| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 877.1 | any good books on the subject? | ZFC::DERAMO | I am, therefore I'll think. | Mon May 23 1988 00:15 | 53 |
| This has nothing to do with answering your question.
The problem itself raised other questions in my mind.
>> Now D can be made into a metric space naturally.
I suppose you just mean that D is "Euclidean n^2 space."
>> Then S is nowhere dense.
I had to look this up to make sure I understood it.
The definition means the closure of S does not contain
a nonempty open set. How obvious is this here? Is the
answer to "is S closed" obvious?
If I understand this right then it is fairly easy using the
characteristic polynomial of a matrix A (p(x) =
determinant(A - xI) where I is the identity nxn matrix) to
show that S is closed and nowhere dense.
If T is a nxn matrix in D that is in the closure of S,
then there is a sequence S1, S2, ... of points of S that
converge to T. A nxn matrix A is singular if and only
if zero is a root of its characteristic polynomial.
The charactersitic polynomials of the {Sn} converge to
the charactersitic polynomial of T, and every one has
zero as a root, and so T has zero as a root: so T is
in S, and S is closed.
An open set containing the point A of S also contains
A - eI for small enough e (e is "epsilon"). This is
singular only if e is a root of the characteristic
polynomial of A, but there are at most n such roots,
so we can find an epsilon that gives a point of the open
set that is not in [the closure of] S. So S is nowhere
dense.
How am I doing so far? If the above is all garbage please
let me know.
>> Now since D is a metric space, a measure {u,D} can be defined.
There is the Lebesgue measure for D as a Euclidean n^2-space,
but I didn't realize that there was a way to define a
measure on an arbitrary metric space. Could you describe
the construction?
>> Is u(S) = 0?
Is part of the problem statement: "... for every measure
{u,D} that can be defined for D"? Or would a proof for
Lebesgue measure suffice?
Dan
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| 877.2 | | HPSTEK::XIA | | Mon May 23 1988 16:13 | 4 |
| Re. .1
I think your proof of S being nowhere dense is fine. Yes, I meant
it to be the L-measure when I say naturally.
Eugene
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| 877.3 | one possible approach | ZFC::DERAMO | I am, therefore I'll think. | Mon May 23 1988 19:21 | 44 |
| I once came across the following, given as a "lemma without
proof":
If A is a subset of R^2, then A is null if and only
if for almost all x, A_x is null.
Terminology:
R^2 is the Euclidean plane
null is "has Lebesgue measure zero"
almost all is "except on a set of Lebesgue measure zero"
A_x is {y | (x,y) is in A}
So a subset A of R^2 is null if and only if each slice
A_x is a null subset of R, except perhaps for a null
[subset of R] set of counterexample x's.
This should have analogs for other R^n. So if you consider
2x2 matrices
[ a b ]
[ c d ]
then the set of all 2x2 matrices with a=0 should be null,
because "A_x" is null for all x except x=0, where here
"A_x" is the three dimensional slice with a common upper
left corner. Call this a 3-1 analog of the lemma.
A 2x2 matrix will be singular if and only if ad - bc = 0.
This can be broken down into [the first two overlap]:
a = 0 and (b = 0 or c = 0) [null]
d = 0 and (b = 0 or c = 0) [null]
a ~= 0 and d ~= 0 and ad = bc
Shouldn't the third set be null, because for each point
(a,d) in R^2 - {(a,d) | ad=0}, the slice A_ad is the
hyperbola ad = bc in R^2, which is a null subset of R^2?
[This would be a 2-2 analog of the lemma.]
Does this look convincing for the n=2 case? How hard
would it be to generalize to larger n?
Dan
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| 877.4 | there he goes again ... | ZFC::DERAMO | I am, therefore I'll think. | Wed May 25 1988 13:54 | 8 |
| Oops, I left part of the lemma out. It should be
If A is a measurable subset of R^2,
^^^^^^^^^^
then A is null if and only if {x | A_x is not null} is null.
Dan
|