# Analysis: proving inverses involving sets

1. Nov 9, 2008

### vikkisut88

Let f: X → Y and A is a subset of Y and B is a subset of Y. Prove that:

a) f⁻¹(A union B) = f⁻¹(A) union f⁻¹(B)
b) f⁻¹(A intersetion B) = f⁻¹(A) intersection f⁻¹(B).

I know that f⁻¹(A) = {x ε X : f(x) ε A}
and so f⁻¹(B) {x ε X : f(x) ε B}

but after that I really don't understand how to prove this.

Last edited: Nov 9, 2008
2. Nov 9, 2008

### HallsofIvy

Staff Emeritus
Any time you are asked to prove X= Y for sets X and Y you do two things:
Prove that X is a subset of Y and then that Y is a subset of X.

To prove X is a subset of Y, you start "if x is a member of X" and then use the definitions of X and Y to show that x must be a member of Y.

For example, here, if x is in f-1(A union B) then f(x) is in A union B. That means either f(x) is in A or f(x) is in B. If f(x) is in A, then x is in f-1(A). If f(x) is in B, then x is in f-1(B). In either case, x is in f-1(A) union f-1(B). That proves that f-1(A union B) is a subset of f-1(A) union f-1(B). Now do it the other way: if x is in f-1(A) union f-1(B), then ...

3. Nov 9, 2008

### CompuChip

Welcome to PF, vikkisut.
It's really just a matter of plugging in the definitions:
$$f^{-1}(A \cup B) = \{ x \in X: f(x) \in A \cup B \} \stackrel{\star}{=} \{ x \in X: f(x) \in A \text{ or } f(x) \in B \} \stackrel{\star}{=} \{ x \in X: f(x) \in A \} \cup \{ x \in X: f(x) \in B \} = f^{-1}(A) \cup f^{-1}(B)$$
That is just a partial proof though; I have marked two of the identities with a $\star$, I suggest you try to prove these by a standard argument (let x be in one of them and show that it is in the other and vice versa).

4. Nov 9, 2008

### vikkisut88

Sorry I don't quite understand what you're suggesting me to do there? Let x be in what exactly? Be part part of A, and then show that it is in B?

5. Nov 9, 2008

### CompuChip

No, let it be in
$$\{ x \in X: f(x) \in A \cup B \}$$
and prove it to be in
$$\{ x \in X: f(x) \in A \} \cup \{ x \in X: f(x) \in B \}$$
Also see HallsOfIvy's post, for more explanation.

You should also prove the converse of course, letting it be in the latter and showing that it is in the former.[/edit]

1500 posts by the way, almost went unnoticed :tongue:[/edit]

6. Nov 9, 2008

### vikkisut88

oh okay, thank you very much :)