# Suppose that f : R -> R is continuous and one-to-one

1. Sep 14, 2006

### ek

Sorry for no description of the type of question, but I'm so out to lunch on this stuff I don't even know what I'd call it. Ok, so I'm not looking for answers here, I'd just appreciate being pushed in the right direction.

Suppose that f : R -> R is continuous and one-to-one (that is, x1 != x2 --> f(x1) != f(x2). Prove that for each interval I = [a,b] contained in R either f(I) = [f(a), f(b)] or f(I) = [f(b), f(a)].

I have no idea what to consider for this problem.

Sorry for the bad notation. I'm not too proficient at latex.

2. Sep 14, 2006

### e(ho0n3

Perhaps a proof by contradiction would work?

3. Sep 14, 2006

### HallsofIvy

Staff Emeritus
It is hard to suggest how to prove something like that without knowing what you have to work with. For example, there is a theorem saying that a continous function maps an interval into an interval. Do you know that theorem and can you use it? It's a special case of a more general theorem that says that a continous function maps a connected set into a connected set- and, of course, in R connected sets are precisely intervals.

If you can use that theorem then you have immediately that f([a, b])= [c, d] for some c, d. Then use "one-to-one" to prove c= f(a), d= f(b) or c= f(b), d= f(a).

4. Sep 15, 2006

### ek

Thanks very much for the help and sorry for the late reply.

This is a third year "advanced calculus" class and I was privy to the connected set theorem.

I suck (relatively anyway, compared to phys/astr) at math, getting through this final major math class will be quite a challenge to get my degree (Astronomy).

I'll probably be calling upon the very knowlegable PF members for some guidance throughout the semester.

Thanks again.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook