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- Newsgroups: sci.math
- Path: sparky!uunet!snorkelwacker.mit.edu!galois!riesz!jbaez
- From: jbaez@riesz.mit.edu (John C. Baez)
- Subject: Re: Assorted questions and problems
- Message-ID: <1992Nov10.025719.10180@galois.mit.edu>
- Sender: news@galois.mit.edu
- Nntp-Posting-Host: riesz
- Organization: MIT Department of Mathematics, Cambridge, MA
- References: <BxDJ8v.DCw@world.std.com> <1992Nov8.181631.13298@Princeton.EDU>
- Date: Tue, 10 Nov 92 02:57:19 GMT
- Lines: 39
-
- In article <1992Nov8.181631.13298@Princeton.EDU> tao@fine.princeton.edu (Terry Tao) writes:
- >I have three questions that I can't do. I hope you can see from the
- >diversity of them that they are not homework.
- >
- >(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.
-
- This sounds like a homework problem. (Just kidding.) This conjecture
- was proved by deBranges a few years ago. It's a famous story because he
- had earlier claimed to have proved it, but had made a mistake; this made
- people not take his second attempt seriously at first. (Perhaps someone
- who knows the whole story may have fun telling it.)
-
- >(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?
-
- Yes, let D be a dense subspace of X that's not all of X. Define f:X ->
- Y to be zero on D but then extend f to a nonzero linear function on all
- of X, which can be done by linear algebra nonsense. In general this may
- use the axiom of choice, but there might be hope for a constructive
- example --- one needs an example of X with dense D such that there is an
- *explicit* basis for a complementary subspace E (i.e., X = D + E as
- vector spaces (not as Banach spaces, though!)).
-
- >(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?
-
- What is "totally ordered"? If you simply mean "linearly ordered" the
- answer is no; let one set be R and the other be R copies of Z stacked on
- top of each other, or the first ordinal with cardinality of the
- continuum. If you mean "well-ordered" the answer is yes and you only
- need AC.
-
-
-