Simple Sets Homework: Q on f[A] & f[B] General Statement

[WannabeNewton]

Homework Statement

All the b's in f should be capitalized for the problem statement and attempt; I had it in the latex but it showed up lower case in the post I don't know why, my apologies =p.
If f:X \mapsto Y and A \subset X, B \subset X, is:
(a) f[A \cap B] = f[A] \cap f in general?
(b) f[A - B] = f[A] - f in general?

The Attempt at a Solution

My ability to write proofs is atrocious at best so bear with me please =D.
For (a), let y\in f[A \cap B], then there is an x\in A \cap B such that (x, y) \in f. Since x\in A and x\in B, y\in f[A]\cap f and f[A\cap B]\subset f[A]\cap f. Now, let y\in f[A]\cap f. For (a, y)\in f, a\in A and (b, y)\in f, b\in B a \neq b in general so even if y\in f[A]\cap f, x\notin A\cap B in general. Therefore, the statement (a) is not true in general. Is this enough?
(b) I have more of a question with this one: if y\in f[A] and y\notin f does that necessarily mean x\in A - B?

 

[CompuChip]

The first part of a) looks OK.
The second part is probably true, but it is not very clear.
Note that if you suspect that a statement is false, in general it is simplest to try and find a counter-example. You have already shown that you cannot find an y \in f(A \cap B) which is not in f[(A) \cap f(B), so you know where to look for the counter-example.

For b), the same advice. Can you think of a set A and a set B, such that x is not in A - B, but f(x) is in f(A) - f(B)?

Both statements are true by the way, when you replace f by f^-1 (defined by f^{-1}(y) = \{ x \in X \mid f(x) = y \}). This should also give you the hint that you won't find any counterexamples for injective functions (for which f^-1(y) is unique for all y).

 

[micromass]

Note that it IS true that

f(A\cup B)=f(A)\cup f(B)

so the union behaves nicely. Unlike the two operations in the OP.

Just throwing that out there

 

[WannabeNewton]

Sorry for responding so late but I'm on the move as it would seem =D.
For (a) if f:x \mapsto x^2 and A = [1 , 2, 3], B = [-5 , -4, -1] then f(A) = [1, 4, 9], f(B) = [25, 16, 1]. In this case, f(A \cap B) = \varnothing so f(A)\cap f(B) \nsubseteq f(A \cap B). I guess that makes it more clear.
For (b), if x\in A and x\in B then y\notin f(A) - f(B) which can't be. So I think the statement is true? I'm still not sure because the question asks me to determine which is more well behaved, f or f^{-1}, in terms of sets so I'm thinking something should be wrong with the statement (b) xD.