# Simple sets question

1. Aug 17, 2011

### WannabeNewton

1. The problem statement, all variables and given/known data
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?
3. 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 $x\in A - B$?

2. Aug 17, 2011

### 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).

Last edited: Aug 18, 2011
3. Aug 17, 2011

### micromass

Staff Emeritus
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

4. Aug 18, 2011

### 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.