Alright, I've got my answers to my questions. Many thanks to the dozen people who responded. I'll now summarize in an attempt to forestall another dozen.
|> >(1) what is the current status of the Bieberbach conjecture, that any
|> >univalent holomorphic function f on the unit disk such that f(0) = 0 and
|> >f'(0) = 1 satisfies the fact that the taylor expansion f(x) = \sum a_n x^n
|> >has the property |a_n| \leq n? The last I heard, it was proved for n up to
|> >7 only, and also for all n sufficiently large |a_n| \leq 1.08 n.
The general consensus seems to be it was proven in controversial circumstances by Louis de Branges somwhere between 1975 and 1984. If someone has the exact date, the reason behind the controversy, and/or a reference, could you please email me. By the way, the book I was using as a reference was printed in 1983, so if it was proven in 1975, there's something wrong with the math grapevine...
|>
|> >(2) Suppose X and Y are Banach spaces. Can one construct a linear mapping
|> >from X to Y which is NOT continous?e.g. a map from L^2 to L^2 which is not
|> >bounded. Is it possible to construct one without AC?
|>
The answer is a resounding YES if axiom of choice is true, though someone pointed out to me that L^2 was a bad choice for an example (yes, I know it's a Hilbert space, but what's the point of this?), and apparently there is a "Soloway model" under which the answer is always NO.
|> >(3) Assume the axiom of choice and the axiom of the continuum. Is it
|> >true that two chains (totally ordered sets) which both have the
|> >cardinality of the continuum have a one-to-one and onto order
|> >preserving mapping betweem them?
|>
OK, so I screwed up this question. I have a dozen different counterexamples now of the original question. I'll rephrase this question into two different ones
(3a) if you assume that the chains are unbounded above and below, and are dense, i.e. between any two elements there is a third, is the above now true?
(3b) is (3) true if you remove the "onto" criterion, i.e. is there an order imbedding from the first chain to the other?
I'm fairly sure that both of these are true.
As soon as my news reader tells me how, I'm going to cancel the original article. Thank you again for the large number of responses.
