# Is f-1(A ∪ B) the same as f-1(A) ∪ f-1(B)?

1. Sep 7, 2011

### brookey86

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

2. Relevant equations

3. The attempt at a solution

Last edited: Sep 7, 2011
2. Sep 7, 2011

### Dick

Yes, you are. But I'd feel more comfortable if you tried to prove it rather than taking as an assumption.

3. Sep 7, 2011

### SammyS

Staff Emeritus
Have you tried to prove it's true?

4. Sep 7, 2011

### brookey86

I can prove it using words, not quite there using mathematical symbols, but that part is out of the scope of my class. Thanks guys!

5. Sep 8, 2011

### HallsofIvy

Staff Emeritus
You prove two sets are equal by proving that each is a subset of the other. You prove "A" is a subset of "B" by saying "let $x\in A$", then show "$x\in B$".

Here, to show that $f^{-1}(A\cup B)\subset f^{-1}(A)\cup f^{-1}(B)$, start by saying "let $x\in f^{-1}(A\cup B)$". Then $y= f(x)\in A\cup B$. And that, in turn, means that either $y\in A$ or $y \in B$. Consider each of those.

Note, by the way, that we are considering the inverse image of sets. None of this implies or requires that f actually have an "inverse".