sci.math Google Group

Re: field generated by the set of roots of unity

galath...@gmail.com (galathaea) — Mon, 06 Oct 2008 20:14:40 UT

products work just like polynomial products
as an example
look at the cyclotomic field of degree 3
example "polynomials" might be
w
3
x + x + 1
and
2
w w
3 3
x + x - 1
where w_3 is the cube root of unity
multiplying these together gives
term-by-term
2 2

Re: Why "meta diagonals" are irrelevant

wpihug...@hotmail.com (William Hughes) — Mon, 06 Oct 2008 20:07:16 UT

Indeed as it is the standard definition of a set having cardinality
aleph_0. You have neither shown it is inconsistent (Demonstrating
that
a particular proof does not show that R is not contained by a
sequence of sequences is a long way from showing that there
is a sequence of sequences that contains R) nor have

Maximum flow computing algorithm with a "meet in middle" method

robert.bad...@gmail.com — Mon, 06 Oct 2008 20:07:03 UT

The meet in middle method for maximum flow calculation
Given the problem to compute the maximum flow between a source and a
destination in a graph, one could use the Ford Fulkerson method with
Edmonds and Karp breadth-first-scanning in O(VE^2). Imagining that we
are computing the flow between 2 cities in the world, say Bucharest

(a_n,b_n) nested open intervals for all n > n_0 show either a_n=A or b_n=A

theg...@nospam.net (TheGist) — Mon, 06 Oct 2008 19:59:38 UT

Let (a_n,b_n) be a sequence of nested open intervals. Assume
that intersection of all (a_n, b_n) is empty.
Let A=Sup a_n=Inf b_n
Show that there exists an n_0 such that for all n > n_0 either a_n=A or
b_n=A.
Proof:
If r,s >= n_0 the x_r, x_s both belong to
(a_n0, b_n0) . therefore |x_r - x_s| <= b_n0 - a_n0

Re: vitali sets

isr...@math.myuniversitysinitials.ca (Robert Israel) — Mon, 06 Oct 2008 19:54:40 UT

... with, of course, x > 0.
Yes to the first version, no to the second (even if you replace measurable
with closed).
1) For each integer n, let F_n be the family of open subsets of [n, n+x]
with measure < x, and let F = {{t}: t in R} union union_n F_n. This has
cardinality c, so well-order it in such a way that every member of F has fewer

Re: New Specie of Ape-Man Discovered, Almost Human

strich.9...@gmail.com — Mon, 06 Oct 2008 19:44:57 UT

That is one angle of looking at it. Judging from your prior angles,
it is not reliable.

Re: Why "meta diagonals" are irrelevant

nom...@invalid (Herbert Newman) — Mon, 06 Oct 2008 19:43:03 UT

Oh, may poor brain. How confusing. Is this math? :-)
Herb

Re: Why "meta diagonals" are irrelevant

mueck...@rz.fh-augsburg.de (WM) — Mon, 06 Oct 2008 19:41:51 UT

This definition is not new. But the result aleph_0 * aleph_0 =
aleph_0 then is false.
False. That would be true only if there was a last sequence of meta-
diagonals. But there isn't.
Regards, WM

Re: Why "meta diagonals" are irrelevant

wpihug...@hotmail.com (William Hughes) — Mon, 06 Oct 2008 19:25:21 UT

Good. Now I say that _by definition_ a set has cardinality
aleph_0 iff it can be put in a sequence. So I say that
R does not have cardinality aleph_0.
Your claim is that R can be put in a sequence of sequences
and thus has cardinality aleph_0.
(Note that you have not shown this, you have only shown

-- boundary of an open set

qu...@null.set (quasi) — Mon, 06 Oct 2008 19:23:38 UT

If U is a nonempty, non-dense open subset of R^n, with n > 1, must the
boundary of U contain a nontrivial path?
quasi

Re: A consideration concerning the diagonal argument of G. Cantor

nom...@invalid (Herbert Newman) — Mon, 06 Oct 2008 19:22:01 UT

I guess one of the many reasons why WM is not able to comprehend
mathematics is the fact that he was never taught real math (i.e. pure
math); and he obviously is not able to teach it to himself (by reading a
good textbook).
Herb

Re: Why "meta diagonals" are irrelevant

nom...@invalid (Herbert Newman) — Mon, 06 Oct 2008 19:17:27 UT

Great. Now WM claims that he has "proven" that IR has the cardinality
aleph_0. With other words, there is a set with aleph_0 (namely IR, the set
of all real numbers) which is unsequencable. Are YOU glad now? :-o
Herb

Re: Yet another disproof of the diagonal argument

nom...@invalid (Herbert Newman) — Mon, 06 Oct 2008 19:14:35 UT

LV: Pardon, that should read [bla bla].
george: Damn, you're stupid.
Yes he's a troll. George (and others) really should stop to answer those
trolling attempts.
Herb

Re: Why "meta diagonals" are irrelevant

mueck...@rz.fh-augsburg.de (WM) — Mon, 06 Oct 2008 19:11:25 UT

Yes.
Regards, WM

Re: Yet another disproof of the diagonal argument

denis.feldmann.sanss...@neuf.fr (Denis Feldmann) — Mon, 06 Oct 2008 19:08:25 UT

george a écrit :
No, he is a troll. How long will yuneed to understand?