# Determine if true or false F(A-B) = F(A) -F(B)

1. Jul 20, 2014

### jonroberts74

1. The problem statement, all variables and given/known data

Let X and Y be sets, and let A and B be any subsets of X.Determine if for all functions from X to Y, F(A-B) = F(A) - F(B) Justify your answer

2. Relevant equations

3. The attempt at a solution

intuition tells me no because the F(A-B) will have a different x values going to a different y values in Y than F(A) - F(B)

also,

the left side will have x values from X such that they are in $$A \cap B^c$$

whereas the right side would have x values from X such that they are in $$A \cup B$$

and clearly $$A \cap B^c \neq A \cup B$$

but I have a feeling that intuition is incorrect.

2. Jul 20, 2014

### jbunniii

No. The left side is a subset of Y, not of X.
No. The right side is a subset of Y, not of X.

Try looking for a counterexample to show that the equality need not be true. It doesn't have to be complicated: I found one where X only contains two points.

3. Jul 20, 2014

### jonroberts74

if I think about the resulting y's in Y

$$F(A-B) = F(A \cap B^c)$$

and

$$F(A) - F(B) = F(A) \cap [F(B)]^c$$

maybe as a simple example

let F be y = x^2

A = {1,2,3} and B = {2,3,4}

so F(A-B) = 1

where F(A) - F(B) = 1,-3 -8, -15, 0, -5, -12, 5, -6

I don't think this is properly done notation wise. I didn't know if it should go in set roster form, ordered pairs or what.

4. Jul 20, 2014

### jbunniii

Right, except it should be F(A-B) = {1}.
How did you get this? F(A) = {1,4,9} and F{B} = {4,9,16}. So F(A) - F(B) = {1}. So unfortunately this isn't a counterexample.

Hint: Try an example where F is not injective.

5. Jul 20, 2014

### jonroberts74

for the second part I did

each value from A minus each value from B after the function was acted upon them

like $$1^2 - 2^2 = -3; 1^2-3^2 = -8$$

6. Jul 20, 2014

### jonroberts74

I used y=x^2 because its not injective and its a simple one

7. Jul 20, 2014

### jbunniii

But doesn't F(A) - F(B) mean the set difference? In other words, as you wrote earlier, $F(A) \cap [F(B)]^c$. This has nothing to do with subtracting values. It means the set of all elements of F(A) which are not in F(B).

8. Jul 20, 2014

### jbunniii

It's not injective, but you only evaluated it at positive values, and it is injective if you restrict its domain to positive values. Try including some negative values as well.

9. Jul 20, 2014

### jonroberts74

A = {-1,1,-2,3} B = {2,3,4}

F(A-B) = {1,4}

F(A) - F(B) = {1,4,9} intersect {4,9,16} = {4,9}

$$\{1,4\} \neq \{4,9\}$$

10. Jul 20, 2014

### jbunniii

So far so good.
No, F(A) - F(B) = stuff in {1,4,9} which is not in {4,9,16}, so F(A) - F(B) = {1}. So this is a valid counterexample, but not for the reason you gave. :tongue:

11. Jul 20, 2014

### jonroberts74

oh right!! intersect with the complement.

thanks!