Theorem. Let $f$ be a mapping from $X$ to $Y$ . If $\{B_i\}_{i\in I}$ is a (possibly uncountable) collection of subsets in $Y$ , then the following relations hold for the inverse image:

(1)

$ \displaystyle f^{-1}\big(\bigcup_{i\in I} B_i\big) = \bigcup_{i\in I} f^{-1}\big(B_i\big) $

(2)

$ \displaystyle f^{-1}\big(\bigcap_{i\in I} B_i\big) = \bigcap_{i\in I} f^{-1}\big(B_i\big) $

(3)

For the set complement, $$\big(f^{-1}(A)\big)^\complement=f^{-1}(A^\complement).$$

(4)

For the set difference, $$f^{-1}(A\setminus B) = f^{-1}(A)\setminus f^{-1}(B).$$

(5)

For the symmetric difference, $$f^{-1}(A \bigtriangleup B)=f^{-1}(A) \bigtriangleup f^{-1}(B).$$

Proof. For part (1), we have \begin{eqnarray*} f^{-1}\big(\bigcup_{i\in I} B_i\big) &=& \Big\{ x\in X\mid f(x) \in \bigcup_{i\in I} B_i\Big\} \\ &=& \left\{ x\in X \mid f(x) \in B_i\ \mbox{for some}\ i\in I\right\} \\ &=& \bigcup_{i\in I}\left\{ x\in X \mid f(x) \in B_i \right\} \\ &=& \bigcup_{i\in I} f^{-1}\big(B_i\big). \end{eqnarray*}Similarly, for part (2), we have \begin{eqnarray*} f^{-1}\big(\bigcap_{i\in I} B_i\big) &=& \big\{ x\in X \mid f(x) \in \bigcap_{i\in I} B_i\big\} \\ &=& \left\{ x\in X \mid f(x) \in B_i\ \mbox{for all}\ i\in I\right\} \\ &=& \bigcap_{i\in I}\left\{ x\in X \mid f(x) \in B_i \right\} \\ &=& \bigcap_{i\in I} f^{-1}\big(B_i\big). \end{eqnarray*}For the set complement, suppose $x\notin f^{-1}(A)$ . This is equivalent to $f(x)\notin A$ , or $f(x)\in A^\complement$ , which is equivalent to $x\in f^{-1}(A^\complement)$ . Since the set difference $A\setminus B$ can be written as $A\cap B^c$ , part (4) follows from parts (2) and (3). Similarly, since $A\bigtriangleup B=(A\setminus B) \cup (B\setminus A)$ , part (5) follows from parts (1) and (4).