Union and Intersections proofs for real analysis

[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?

 

[kurt.physics]

For the other three problems, I am not sure how to approach them. Any help would be greatly appreciated. Thanks! For problem 1, you can show that f(A1 U A2) C= f(A1) U f(A2) by letting x € f(A1) U f(A2). Since (A1 U A2) >=A1, x€f(A1), and since (A1 U A2) >=A2, x€f(A2). Therefore, x € f(A1 U A2). This shows that f(A1 U A2) C= f(A1) U f(A2). To show the other direction, let x € f(A1 U A2). Since (A1 U A2) <=A1, x€f(A1), and since (A1 U A2) <=A2, x€f(A2). Therefore, x € f(A1) U f(A2). This shows that f(A1) U f(A2) C= f(A1 U A2). Since both directions have been established, it follows that f(A1 U A2) = f(A1) U f(A2).For problem 3, you can show that g(-1)(B1 U B2) C= g(B1) U g(B2) by letting x € g(B1) U g(B2). Since (B1 U B2) >=B1, x€g(B1), and since (B1 U B2) >=B2, x€g(B2). Therefore, x € g(-1)(B1 U B2). This shows that g(-1)(B1 U B2) C= g(B1) U g(B2). To show the other direction, let x € g(-1)(B1 U B2). Since (B1 U B2) <=B1, x€g(B1), and since (B1 U B2) <=B2, x€g(B2). Therefore, x € g(B1) U g(B2). This shows that g(B1) U g(