Collections of Sets, Problem 1(a)


Suppose that y is an arbitrary element of \(f(A \cap B)\). By the definition of the image of a set, there is an element \(x \in A \cap B\) such that \(f(x)=y\). Since \(x \in A \cap B\), we know \(x \in A\) and \(x \in B\). Thus, \(y \in f(A)\) and \(y \in f(B)\). So \(y \in f(A) \cap f(B)\). Since y was chosen arbitrarily, we conclude that \(f(A \cap B) \subseteq f(A)\cap f(B)\).