From: ikastan@uranus.uucp (Ilias Kastanas)
Subject: Re: Question from Rudin
Date: 20 May 2000 20:54:37 GMT
Newsgroups: sci.math
Summary: A Cantor set whose members are linearly independent pver Q

In article <01bfbe79$a19b6820$04cd8ad1@daves-dell>,
David C. Ullrich <ullrich@math.okstate.edu> wrote:
@
@
@Ilias Kastanas <ikastan@uranus.uucp> wrote in article
@<8fmf54$e8m2$1@hades.csu.net>...
@[...] 
@> 	A slightly more challenging result: there is a Cantor set
@> whose members are linearly independent over  Q. 
@
@	Hah: you can do this by just "constructing" a very thin (borderline
@anorexic) Cantor set.
@
@	Writing down all the details would take some space - the main
@lemma is a sugar-honey fruit:
@
@Lemma: If K is the union of n disjoint compact intervals I_1, ... I_n
@and S is a finite set of 2n-tuples of rationals then there exist 2n
@disjoint compact intervals J_1, ... J_2n, with J_(2j-1) and J_2j
@subsets of I_j for all j, such that:
@
@	If x_j is in J_j (j=1, ... 2n) and q = (q_j) is any element of S 
@then
@
@(*)	q_1 * x_1 + ... + q_2n * x_2n <> 0.
@
@(Let's agree here that an "interval" has positive length; [a,a]
@doesn't count.)
@
@Proof: Begin by choosing a Q-linearly-independent set of 2n 
@points a_1, ... a_2n with a_(2j-1) and a_2j in I_j . Then we have
@
@(**)   q_1 * a_1 + ... + q_2n * a_2n <> 0
@
@for all q in S. Since S is finite it follows that (*) holds for
@all q in S if x_j is sufficiently near a_j. Let J_j be a 
@sufficiently thin interval centered at a_j. QED.
@
@	You apply the lemma repeatedly with an appropriately
@huge set S at each stage and the intersection of the sets
@you get is a Cantor set with no rational linear dependencies
@left. 
@
@Mother Teresa


	Hey, this mother can fly!

	[Obscure reference to a Los Angeles Times tidbit a few years
ago; a group of Van Nuys housewives taking pilot lessons had T-shirts
made emblazoned with the motto  "These mothers can fly"].


@	(There's a detail here: You can rule out any finite set of
@dependence relations at each stage. Now you need to
@note that a dependence relation you did not rule out
@at the current stage can be ruled out at the next stage
@by including enough q's in the next S.
@	A more explicit (although not quite general enough)
@version of what I just said: Suppose that I_1, ... I_n
@are as in the lemma. Suppose that q is an n-tuple of
@rationals (not a 2n-tuple - q is a dependence that we
@didn't get around to ruling out at the previous stage.)
@Then there exist 2^n 2n-tuples r_1, ... r_(2^n) such that
@if J_1, ... J_2n are as in the proof of the lemma, with
@{r_1, ... r_(2^n)} contained in S, then
@
@   (*)	q_1 * x_1 + ... + q_n * x_n <> 0
@
@whenever x_j is in the union of J_(2j-1) and J_2j. The
@r_k are all the 2n-tuples with the property that for all
@j, one of (r_k)_(2j-1) and (r_k)_2j is q_j and the other
@is 0.
@
@	The last two paragraphs may not be so clear. The
@result _does_ follow from the lemma - if the reader sees
@sort of how to get the result from the lemma, tries to
@write it down exactly and gets stuck on a detail, the last
@two paragraphs may make sense at that point.)



	Considering David's solution, one might well suspect that
something more general is going on.  Indeed, we can bring it out
in terms of Baire category.  All we need about linear independence
is this: the set  LI_2  of (a,b),  a, b  linearly independent over  Q,
is comeager in  R^2  (it's a dense G_delta... complement = countable
union of lines); likewise for the set  LI_3  of (a,b,c)'s... and so
on, for each  LI_n.

	Sketch: the space  K(R)  of compact subsets  K  of  R  (with the
Hausdorff metric) is complete and separable.  Key fact: if  X  is
comeager in  R^2,  the set  {K: for all (distinct!)  k_0, k_1  in  K,
(k_0,k_1) is in  X}  is comeager  (in  K(R)); the analogous statement
holds for, uh, all higher values of  2.  Apply this to  X = LI_2, LI_3,
... , intersect the resulting comeager sets, and you have shown that
{K:  K  is linearly independent}  is comeager.  We want a  K  that is
perfect, too; fortunately, {K:  K  is perfect}  is a dense G_delta as
well, so intersecting we obtain comeagerly many such  K, and thus an
abundance of sets satisfying the requirements.
 
 

 
 						Ilias