Show that (b), (c), (f), and (g) of Exercise 2 hold for arbitrary unions and intersections.
Solution
Let \(\{B_i\}_{i\in I}\) be subsets of \(B\) and \(\{A_i\}_{i\in I}\) subsets of \(A\).
Preimages commute with arbitrary unions: \[ \begin{aligned} x\in f^{-1}\!\left(\bigcup_i B_i\right) &\iff f(x)\in B_i\text{ for some }i\\ &\iff x\in\bigcup_i f^{-1}(B_i). \end{aligned} \] Hence \[ f^{-1}\!\left(\bigcup_iB_i\right)=\bigcup_i f^{-1}(B_i). \] Similarly, \[ f^{-1}\!\left(\bigcap_iB_i\right)=\bigcap_i f^{-1}(B_i). \]
Images commute with arbitrary unions: \[ f\!\left(\bigcup_i A_i\right)=\bigcup_i f(A_i). \] For intersections one always has \[ f\!\left(\bigcap_i A_i\right)\subset\bigcap_i f(A_i), \] since a point of the common domain intersection maps into every \(f(A_i)\). Equality can fail for noninjective \(f\): the two-set counterexample of E-MUN-2-2(g) is a family indexed by \(I=\{0,1\}\). If \(f\) is injective and the index set is nonempty, equality holds: if \(y\in\bigcap_i f(A_i)\), then for each \(i\) there is \(a_i\in A_i\) with \(f(a_i)=y\); injectivity forces all \(a_i\) to be the same point, lying in \(\bigcap_iA_i\).