Real analysis - unions and intersections

[badsis]

Hi,

I have four similar problems that I am not sure how to do: Given: A1 and A2 are in X, B1 and B2 are in Y f: X->Y, g - inverse of f
I have to either prove or if false find counterargument
1. f(A1 U A2) = f(A1) U f(A2)
2. f(A1 n A2) = f(A1) n f(A2)
3. g(-1)(B1 U B2) = g(B1) U g(B2)
4. g(B1 n B2) = g(B1) n f(B2)

I started doing 2. I was able to show that f(A1 n A2) C=(is contained in) f(A1) n f(A2):
let x € f(A1) and x € f(A2)
since (A1 n A2) <=A1, x€f(A1)
since (A1 n A2) <=A2, x€f(A2)
=> x € f(A1 n A2), x € f(A1) n f(A2), i.e. (A1 n A2) C= f(A1) n f(A2)

But I am not sure how to show the other way, i.e. that f(A1) n f(A2) C= (A1 n A2), in order to conclude that both expressions are equal. Or are they equal at all?

 

[HallsofIvy]

Are you saying that you are given that f has an inverse- that f is one-to-one and onto- or are you saying that (3) and (4) ask about f^-1 of the sets? Those are very different things!

 

[snipez90]

This is set theory, not real analysis. The first one should be straightforward. Your notation is sort of hard to follow for 2, but that is the correct containment direction you concluded. Can you try coming up with a simple counterexample for the reverse direction? Consider perhaps a set A containing distinct elements a1 and a2, and let A1 = {a1} and A2 = {a2} and think of the simplest functions possible.

 

[badsis]

Ok,
That worked.

Thanks!