Let be increasing. Show that there exists such that .
I have not given this an extreme amount of thought, but my first guess is to consider . We first can note that since that and so since is increasing we have that and so and thus by induction . Thus, is a bounded monotone sequence and so by the monotone convergence theorem it is convergent with limit . So, I have absolutely no proof (yet) but it may be true that is the fixed point. If were continuous this would obviously be true since . But, from there I'm not sure yet.
That said, I feel that can't be right. The fact that is compact seems like it should come into play.
Just some input.
I'm wondering whether this more general proposition is also true:
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!
I'd say it's definitely true, and I'll see if I can adapt my proof to this beautiful generalization!
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...
Let
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 .
Therefore .
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.