Prove inverse image of function

[simmonj7]

Homework Statement

When f: A -> B and S ⊆ B, we definte I_f(S) = { x ∈ A: f(s) ∈ S}. Let X and Y be subets of B.
a) Determine whether must equal I_f(X) ∪ I_f(Y).
b) Determine whether I_f(X ∩ Y) must equal I_f(X) ∩ I_f(Y).

Homework Equations

The definition of an inverse image is stated in the problem.

The Attempt at a Solution

So far I have tried to draw the schematic representation of the problem and didn't really get anything out of that.
Then I tried messing around with some proofs by saying "Choose an arbitrary x in I_f(X ∪ Y). Then, x = f(a) for some a in X ∪ Y." And then from there are I start questioning myself and don't know if I should continue/where I should go.

Thanks

 

[hgfalling]

Well, you seem to be kind of on the right track with your second attempt.

Suppose you have an arbitrary x in I_f(X $$\cup$$Y). Then a = f(x) for some a in X $$\cup$$Y.

Now break that union sign apart, and see what you get.

The same type of thinking will work for the "and" part of the question too.

Think about what this all means: you have two domains. In domain #1, you have "source" points. In domain #2, you have "target" points, like X and Y.

Now suppose you have a bunch of "target" points in domain #2. The preimage I_f is the set of "source" points that give you those target points.

 

[simmonj7]

Ok so I see that you break up the a∈ X∪Y to get a∈X or a∈Y but I don't see quite how to go from there.
I know when I prove this problem with images rather than inverse images, at this point I would say (assuming the problem and variables were in the right spots) f(a)∈ f(X) or f(a)∈ f(Y) and then x would be equal to f(A) but since I am doing inverse images, I don't know how to get back to x.
Hope what I'm trying to say kinda makes sense.

 

[hgfalling]

Ok so I see that you break up the a∈ X∪Y to get a∈X or a∈Y but I don't see quite how to go from there.

What have you shown so far? You picked an arbitrary x value in the preimage of X∪Y (the "source"). So that means that a = f(x) for some a in X∪Y; that is, there's a "target" a in
X∪Y. That means that the target is either in X or in Y or both.

If the target is in X, then the source is in I_f(X), right? That's the definition of a preimage. And if the target is in Y, then the source is in I_f(Y).

Since we know that the target is either in X or Y or both, then the source is in the preimage of X or the preimage of Y or both. Hence we have shown that part 1 is true.

Now take a look at the second part. You can use the same ideas of sources and targets to solve that; you probably won't be surprised that part 2 is NOT true, since part 1 WAS true. Funny how problems often work that way.

 

[simmonj7]

Thank you for your help!
:)