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

[jonroberts74]

Homework Statement

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

Homework Equations

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.

 

[jbunniii]

intuition tells me no

Your intuition is correct but your reasoning is not.

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

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

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

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.

 

[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.

 

[jbunniii]

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

Right, except it should be F(A-B) = {1}.

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

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.

 

[jonroberts74]

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.

for the second part I did

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

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

 

[jonroberts74]

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

 

[jbunniii]

for the second part I did

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

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

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).

 

[jbunniii]

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

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.

 

[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}

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

 

[jbunniii]

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

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

So far so good.

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

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.

 

[jonroberts74]

oh right! intersect with the complement.

thanks!