Let be increasing. Show that there exists such that .
That said, I feel that can't be right. The fact that is compact seems like it should come into play.
Just some input.
Let be increasing, and let be continuous and increasing, with and . Then there exists such that .
I don't think I'm mathematically equipped for these kinds of proofs presently, though. Need to study more!
Tomorrow though. There's a time for sleep and a time for math - the boundaries of both often blend together, but healthy living requires there be a time of the day where they unambiguously intersect!
Hmm I was just going to use the Intermediate value theorem but then I saw you haven't stated f is continuous so trouble...
We have that and .
Then at this point I would've said. By IVT there exists an etc... But I need continuity... Hmm will think some more.
If f(0) = 0 or f(1) = 1 there is nothing to prove. So we may assume that f(0) > 0 and f(1) < 1.
Let . Then A is nonempty because . Let , and notice that u > 0 because Given with , (since f is increasing). Therefore . If u = 1 that would imply f(1) = 1, so we may assume that u < 1.
Then given there exists a point with such that and therefore there exists w with such that . So . Since this holds for all it follows that .
I haven't checked carefully, but I think that this argument will extend to deal with undefined's generalisations of the problem.
Just my few cents here. Please correct me where I'm wrong, but I believe the idea is ok.
Let increasing and suppose there's no such then we either have or and we can partition where and . Observe that and are non-empty since and
From this we can conclude that and must at least consist of a countable set of points, since . Idem for .
To show this is true, let and suppose that then . Then we either have or and this contradicts the fact that increases. (for increasing we'd have ) Similar things can be said for .
This is where I feel a nice topological argument would be in it's place, although I couldn't think of any good one.
Anyway, we can safely assume there's a . Suppose then by the above observations and we can find such that is close enough to that . And since this contradicts that increases.
Similar things can be said for . I believe this is the desired contradiction.
If then by definition of we have and since we obtain .
But if then and and we get
Then we make the reversed argument: we can find close enough to p such that . Again since this contradicts that f increases.
But maybe you're right. It doesn't feel as water-tight proof ;p
About you're second question. I don't see where I made such an assumption. Maybe implicitly?
I just want to mention, that this is a special case of Knaster-Tarski theorem, which holds in complete lattices. It says that every order-preserving function f:L->L has a fixpoint, if L is a complete lattice. (In fact, it claims even more, but this is what is needed here.)
In fact, the proof of this theorem at wikipedia is similar to Opalg's one.
Knaster-Tarski theorem can be used to make a nice and short proof of Cantor-Bernstein theorem.