# Prove f continuous given IVP and 1-1

1. Apr 18, 2012

1. The problem statement, all variables and given/known data
Let f : (-1, 1) → ℝ. f satisfies the intermediate value property and is one-to-one on (-1, 1). Prove f is continuous on (-1, 1)

3. The attempt at a solution
I was thinking that the IVP and one-to-one implies that f should be strictly monotonic and that a strictly monotonic one-to-one function is continuous. Both of these seem very intuitive to me and yet I have no idea how to do the rigorous proof. For example if f is not strictly monotonic then it seems there would be a contradiction in f one-to-one but does that follow directly? Since you don't know anything else about the function..

2. Apr 18, 2012

### Poopsilon

You know it's funny, on wikipedia both versions of the Intermediate Value Theorem require f to be continuous as a hypothesis, are you sure your question isn't asking you to prove $f^{-1}$ is continuous?

3. Apr 18, 2012

Thus choose a point $x \in (-1,1)$ and use the IVP to find another suitable point $y$ (this choice will depend on epsilon). Think about what to do with |f(x) - f(y)| and with |x-y|. Then consider an arbitrary point w with a certain bound on |x-w| and its implications for |f(x) - f(w)|; I believe proving these implications will require applying the IVP and one-to-one in coordination, but I haven't formally worked out the details.