Do I need induction to prove this? Show that if $B_1, B_2, \ldots , B_n \subseteq T,$ then $$f^{-1}(B_1 \cap B_2 \cap \cdots \cap B_n) = f^{-1}(B_1) \cap f^{-1}(B_2) \cap \cdots \cap f^{-1}(B_n).$$
My question is:
I got a hint to use induction to prove this in which I should prove the base case for $n=2,$ but my opinion is I do not need induction and I will just prove this by the same way I proved the statement for the case $n = 2.$ Could anyone tell me which is the correct idea please?
 A: I wouldn't use induction here.
\begin{align}
& w \in f^{-1} \left( B_1\cap \cdots \cap B_n \right) \\[8pt]
\text{iff } & f(w) \in B_1\cap \cdots \cap B_n \\[8pt]
\text{iff } & \text{ for all } i\in\{1,\ldots,n\},\,\, f(w)\in B_i \\[8pt]
\text{iff } & \text{ for all } i\in\{1,\ldots,n\},\,\, w\in f^{-1}(B_i) \\[8pt]
\text{iff } & w \in f^{-1}(B_1) \cap \cdots \cap f^{-1}(B_n).
\end{align}
In fact,
$$
f^{-1} \left( \bigcap_{i\,\in\,\mathcal I} B_i \right) = \bigcap_{i\,\in\,\mathcal I} f^{-1}(B_i),
$$
with no assumption that $\mathcal I$ is finite.
Proof:
\begin{align}
& w \in f^{-1} \left( \bigcap_{i\,\in\,\mathcal I} B_i \right) \\[8pt]
\text{iff } & f(w) \in \bigcap_{i\,\in\,\mathcal I} B_i \\[8pt]
\text{iff } & \text{ for all } i\in\mathcal I,\,\, f(w)\in B_i \\[8pt]
\text{iff } & \text{ for all } i\in\mathcal I,\,\, w\in f^{-1}(B_i) \\[8pt]
\text{iff } & w \in \bigcap_{i\,\in\,\mathcal I} f^{-1}(B_i).
\end{align}
A: No, induction is not needed nor does the number of B's have to be finite.
Let C be a collection of sets.
Theorem.
f$^{-1}$($\cap$C) = $\cap${ f$^{-1}$(B) : B in C }.
Proof.
x in f$^{-1}$($\cap$C) iff
f(x) in $\cap$C iff
for all B in C, f(x) in B iff
for all B in C, x in f$^{-1}$(B) iff
x in $\cap${ f$^{-1}$(B) : B in C }.  
Exercise.  Prove, in a similar fashion,
f$^{-1}$($\cup$C) = $\cup${ f$^{-1}$(B) : B in C }. 
