From: "David C. Ullrich" Subject: Re: Simple several complex variables question Date: Wed, 23 Feb 2000 10:51:46 -0600 Newsgroups: sci.math Summary: [missing] Daniel Giaimo wrote: > Is it possible for a map f:C^2->C to be holomorphic in each variable > separately, but fail to be holomorphic? No. If you find this surprising you should, but it's true - a separately holomorphic function in several variables is holomorphic. This is due to Hartogs - there's a proof in Narasimhan's little book "Several Complex Variables". > Can it fail to be continuous? Of > course, if C is replaced by R then the answer to both is yes, but I don't > think(?) that the standard examples will extend to C^2 in a holomorphic way. > > --Daniel Giaimo ============================================================================== From: israel@math.ubc.ca (Robert Israel) Subject: Re: Simple several complex variables question Date: 23 Feb 2000 19:33:23 GMT Newsgroups: sci.math In article <88vktu$n8l$1@nntp8.atl.mindspring.net>, "Daniel Giaimo" writes: > Is it possible for a map f:C^2->C to be holomorphic in each variable > separately, but fail to be holomorphic? Can it fail to be continuous? Of > course, if C is replaced by R then the answer to both is yes, but I don't > think(?) that the standard examples will extend to C^2 in a holomorphic way. No. Hartogs's Theorem says if a complex-valued function f is defined in an open subset of C^n and is holomorphic in each variable separately, then it is continuous and therefore (by definition) holomorphic. See Hormander, "An Introduction to Complex Analysis in Several Variables", for a proof. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ============================================================================== From: ags@seaman.cc.purdue.edu (Dave Seaman) Subject: Re: Equinumerosity of open,closed,open & closed, unit intervals. Date: 17 Oct 2000 10:16:26 -0500 Newsgroups: sci.math Summary: [missing] In article <39ec46c9.857617912@nntp.sprynet.com>, David C. Ullrich wrote: >On Mon, 16 Oct 2000 22:01:50 +0300, glenn wrote: >>> What is the definition of cardinality, _exactly_? >>First of all two sets A and B are called equinumerus if there exists >>an 1-1 and onto correspondence between them. Then we write A~B or >>|A|=|B|. The notation |A|<=|B| means that there exists an embeeding >>from A in B. The notation |A|<|B| means that there exists an embeeding >>from A in B but not from B into A. Now, take the class H(A)={a \in >>On:|a|<=|A|}, where On is the class of all ordinals. Then H(A) is an >>ordinal too. This can be proved _without_ choice _or_ Cantor - >>Bernstein (CB). > You may well be right. I find this very very surprising. >Exactly how do you prove that H(A) is an ordinal? (I'll even >tell you what bit I'm stuck on: showing that there is an >ordinal which is not an element of H(A).) Ouch. >> The number H(A) is called the Hartogs number for A. >>Now, for every well ordered set A, the following implication is valid: >>b|b|<|H(A)|. Indeed if b>numbers having this property are called initial ordinals, or >>cardinals. What if A is not well ordered? Besides David Ullrich's objection, there is the possibility that H(A) may exist and be an ordinal, but |A| would not be an ordinal. For example, A could be an infinite set, each of whose well ordered subsets is finite. Then H(A) = omega, but A is not well ordered and therefore |A| is not an ordinal. For all we know (or at least for all *I* know), it could be that the real numbers have no uncountable well ordered subsets. Then H(R) = omega_1, but what is |R|? It's 2^aleph_0, or course, but is that an ordinal? I won't even ask which one. -- Dave Seaman dseaman@purdue.edu Amnesty International calls for new trial for Mumia Abu-Jamal