I am trying to solve the below problem.
Let $f:X\to Y$ be a map of set.
For $B\subset Y$ we define $f^{-1}(B)=\{x\in X\,|\,f(x)\in B\}$.
Show that $$f^{-1}(Y-B)=X-f^{-1}(B)\;\;,\quad \textstyle f^{-1}\Bigl(\bigcup_{i\in I} B_i\Bigr)=\bigcup_{i\in I} f^{-1}(B_i)\;\;, \quad \text{ and }\;\; f^{-1}\Bigl(\bigcap_{i\in I} B_i\Bigr)=\bigcap_{i\in I} f^{-1}(B_i)\;\;.$$
Here is my attempt.
(a) For any $x \in X$, we have: \begin{align*} x \in f^{-1} (Y - B) & \iff f(x) \in Y - B \\ & \iff f(x) \not \in B \\ & \iff x \not \in f^{-1} (B) \\ & \iff x \in X - f^{-1} (B) \end{align*} So $f^{-1} (Y - B) = X - f^{-1} (B)$.
Furthermore, we have: \begin{align*} x \in f^{-1} \left(\bigcup\limits_{i \in I} B_i\right) & \iff f(x) \in \bigcup\limits_{i \in I} B_i \\ & \iff \exists i \in I, \; f(x) \in B_i \\ & \iff \exists i \in I, \; x \in f^{-1} (B_i) \\ & \iff x \in \bigcup\limits_{i \in I} f^{-1} (B_i) \end{align*} So $f^{-1} \left(\bigcup\limits_{i \in I} B_i\right) = \bigcup\limits_{i \in I} f^{-1} (B_i)$.
Finally, we have: \begin{align*} x \in f^{-1} \left(\bigcap\limits_{i \in I} B_i\right) & \iff f(x) \in \bigcap\limits_{i \in I} B_i \\ & \iff f(x) \in B_i, \; \forall i \in I \\ & \iff x \in f^{-1} (B_i), \; \forall i \in I \\ & \iff x \in \bigcap\limits_{i \in I} f^{-1} (B_i) \end{align*} So $f^{-1} \left(\bigcap\limits_{i \in I} B_i\right) = \bigcap\limits_{i \in I} f^{-1} (B_i)$, as desired.
How does this look?
