1. ## Inverse of intersections

Prove that $f^{-1} ( \bigcap _ { \alpha \in A } ) = \bigcap _{ \alpha \in A } f^ {-1} (E_ \alpha )$

My proof.

Pick $x \in f^{-1} ( \bigcap _ { \alpha \in A } E_ \alpha)$

implies that $\exists y \in \bigcap _{ \alpha \in A } E_ \alpha$ such that $f(y)=x$

Note that $y \in E_ \alpha \ \ \ \ \ \forall \alpha \in A$

implies that $x \in f^{-1} (E_ \alpha ) \ \ \ \ \ \forall \alpha \in A$

Therefore we have $x \in \bigcap _ { \alpha \in A } f^{-1} (E_ \alpha )$

On the other hand, pick $x \in \bigcap _ { \alpha \in A } f^ {-1} (E _ \alpha )$

implies that $x \in f^{-1} (E _ \alpha ) \ \ \ \ \ \forall \alpha \in A$

so $\exists y \in E_ \alpha$ such that $f(y) = x$

Note that we can find such y for every $\alpha \in A$, therefore, $x \in f^{-1} ( \bigcap _ { \alpha \in A } E_ \alpha )$

I have doubts about the second last line of my proof, is it reasonable for me to state that? Thanks!

2. $\begin{gathered}
\hfill \\
x \in f^{ - 1} \left( {\bigcap\limits_n {E_n } } \right)\, \Rightarrow \,f(x) \in \bigcap\limits_n {E_n } \, \Rightarrow \,\left( {\forall n} \right)\left[ {f(x) \in E_n } \right]\, \Rightarrow \,\left( {\forall n} \right)\left[ {x \in f^{ - 1} \left( {E_n } \right)} \right] \hfill \\
\end{gathered}$