1. Analysis, Borel sets

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.

2. Originally Posted by JJMC89
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.

3. Originally Posted by southprkfan1
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. $\displaystyle f^{-1}(E^c) = (f^{-1}(E))^c$

I don't know how to show A.

4. Originally Posted by JJMC89
I don't know how to show A and B.
Come on. Why don't you try?

5. Originally Posted by Drexel28
Come on. Why don't you try?
Come on. Why don't you help a little ?

To the OP : consider an element in the set that is LHS of the equation, and prove it belongs to the set that is RHS of the equation.
And vice-versa

6. Originally Posted by Moo
Come on. Why don't you help a little ?
Because someone doing measure theory and algebraic topology should be expected to just know let alone be able to prove a common fact about mappings.