Struggling on what steps to take to construct a successful proof for this. This will be the last. Can I get a verification? I think the approach and logic is wrong.
Prove:
If $f:A \rightarrow B$ is a function with domain $A$ and $T_i$ with $i \in I$ is a family of sets where $\forall i \in I$,$T_i \subseteq B$ then
$f^{-1}( \bigcup\limits_{i \in I} T_{i})= \bigcup\limits_{i \in I}f^{-1} (T_{i})$
Proving Subset both ways
Prove $f^{-1}( \bigcup\limits_{i \in I} T_{i}) \subseteq \bigcup\limits_{i \in I}f^{-1} (T_{i})$
Assume $s \in f^{-1}( \bigcup\limits_{i \in I} T_{i})$
$\implies$ $f(s) \in \bigcup\limits_{i \in I} T_{i}$
$(\exists i \in I)$ s.t. $f(s) \in T_{i}$
$(\exists i \in I)$ s.t. $s \in f^{-1} (T_{i})$
So
$f^{-1}( \bigcup\limits_{i \in I} T_{i}) \subseteq \bigcup\limits_{i \in I}f^{-1} (T_{i})$
Next Prove
$f^{-1}( \bigcup\limits_{i \in I} T_{i}) \supseteq \bigcup\limits_{i \in I}f^{-1} (T_{i})$
Assume
$s \in \bigcup\limits_{i \in I}f^{-1} (T_{i})$
$\implies (\exists i \in I)$ s.t. $s \in f^{-1}(T_i)$
$\implies (\exists i \in I)$ s.t. $f(s) \in T_i$
$\implies f(s)\in \bigcup\limits_{i \in I}T_{i}$
$\implies s \in f^{-1}(\bigcup\limits_{i \in I}T_{i})$