# One-to-one function

1. Feb 7, 2008

### mathboy

[SOLVED] One-to-one function

Let f: X -> Y. Suppose f(X-A) = Y-f(A) for any A in X. Prove that f is one-to-one.

Suppose f(a)=f(b). I tried letting A = {a,b}, A= X-{a,b}, but I can't manage to prove that a=b. Can someone help?

2. Feb 7, 2008

### NateTG

You need to clean up your notation. What are $X$ and $Y$? Sets?

3. Feb 7, 2008

### mathboy

X is the domain of f and Y is the range of f, so they are sets. A is any subset of X. So the question is asking to prove that if f(complement of A)= complement of f(A), then f is one-to-one.

4. Feb 7, 2008

### morphism

Let a be in X and set A=X-{a}. What's f(X-A) now?

5. Feb 7, 2008

### HallsofIvy

Staff Emeritus
You might want to try proving the contrapositive: if f is not one to one, then there exist a set A such that f(complement of A) is not complement of f(A).

If f is not one to one, then there exist x1 and x2 such that f(x1)= f(x2). What if A= {x1}?

6. Feb 7, 2008

### NateTG

In situations like this, you can always try assuming that $a \neq b$ by contradiction. The proof might not require it, but you can sometimes find direction that way.