I decided to take a Real Analysis course, and I'd appreciate some help throughout this semester on questions I have if possible. I'm alrdy having trouble w/ one of the 1st problems..

1.) Given a function such that f : D ->Rand given that a subset B "is a subset of"R, let f^(-1)(B) be the set of all the points from domain D that will get mapped into B;

IE: f^(-1)(B) = {x "is an element of" D : f(x) "is an element of B}. Note that this is referred to as the preimage of B.

a.) Let f(x) = x^2. Suppose A is the interval [0, 4] and B is the interval [-1, 1] (note both of these are closed), what is f^(-1)(A) and f^(-1)(B).

So this seems elementary, and perhaps I'm overlooking it.

We know the domain D maps to the reals..

So for A, we'd have [0,0], [1,1], [2,4], [3,9], [4,16]

And for B, we'd have [-1,1], [0,0], [1,1]

I'm not quite sure what they're wanting.

b.) Does f^(-1)(A^B) = f^(-1)(A)^f^(-1)(B) in this case? Does f^(-1)(AvB) = f^(-1)(A)vf^(-1)(B)?

(**The v and ^ by itself means union and intersection, respectively)

Well, A^B would be [0,0],[1,1] if im not mistaken, and

AvB would be [-1,1], {A} => [-1,1], [0,0], [1,1], [2,4], [3,9], [4,16]

But if that's f^(-1)(A^B) and f^(-1)(AvB), what is it by component (ex. f^(-1)(A)^f^(-1)(B)

c.) The good behavior, as showed above, of these preimages is general. Show that for some arbitrary function g :R->R, its always true that g^(-1)(A^B) = g^(-1)(A)^g^(-1)(B) and g^(-1)(AvB) = g^(-1)(A)vg^(-1)(B) for all sets A, B "is a subset of"R

Not sure how to show this if I can't even show it for a specific function.

Thanks for your help and clarification.