Prove f continuous given IVP and 1-1

[luxade]

Homework Statement

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)

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..

 

[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?

 

[luxade]

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?

The question as stated is correct. It states that f satisfies the intermediate value property: the result of the theorem rather than the theorem itself.

 

[Poopsilon]

Ah yes, I apologize. This problem is tough. You won't be able to prove it's uniformly continuous, thus recall that proving point-wise continuity allows delta to depend not only on epsilon but also on the point at which you are trying to prove continuity.

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.