mm-1919
===
operator and surjective T maps open sets to open sets. What about if
you drop the hypothesis that T is surjective, is T(0), O open in X,
Borel?
Jos.8e Gasc.97n
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
No.
from one Banach space to another such that the image of
the closed unit ball is not a Borel set.  It then follows that the
image of the open ball is not Borel, since any closed
ball is a countable intersection of open balls.  Taking
example mapping a Banach space to itself.
Let Y be l^2, say, nicely rotund.  Let S be the unit sphere in Y,
S = { y : ||y|| = 1 }, a closed set.  Let Q be a nasty subset
of S, non-Borel.  Let X = l^1(Q), the space of functions
absolutely.  T is linear and ||T|| = 1.  Now I claim the closed ball
B = { f in X : ||f|| <= 1 } maps onto a non-Borel set in Y.
Indeed, the intersection of T(B) with the closed set S is exactly
Q, so T(B) is not Borel.
-- 
G. A. Edgar                               
http://www.math.ohio-state.edu/~edgar/
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
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operator and surjective T maps open sets to open sets. What about if
you drop the hypothesis that T is surjective, is T(0), O open in X,
Borel?
Jos,bi(B Gasc,bs(Bn
think your assertion is right, if in addition you assume X to be
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X-ASG-Orig-Subj: Re: skew fields
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
all
This is too complicated: the multiplicative group of the field R (real
numbers) is not residually finite. Thus you do not need non-commutativity.
Here is a proof that R^* is not residually finite. Suppose that there 
exists
a homomorphism f from R^* to a finite group separating 2 from 1. Suppose
that a root a of 2 of degree k^m and a root b of 2 of degree k^n have the
same images under f. Thus f(a)=f(b). Raising to the power k^m gives
f(2)=f(b^(k^m))=f(2^(k^(m-n)))=1, a contradiction.
Perhaps a more appropriate question is: are there skew fields K for which
not every finitely generated subgroup of K^* is residually finite?
Mark
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
Kadec-Klee property is the more modern term.  For a history of the
development of the concept, starting with Radon, see the PDF file at
 
<http://www.ams.org/tran/1996-348-12/S0002-9947-96-01782-5/S0002-9947-96-
(careful that this line doesn't get truncated).
Uniformly Kadec-Klee is perhaps a more useful concept.
--Ron Bruck
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
I can say a lot of things you
have probably already thought of.
Vertex v must be in the cover or
the neighbor set of v, NS(v), must
be in the cover. If v is in set A
then NS(v) must be in C.
(Converse, if NS(v) is in set C
then v must be in A.)
It is easy to come up with graphs
where every vertex is in set B.
A graph where every vertex is of degree 2
is an example.
I think the following is true.
For bipartite graphs, all vertices are
in set B or no vertices are in set B.
A bipartite graph can be broken into two
partitions such that no vertex in
a partition shares an edge with another
vertex in that partition.
Each partition is a vertex cover and at
least one of them is a minimal vertex cover.
If the two partitions are the same size
then all vertices are in your set B.
Otherwise, the smaller partition is in set C
and the larger partition is in set A.
To complete the proof I would have to
show there is only one way to partition
a bipartite graph.
Your method is similar to somthing I have
been studying.
We start with a Boolean Satisfaction instance, F(),
derived from a set of variables, V().
Consider the set of satisfying assignments for F(),
We classify the variables in V() into three categories:
Fixed: Variables that have the same assignment
in every satisfying assignment.
Dependent: A variable where the assignment depends on
the assignment of another variable (The other variable
can't be a fixed variable)
Independent: For every satisfying assignment where v
is True, there is an identical satisfying assignment
except v is False.
Fixed variables are similar to your set C.
Since every variable has an assignment there is
no classification corresponding to your set A.
Independent variables are a subset of dependent
variables. Think of an indepedent variable as
a dependent variable that depends on the empty set.
If a SAT problem were reduced to the minimum
number of variables all of the indepedent
vriables would be removed.
SAT instances can be converted to minimal vertex
cover problems. So, both our methods could be used to
describe graphs representing SAT problems.
Russell
- 2 many 2 count
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
No. Consider the following example: g is spanned by one element H and V is 
spanned by one vector X. Assume H.X=X.
Then the semi-direct product is spanned by H and X and one has 
    [H,X]=X
Let b be an invariant bilinear form, then
   b(X,X) = b(X,[H,X]) = b([X,H],X) = -b(X,X),
so b(X,X)=0. Next
   b(H,X) = b(H,[H,X]) = b([H,H],X) = b(0,X) = 0
and finally,
   b(X,H) = b([H,X],H) = b(H,[X,H]) = -b(H,X) = 0
So b is degenerate.
@nton
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
in g one can define an invariant non-degenerate bilinear product, but
your answer covers that case too.
What if one alse assumes that g is a simple Lie algebra?
Jose Carlos Santos
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
The following paper has been published:
Algebraic and Geometric Topology
Volume 5 (2005), paper no. 10, pages 183--205
URL:
http://www.maths.warwick.ac.uk/agt/AGTVol5/agt-5-10.abs.html
Title:
The 3-cocycles of the Alexander quandles F_q[T]/(T-omega)
Author(s):
Takuro Mochizuki
Abstract:
We determine the third cohomology of Alexander quandles of the form
F_q[T]/(T-omega), where F_q denotes the finite field of order q and
omega is an element of F-q which is neither 0 nor 1. As a result, we
obtain many concrete examples of non-trivial 3-cocycles.
Secondary: 55A25, 57Q45
Keywords:
Quandle, cohomology, knot
Published: 23 March 2005
Author(s) address(es):
Department of Mathematics, Kyoto University 
Kyoto, 606--8502, Japan
Email: takuro@math.kyoto-u.ac.jp
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
Hi! Ted Slaman explained to me today that the sets constructed in the
proof of Friedberg-Muchnik theorem about uncomparable c.e. degrees are
what I need up to some minor modifications. So the question has been
answered.
Mark
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
I'm not sure why you bring real numbers into it,
since all the numbers you are dealing with appear to be integers.
Looking again at this again, I am not sure what you mean by accepting 
y.
I think the word acceptor is normally used for a Turing machine
which accepts or recognises a recursive set S,
ie outputs 1 or 0 according as the input is or is not in S.
With that definition, a non-recursive set could not have an acceptor.
But from the parallel with the polynomial case,
I assume you mean the length of the smallest input which _outputs_ y,
which as I mentioned is called (after Chaitin) the algorithmic entropy of y
with respect to T, and is denoted (at least by Chaitin) by H_T(y).
That is not clear to me.
In fact I am not sure what you mean.
Presumably you are now restricting your good numbers to integers,
since it does not make sense to me to speak of a set of real numbers
being recursively enumerable,
particularly in this case since as far as I can see if r is good
then so are all the numbers in the interval [[r], [r] + 1).
As far as I can see, if you have a set S of natural numbers
such that H_T(s) <= s for all s in S
then all real numbers will be good with respect to S.
It seems to me that one can construct such a set S
containing arbitrarily long sequences of numbers
by the argument that I gave.
But probably I have misunderstood the question.
-- 
Timothy Murphy  
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
Not as a density. Take L = 0, i.e. the case of a deterministic ode. Then Q' 

is supported in a single trajectory.
Again the answer is no. Take L = [1 0; 0 0] in R^2, i.e. P' corresponds to 
1-D Brownian motion embedded in R^2, and
f(X_t,t)= [0 1]^T, i.e. drift in an orthogonal direction. The two laws have 

disjoint support in this case.
It appears that some form of compatibility is needed to get a formula for 
dQ'/dP', e.g. f should leave the column space of  L  invariant.
Hope this helps
Hans
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
The Fourth International Conference on Quality and Reliability
9 - 11 August, 2005
Beijing, P.R. China
www.Quality-Reliability.cn
ICQR 2005 ~ Quality and Reliability in theory and practice
        CALL FOR PAPERS
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mail@quality-reliability.cn
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
I am writing my Master Thesis about the free resolutions of canonical
k-gonal curves for various k. For this I am using a lot of information
special linear series where he gives a not necessarily minimal free
resolution of a canonical pentagonal curve. I would like to have more
information about where this resolution is not minimal or more
precisely how the minimal resolution looks like, especially for a
topic I am very grateful if you could tell me where I can find more
information.
Andrea Hofmann
Originator: bergv@math.uiuc.edu (Maarten Bergvelt)
Hi 
Georges