \cap C = A \cap (B \cap C)$

$A \cap A = A$

$(A \subset B) \iff (A \cap B = A)$.