Show that if X and Y are topological spaces, and f : X --> Y is a continuous mapping, then f^-1(B) is a Borel set in X whenever B is a Borel set in Y.
see attached image for hints.
As the hint suggests, it is easy to show that for all $\displaystyle E, E_n \in B $:
A. $\displaystyle f^{-1}(\bigcup E_n) = \bigcup (f^{-1}(E_n)) $
B. $\displaystyle f^{-1}(E^c) = (f^{-1}(E))^c $
Recall that a sigma algebra is a set closed under compliments and countable unions; and a Borel sigma algebra is the "smallest" sigma algebra containing every open set; thus, you have to show 3 things.
1. If B is open or closed then so is $\displaystyle f^{-1}(B) $. This follows from continuity of f.
2. If B = union of open and/or closed sets, then $\displaystyle f^{-1}(B) $ is Borel. This follows from 1 and A.
3. If B is a compliment of a set from 1 or 2, then $\displaystyle f^{-1}(B^c) $ is borel. This follows from 1/2 and B.