Subject: Re: Infinitude of Primes in P-adics
From: Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium)
Date: 7 Jan 1997 03:04:03 GMT
In article <5asb6a$ma6$1@dartvax.dartmouth.edu>
Archimedes.Plutonium@dartmouth.edu (Archimedes Plutonium) writes:
> Why do this program? Answer is threefold. To connect adics with
> geometry. And another reason is to find, or discover what the
> composite-adics are. Third reason: in a way, algebraic structures
> fields, rings, groups should be connected with simply a function and a
> graph.
One should be able to take a graph, a function and point to this graph
and point out various algebraic structures in the graph itself. Take
for example a linear function. Then the line of that function
represents a Field. Then, some of the rest of the graph is a Ring,
another portion is a Group etc.
What I am looking for is what is the most Complex and Complicated yet
meaningful Algebraic structure when I amass all of the prime-adics and
composite-adics together. Note the word meaningful. I suppose it is a
Geometry. That the highest algebraic structure is a geometry and we
have only three of those-- Riemannian, Lobachevskian, and Euclidean. A
Field, a Ring, a Group, etc all of these algebraic structures are parts
of a geometry.
I need to know what is the highest meaningful algebraic geometry
when I pile all the adic forms together, both prime and composite
adics. I think the answer is that it forms a geometry.
Now, let us look at Euclidean geometry since we know it well. Do the
points as Reals+i+j do those points form a Field? Can some of them
form a Ring? Can some of them form a Group? But the highest algebraic
structure of Euclidean 3-space is Euclidean geometry itself.
Subject: Re: Re Difference between zero and nought
From: hetherwi@math.wisc.edu (Brent Hetherwick)
Date: 7 Jan 1997 06:57:44 GMT
John Edser (edser@ans.com.au) wrote:
: It appears to me that you cannot have less than nothing (nought)
: but you can have less than zero.
It appears to me that you haven't yet met my accountant.
: Do mathematicians discriminate between these two concepts?
That would be racism, now wouldn't it?
: This means to me that nought is a non reversable concept
: and zero is reversable.
Just because a sweater has three holes doesn't mean you can wear it as
pants.
: This also means to me that nought is an absolute concept
: while zero is a relative concept and as such is deducible from nought.
Don't forget about the in-law concepts zip, zilch, and not-a-damn-thing.
: We must assume nought to have zero.
Hm. No wonder I can't balance my checkbook. Each time I sit down to
square my books, I begin by writing, "let my balance > 0".
: This is why mathematics cannot describe itself (Godel)?
No. It's because mathematics is written in pea-addicts, and
mathematicians think that finite topologies = Lie algebras, which is
wrong, since actually artinian rings = Lebesgue measure.
: Is there a non reversable mathematics?
Not unless you don't wash first in Woolite.
$$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666
hetherwi@math.wisc.edu
$$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666
Subject: Matrix algebra problem
From: ptitjean@ccr.jussieu.fr (Michel PETITJEAN)
Date: 7 Jan 1997 10:03:28 +0100
Ref:
Wicher Bergsma (wbergsma@kub.nl) wrote:
>With A and B full column rank matrices, and B the orthogonal
>complement of A, the following identity holds:
>
> -1 -1
> A (A'A) A' = I - B (B'B) B'
>
>with I the identity matrix.
>
>How to prove this?
May I suggest the following:
-1 -1
PA = A (A'A) A' and PB = B (B'B) B' are the respective projectors
on the two complementary subspaces.Thus, PA*PA=PA, PB*PB=PB, and PA*PB=0
and PB*PA=0 because A'B=0 and B'A=0 (the "0" matrices have not the same
dimensions).
(PA-PB) = PA*PA-PB*PB = (PA+PB) * (PA-PB)
(PA+PB-I) * (PA-PB) = 0
There is no non-zero vector v such that (PA-PB)*v=0, because PA*v=PB*v is
impossible (PA*v orthogonal to PB*v).Thus (PA-PB) is inversible and PA+PB=I.
Michel Petitjean, Email: petitjean@itodys.jussieu.fr
ITODYS (CNRS, URA34) ptitjean@ccr.jussieu.fr
Subject: Re: Why is a finite set not an open set?
From: hetherwi@math.wisc.edu (Brent Hetherwick)
Date: 7 Jan 1997 07:41:14 GMT
Jose Unpingco (u13839@pauline.sdsc.edu) wrote:
: A = {1,2,3} is an example of a finite set. Now, for every point in
: this set I can wrap a neighborhood about it of radius < 1/2. This
: means that the neighborhood of the point contains only itself and the
: neighborhood is a subset of A. Thus, every point is an interior point
: and the set A is open.
This depends upon the topology you're using. In the standard topology of
the reals, 3/4 is a point in your open nbd of 1 not in the set. In the
subspace topology (of the integers, say), {1, 2, 3} is open, and your
argument is "essentially" correct.
$$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666
hetherwi@math.wisc.edu
$$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666 $$$ 666
Subject: Re: Why is a finite set not an open set?
From: Marzocchi Marco
Date: Tue, 7 Jan 1997 10:39:14 +0100 (MET)
On 7 Jan 1997, Jose Unpingco wrote:
> hi,
>
> This should be simple.
>
> A = {1,2,3} is an example of a finite set. Now, for every point in
> this set I can wrap a neighborhood about it of radius < 1/2. This
> means that the neighborhood of the point contains only itself and the
> neighborhood is a subset of A. Thus, every point is an interior point
> and the set A is open.
>
> What's wrong with this argument? There is not restriction of the
> definition of neighborhood which says that it must contain a point
> other than the point at its center.
>
> I'm confused. Help.
>
>
It depends on your ambient space. You are saying that your set A is open
when considered as a subset of itself, considered as a metric space,
with the topology induced by the standard topology of the reals.
This is trivial, since in any topological space the empty set and the
whole space are always open and closed.
However, if you consider R (the reals) as your ambient space, with the
standard topology, you should find a neighborhood IN R contained in your
set A, that is a little interval around the singleton {n} with n=1,2,3 ,
which is not possible.
Hope this helps,
Marco
''~``
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