# Topology - Darboux Theorem

• Feb 26th 2009, 01:41 AM
Amanda1990
Topology - Darboux Theorem
Let I be an open interval in $\Re$ and let $f:I \rightarrow \Re$ be a differentiable function.

Let T be the set $T = \{(x,y) \in I \times I : x. I can show that this is a connected subset of $\Re^2$ with the standard topology.

Let $g: T \rightarrow \Re$ be a function defined by $g(x,y) = (f(x) - f(y)) / (x-y)$. Prove that $g(T) \subset f'(I) \subset \overline{g(T)}$.

Show that $f'(I)$ is an interval.

This last deduction would be trivial if g(T) was a connected set, but I don't see why this should necessarily be true. (It would be true if g was a continuous function but we are not given this. Nevertheless, I think there may be a way of showing it).
• Feb 26th 2009, 02:04 AM
tah
Hi, if you want g(T) to be connect, isn't it sufficient to have g continuous on T?
• Feb 26th 2009, 02:05 AM
tah
Hi, if you want g(T) to be connected, isn't it sufficient to have g continuous on T?
• Feb 26th 2009, 02:07 AM
Amanda1990
Sure, but can you actually show this?!
• Feb 26th 2009, 02:17 AM
tah
The derivative of f may be discontinuous then!! hummm
• Feb 26th 2009, 02:19 AM
Amanda1990
I think this is the whole point of the Darboux Theorem. The derivative of f satisfies the intermediate value theorem even if f' is not continuous. In other words, there is nothing to stop f' being discontinuous.
• Feb 26th 2009, 02:40 AM
tah
So instead of the continuity, you may prove certain "convexity" of g !
• Feb 26th 2009, 02:53 AM
Amanda1990
Unless you can actually prove this and why it helps, then I'm not totally convinced by this idea!
• Feb 26th 2009, 03:14 AM
tah
I wanted to show something like the image by g of an interval in T is an interval in R.

But $h(x,y) = f(x) - f(y)$ is continuous on $I\times I$ and $\frac{1}{x-y}$ continuous on T. So why g shouldn't be continuous on T ?
• Feb 26th 2009, 03:33 AM
Amanda1990
Yes, to be honest this was my concern exactly, even though it seems to contradict the "point" of the question! (in that the derivative doesn't have to be continuous). I think it's because of the form of T...
• Feb 26th 2009, 03:38 AM
tah
But the derivative is obtained only at the limit where g could be not defined or discontinuous!