# Function proof

1. Oct 26, 2005

### Kamataat

I had this question on a test today.
Prove that if a function f:X-->Y is injective, then $f(X\setminus A) \subset Y\setminus f(A), \forall A \subset X$.

This is how I did it:
If x_1 is in A, then y_1=f(x_1) is in f(A). Because the function is injective, we can pick (cut Y into pieces) f(A) and f(X\A) so, that their intersection is empty (adjust A if needed). So, if x_2 is not in A = if x_2 is in X\A, then y_2=f(x_2) in f(X\A). From this we get that y_2 is in Y\f(A), since Y\f(A)=f(X\A). So, I've shown using the injective property, that y_2 in Y\f(A) follows from y_2 in f(X\A).

Right? Wrong?

- Kamataat

2. Oct 26, 2005

### Muzza

At first glance, it appears to be wrong. You say that you might adjust A, but that's not "legal". You were asked to prove the given relation for /all/ subsets A of X.

It should be clear that any element of f(X\A) is in Y. To prove that any element of f(X\A) is not in f(A), one might try to derive a contradiction...

3. Oct 27, 2005

### Kamataat

So if any element of f(X\A) were also in f(A), then that element must have two originals, one in A and one in X\A, and hence the function is not injective?

- Kamataat

4. Oct 27, 2005

### Muzza

Yep, that's it.