Question from Rudin
paul mitchell
has been stumping the undergrad real
analysis class at UCSC:
Is there a non-empty perfect set in R
which contains no rational number?
We've been trying to construct one
by a translation of the Cantor set, but so
far no one has found a way to preserve
closedness and not pick up any rationals.
Hints anyone?
paul mitchell
class at UCSC: Is there a non-empty perfect
various translations of the Cantor set, but haven't
been able to demonstrate both closedness and the
absence of rationals. Any hints?
Zdislav V. Kovarik
paul mitchell <pa...@cats.ucsc.edu> wrote:
:This question from Rudin's textbook, p44
:has been stumping the undergrad real
:analysis class at UCSC:
:Is there a non-empty perfect set in R
:We've been trying to construct one
:far no one has found a way to preserve
:closedness and not pick up any rationals.
:Hints anyone?
If they cannot be avoided that way, try to throw them out deliberately.
Number the rationals (all of them, repetitions allowed), pick h>0, and
cover the n-th rational with an open interval of length h/2^n. Let M be
the set of points that were not covered. If M is not already perfect, can
you find a subset of M which is perfect?
Good luck, ZVK(Slavek).
Michael van Opstall
I say there is.
Try this:
Take a sequence {q_n} which contains all the rationals in the interval
[0,1]. Remove an interval of length 1/3^n around each q_n, so there
are no rationals left. The remaining set A is closed and has measure 1/2
so it is nonempty (sum the geometric series). Take a point p in the
complement. Suppose it is not a limit point. Then there exists an open
interval around p containing no other points of A. By our construction,
this means that p is the common endpoint of two intervals which have been
removed. But each interval removed has rational radius and is centered at
a rational, hence p is rational. This contractiction shows that every
point of A is a limit point.
======================================================================
Michael A. Van Opstall
Padelford C-113
ops...@math.washington.edu
On Wed, 10 May 2000, paul mitchell wrote:
> This question from Rudin's textbook, p44
> has been stumping the undergrad Real Analysis
> class at UCSC: Is there a non-empty perfect
Krunoslav Sever
: [0,1]. Remove an interval of length 1/3^n around each q_n, so there
: are no rationals left. The remaining set A is closed and has measure 1/2
: so it is nonempty (sum the geometric series). Take a point p in the
: complement. Suppose it is not a limit point. Then there exists an open
: interval around p containing no other points of A. By our construction,
: this means that p is the common endpoint of two intervals which have been
: removed. But each interval removed has rational radius and is centered at
: a rational, hence p is rational. This contractiction shows that every
: point of A is a limit point.
If you label the rational intervals by B_n=(q_n-h_n,q_n+h_n) with
h_n=1/3^n, and let B=(p-h,p)+(p,p+h) be your interval around p, then by
assumption B is contained in the union of the B_n. But I see no reason to
conclude that one of the B_n's must have left (or right) endpoint p. The
set (p,p+h), for example, is possibly removed sequentially from the right,
by intervals B_n with rational left endpoint converging to p. In this
case, p needn't be rational.
Or am I missing the point of your argument here ?
Kruno
ull...@math.okstate.edu
If you do exactly what he said you certainly can get
isolated points. (I think the error was probably assuming that
the intervals we remove are disjoint).
But I bet you can fix it. Like what if you take the
construction above and then delete the isolated points?
The set is not going to become connected, it's still closed
(since removing an isolated point is the same as removing
a certain open interval about the point), it's still closed
(for the same reason), it's still nonempty, because there
are only countably many isolated points...
> Kruno
>
Sent via Deja.com http://www.deja.com/
Before you buy.
Dave L. Renfro
[sci.math Wed, 10 May 2000 12:33:09 GMT]
<http://forum.swarthmore.edu/epigone/sci.math/flarzhozal>
wrote
> This question from Rudin's textbook, p44
> has been stumping the undergrad Real Analysis
> class at UCSC: Is there a non-empty perfect
> set in R that contains no rational? We've tried
> various translations of the Cantor set, but haven't
> been able to demonstrate both closedness and the
> absence of rationals. Any hints?
Any translation of the Cantor set is closed. There are many
ways to prove this. One way is to observe that the Cantor set
is compact, any translation is a continuous map, any continuous
image of a compact set is compact, and any compact set is
closed. [Or: The Cantor set is closed, any translation is a
homeomorphism, and any homeomorphic image of a closed set
is closed.]
In 1884, Scheeffer [4] (pp. 291-293) published a proof of
the following:
Let C be a perfect nowhere dense set of reals and Z be a
countable set of reals. Then for each pair of real numbers
a1 < a2, there exists a real number b such that a1 < b < a2
and the b-translate of C has empty intersection with Z.
For a proof see pp. 52-53 of Young/Young [5] (these page numbers
are for the Chelsea edition). Young/Young's proof actually shows
that the set of real numbers b for which the b-translate of
C contains no points of Z is the complement of a first category
set of real numbers. In particular, there are continuum many such
b's in every open interval.
Boes/Darst/Erdos prove in [2] that given any countable set Z, the
set of b's in (0,1] such that the symmetric Cantor set, formed in
the following manner, is disjoint from Z is the complement of a
first category subset of (0,1]: Remove from [0,1] a segment of
length b/3 so as to leave two intervals of equal length; from
each of these two intervals remove a segment of length b/9 to
leave 4 intervals of equal length; continue in this manner.
In 1959, Bagemihl proved the following stronger version of
Scheeffer's theorem. [See also Morgan [3], pp. 194-195.]
Let F be a first category set of reals and Z be a countable set
of reals. Then the set of real numbers b for which the b-translate
of F contains no points of Z is the complement of a first category
set of real numbers.
[1] Frederick Bagemihl, "A note on Scheeffer's theorem", Michigan
Math. J. 2 (1954), 149-150.
[2] Duane Boes, Richard Darst, and Paul Erdos, "Fat, Symmetric,
Irrational Cantor Sets", The American Mathematical Monthly
88 (1981), 340-341.
[3] John C. Morgan II, POINT SET THEORY, Pure and Applied
Mathematics 131, Marcel Dekker, 1989. [QA 603 .P67]
[4] L. Scheeffer, "Zur Theorie der stetigen Functionen einer
reellen Veränderlichen", Acta Math. 5 (1884), 279-296.
[JFM 16.0340.01 at <http://www.emis.de/MATH/JFM/JFM.html>]
[5] Grace C. Young and William H. Young, THE THEORY OF SETS OF
POINTS, Cambridge, 1906. [2'nd edition (actually, just a
revised reprint with an appendix of additional notes) was
published by Chelsea in 1972, and can be found in most college
libraries in the QA 248's.]
Dave L. Renfro
Ed Hook
|> analysis class at UCSC:
|> Is there a non-empty perfect set in R
|> We've been trying to construct one
|> by a translation of the Cantor set, but so
|> far no one has found a way to preserve
|> closedness and not pick up any rationals.
|> Hints anyone?
_Translating_ the Cantor set isn't the way to
do it. Instead, you need to imitate the construction
of the Cantor set.
Start with a closed interval having irrational
endpoints and choose a fixed enumeration of the
rationals contained in your interval. Then, at
each stage of the construction, remove slightly
more than the middle third of each component of
whatever is left over from the previous stage,
being sure to excise intervals having irrational
endpoints and doing it in such a way that the
i'th rational point is removed at the i'th stage.
If you can figure out what that paragraph actually
_says_ (:-)), you'll have a recipe for constructing
a perfect, nowhere-dense subset of the real line
containing only irrational points ...
--
Ed Hook | Copula eam, se non posit
Computer Sciences Corporation | acceptera jocularum.
NAS, NASA Ames Research Center | All opinions herein expressed are
Internet: ho...@nas.nasa.gov | mine alone
Miguel A. Lerma
:
: I say there is.
:
: Try this:
:
: [0,1]. Remove an interval of length 1/3^n around each q_n, so there
: are no rationals left. The remaining set A is closed and has measure 1/2
: so it is nonempty (sum the geometric series). Take a point p in the
: complement. Suppose it is not a limit point. Then there exists an open
: interval around p containing no other points of A. By our construction,
: this means that p is the common endpoint of two intervals which have been
: removed.
I do not see this. The reduced open interval around p could be
an infinite union of intervals of length 1/3^n around a subsequence
of q_n's rapidly converging to p.
: But each interval removed has rational radius and is centered at
: a rational, hence p is rational. This contractiction shows that every
: point of A is a limit point.
Miguel A. Lerma
David G Radcliffe
: This question from Rudin's textbook, p44
: has been stumping the undergrad real
: analysis class at UCSC:
: Is there a non-empty perfect set in R
: which contains no rational number?
S = {sum_{n>0} (a_n)/(n+2)! | a_n = 1 or 3 for all n} is such a set.
The usual proof that e is irrational can be modified to show that
each element of S is irrational. S is homeomorphic to the standard
middle-thirds Cantor set C via the following map.
sum_{n>0} a_n/(n+2)! --> sum_{n>0} (a_n - 1)/3^n
Another argument is the following. The Cantor set C is homeomorphic
to CxC. Fix a homeomorphism h: CxC -> C. Then {h(Cx{c}) : c in C}
is an uncountable disjoint collection of Cantor sets. Since the
rationals are countable, one of these Cantor sets must avoid the
rational numbers.
--
David Radcliffe radc...@alpha2.csd.uwm.edu
Michael van Opstall
Yes, all of you are correct in catching my error. How immature of
me. Thanks.
======================================================================
Michael A. Van Opstall
Padelford C-113
ops...@math.washington.edu
ull...@math.okstate.edu
ull...@math.okstate.edu wrote:
[...]
>
> If you do exactly what he said you certainly can get
> isolated points. (I think the error was probably assuming that
> the intervals we remove are disjoint).
>
> But I bet you can fix it. Like what if you take the
> construction above and then delete the isolated points?
> The set is not going to become connected, it's still closed
> (since removing an isolated point is the same as removing
> a certain open interval about the point), it's still closed
> (for the same reason), it's still nonempty, because there
> are only countably many isolated points...
This is all wrong, by the way. Surprised Pertti didn't
catch it; he could give a counterexample and be famous.
The problem is that if you remove the isolated points
more pop up (points that were limits of isolated points
can become isolated after the original isolalated points
are removed.)
Pertti Lounesto
> This is all wrong, by the way. Surprised Pertti didn't
> catch it; he could give a counterexample and be famous.
There is no point catching mistakes of poor mathematicians,
or mistakes in unpublished works. I only catch mistakes
of good mathematicians and in their published works,
which has obtained favrourble reviews, see
http://www.hit.fi/~lounesto/counterexamples.htm
As for becoming famous: I cannot become famous, since
I am already famous.
Richard Carr
:Date: Fri, 12 May 2000 20:10:30 +0200
:From: Pertti Lounesto <Pertti....@hit.fi>
:Newsgroups: sci.math
:Subject: Re: Question from Rudin
:
:
Pertti, your English has let you down. You should have said that you are
infamous.
PF
<ho...@nas.nasa.gov> wrote:
> In article <38f073b...@news.mindspring.com>,
> pa...@cats.ucsc.edu (paul mitchell) writes:
> |> This question from Rudin's textbook, p44
> |> has been stumping the undergrad real
> |> analysis class at UCSC:
> |> Is there a non-empty perfect set in R
> |> which contains no rational number?
> |> We've been trying to construct one
> |> by a translation of the Cantor set, but so
> |> far no one has found a way to preserve
> |> closedness and not pick up any rationals.
> |> Hints anyone?
>
> _Translating_ the Cantor set isn't the way to
> do it. Instead, you need to imitate the construction
> of the Cantor set.
Translating randomly does work. Let C be the Cantor ternary set and
let U be a random variable uniformly distributed over [0,1]. Then if q
is a rational number, Prob[ q in C+U ] = 0 (Fubini). Since the set Q
of rationals is countable, countable subadditivity of Prob leads to the
conclusion that Prob[ Q meets K+U ] = 0. In other words, for
(Lebesgue) almost every number u between 0 and 1, the translated Cantor
set C+u contains no rational numbers. Of course, C+u is perfect for
every u.
This is probably not the solution Rudin had in mind.
P.F.
Edward C. Hook
|> In article <8feicm$4tp$1...@nnrp1.deja.com>,
|> ull...@math.okstate.edu wrote:
|> [...]
|> >
|> > If you do exactly what he said you certainly can get
|> > isolated points. (I think the error was probably assuming that
|> > the intervals we remove are disjoint).
|> >
|> > But I bet you can fix it. Like what if you take the
|> > construction above and then delete the isolated points?
|> > The set is not going to become connected, it's still closed
|> > (since removing an isolated point is the same as removing
|> > a certain open interval about the point), it's still closed
|> > (for the same reason), it's still nonempty, because there
|> > are only countably many isolated points...
|>
|> catch it; he could give a counterexample and be famous.
|>
|> more pop up (points that were limits of isolated points
|> can become isolated after the original isolalated points
|> are removed.)
Maybe the following is a reasonable approach:
Start with [0,1] (say) and an enumeration of
the rationals in there. At the k'th stage,
remove the points in an open interval of length
<whatever>/3^k centered at the k'th rational.
When you're done, you're left with a compact
set containing nothing but irrationals and of
positive measure, hence uncountable. Then appeal
to the Cantor-Bendixson theorem (which is yet
another problem in Rudin -- or, at least, it
was when I had it as a textbook 25+ years ago):
Any closed set in a separable metric space can
be written as the disjoint union of a (possibly
empty) perfect set and a set that is at most
countable.
So just cite that theorem and use the perfect
subset as the example ...
ull...@math.okstate.edu
Pertti Lounesto <Pertti....@hit.fi> wrote:
> ull...@math.okstate.edu wrote:
>
> > This is all wrong, by the way. Surprised Pertti didn't
> > catch it; he could give a counterexample and be famous.
>
> or mistakes in unpublished works. I only catch mistakes
> of good mathematicians and in their published works,
Heh-heh. Not sure whether the reason you didn't catch
my error is I'm not a good mathematician (haha, glad to
have input from the world's narrowest specialist on that)
or because my post was not published. But it doesn't
matter, in either case this is inconsistent with the
hoo-haw you've made about "errors" of mine in the past.
You've become inconsistent, hence you no longer exist.
Poof, no more Pertti. That's too bad.
ull...@math.okstate.edu
ho...@nas.nasa.gov (Edward C. Hook) wrote:
> In article <8fhcj9$99t$1...@nnrp1.deja.com>, ull...@math.okstate.edu
writes:
> |> In article <8feicm$4tp$1...@nnrp1.deja.com>,
> |> ull...@math.okstate.edu wrote:
> |> [...]
> |> >
> |> > If you do exactly what he said you certainly can get
> |> > isolated points. (I think the error was probably assuming that
> |> > the intervals we remove are disjoint).
> |> >
> |> > But I bet you can fix it. Like what if you take the
> |> > construction above and then delete the isolated points?
> |> > The set is not going to become connected, it's still closed
> |> > (since removing an isolated point is the same as removing
> |> > a certain open interval about the point), it's still closed
> |> > (for the same reason), it's still nonempty, because there
> |> > are only countably many isolated points...
> |>
> |> catch it; he could give a counterexample and be famous.
> |>
> |> more pop up (points that were limits of isolated points
> |> can become isolated after the original isolalated points
> |> are removed.)
>
> Maybe the following is a reasonable approach:
>
> Start with [0,1] (say) and an enumeration of
> the rationals in there. At the k'th stage,
> remove the points in an open interval of length
> <whatever>/3^k centered at the k'th rational.
> When you're done, you're left with a compact
> set containing nothing but irrationals and of
> positive measure, hence uncountable. Then appeal
> to the Cantor-Bendixson theorem (which is yet
> another problem in Rudin -- or, at least, it
> was when I had it as a textbook 25+ years ago):
>
> Any closed set in a separable metric space can
> be written as the disjoint union of a (possibly
> empty) perfect set and a set that is at most
> countable.
>
> So just cite that theorem and use the perfect
> subset as the example ...
Or instead of citing the theorem prove it (there's
my bad attitude showing up). What happens if you remove
all the isolated points, then remove all the new ones, do
that infinitely many times if needed, then do it again...
Is it clear that after some countable-ordinal number
of steps there must be no isolated points left?
I bet it is. Let's see, suppose not: If K_alpha
is what's left after removing all the isolated points
alpha times (taking intersections at limits), and if
for any countable alpha we can find a x_alpha which is
in K_alpha but not in any K_beta for beta > alpha
then for every alpha there is d_alpha > 0 such that
d(x_alpha, x_beta) > d_alpha for all beta > alpha.
Now there's an uncountable set of ordinals S and
a number r > 0 such that d_alpha > r for all
alpha in S. Hence d(x_alpha, x_beta) > r for all
alpha, beta in S, so we have uncountably many disjoint
balls of radius r/2, unlikely in a compact metric
space. Hah, our proof is saved.
I wish I was a good mathematician like Pertti.
He could probably find a better proof using Clifford
algebras. (There are plenty of people in the parallel
thread on the same question who've given very nice
proofs, but none of them involve Clifford algebras -
bad bad bad.)
> --
> Ed Hook | Copula eam, se non posit
> Computer Sciences Corporation | acceptera jocularum.
> NAS, NASA Ames Research Center | All opinions herein
expressed are
> Internet: ho...@nas.nasa.gov | mine alone
>
Ilias Kastanas
@In article <8fi4he$js4$1...@sun500.nas.nasa.gov>,
@ ho...@nas.nasa.gov (Edward C. Hook) wrote:
@> In article <8fhcj9$99t$1...@nnrp1.deja.com>, ull...@math.okstate.edu
@writes:
@> |> In article <8feicm$4tp$1...@nnrp1.deja.com>,
@> |> ull...@math.okstate.edu wrote:
@> |> [...]
@> |> >
@> |> > If you do exactly what he said you certainly can get
@> |> > isolated points. (I think the error was probably assuming that
@> |> > the intervals we remove are disjoint).
@> |> >
@> |> > But I bet you can fix it. Like what if you take the
@> |> > construction above and then delete the isolated points?
@> |> > The set is not going to become connected, it's still closed
@> |> > (since removing an isolated point is the same as removing
@> |> > a certain open interval about the point), it's still closed
@> |> > (for the same reason), it's still nonempty, because there
@> |> > are only countably many isolated points...
@> |>
@> |> This is all wrong, by the way. Surprised Pertti didn't
@> |> catch it; he could give a counterexample and be famous.
@> |>
@> |> The problem is that if you remove the isolated points
@> |> more pop up (points that were limits of isolated points
@> |> can become isolated after the original isolalated points
@> |> are removed.)
@>
@> Maybe the following is a reasonable approach:
@>
@> Start with [0,1] (say) and an enumeration of
@> the rationals in there. At the k'th stage,
@> remove the points in an open interval of length
@> <whatever>/3^k centered at the k'th rational.
@> When you're done, you're left with a compact
@> set containing nothing but irrationals and of
@> positive measure, hence uncountable. Then appeal
@> to the Cantor-Bendixson theorem (which is yet
@> another problem in Rudin -- or, at least, it
@> was when I had it as a textbook 25+ years ago):
@>
@> Any closed set in a separable metric space can
@> be written as the disjoint union of a (possibly
@> empty) perfect set and a set that is at most
@> countable.
@>
@> So just cite that theorem and use the perfect
@> subset as the example ...
@
@ Or instead of citing the theorem prove it (there's
@my bad attitude showing up). What happens if you remove
@all the isolated points, then remove all the new ones, do
@that infinitely many times if needed, then do it again...
@Is it clear that after some countable-ordinal number
@of steps there must be no isolated points left?
@
@ I bet it is. Let's see, suppose not: If K_alpha
@is what's left after removing all the isolated points
@alpha times (taking intersections at limits), and if
@for any countable alpha we can find a x_alpha which is
@in K_alpha but not in any K_beta for beta > alpha
@then for every alpha there is d_alpha > 0 such that
@d(x_alpha, x_beta) > d_alpha for all beta > alpha.
@Now there's an uncountable set of ordinals S and
@a number r > 0 such that d_alpha > r for all
@alpha in S. Hence d(x_alpha, x_beta) > r for all
@alpha, beta in S, so we have uncountably many disjoint
@balls of radius r/2, unlikely in a compact metric
@space. Hah, our proof is saved.
Right. We can also streamline the process a bit.
Consider the rational intervals (more generally: members of
a countable base) such that each contains at most countably
many points of K. Remove all such points from K, and what
is left is a perfect set.
Slick... but not as informative as the K_alpha's regarding
the structure of closed sets.
Here is another way of showing that the length of <K_alpha>
is a countable ordinal. Fix a countable base I_0, I_1, ... .
Associate with each V_alpha, complement of K_alpha, the set
{k: I_k subset of V_alpha}... and you have an increasing sequence
of subsets of w; fresh integers appear at every step.
Ilias
Ilias Kastanas
PF <ir...@sdcc3.ucsd.edu> wrote:
@In article <8feilp$bed$2...@sun500.nas.nasa.gov>, Ed Hook
@<ho...@nas.nasa.gov> wrote:
@
@> In article <38f073b...@news.mindspring.com>,
@> pa...@cats.ucsc.edu (paul mitchell) writes:
@> |> This question from Rudin's textbook, p44
@> |> has been stumping the undergrad real
@> |> analysis class at UCSC:
@> |> Is there a non-empty perfect set in R
@> |> which contains no rational number?
@> |> We've been trying to construct one
@> |> by a translation of the Cantor set, but so
@> |> far no one has found a way to preserve
@> |> closedness and not pick up any rationals.
@> |> Hints anyone?
@>
@> _Translating_ the Cantor set isn't the way to
@> do it. Instead, you need to imitate the construction
@> of the Cantor set.
@
@Translating randomly does work. Let C be the Cantor ternary set and
@let U be a random variable uniformly distributed over [0,1]. Then if q
@is a rational number, Prob[ q in C+U ] = 0 (Fubini). Since the set Q
@of rationals is countable, countable subadditivity of Prob leads to the
@conclusion that Prob[ Q meets K+U ] = 0. In other words, for
@(Lebesgue) almost every number u between 0 and 1, the translated Cantor
@set C+u contains no rational numbers. Of course, C+u is perfect for
@every u.
Indeed, this applies to any countable set P and any set of
measure zero D; for any p in P the set T_p of x such that
p is in D+x has measure zero, hence so does the union of all T_p.
Thus for almost all x, D+x is disjoint from P.
@This is probably not the solution Rudin had in mind.
@
@
Who knows. Yet another solution is: the irrationals are a
complete metric space under the Baire metric (same topology), so
the usual decreasing-closed-balls construction works.
A slightly more challenging result: there is a Cantor set
whose members are linearly independent over Q. (The ternary set C
and its translates don't work for this one).
Ilias
ull...@math.okstate.edu
Yeah. I don't know anything about this stuff - I was just
irritated at having said something stupid and fumbling for a
save, which Richard suggested.
It's a more interesting question than I realized at first.
Since you've appeared I don't have to worry about trying to
figure it out<g>: I imagine it's easy to construct examples
of compact subsets of the line such that the K_alpha become
constant at precisely any given alpha_0?
Assuming yes, that's sort of interesting - makes the
countable ordinals a little more "real"...
> Here is another way of showing that the length of <K_alpha>
> is a countable ordinal. Fix a countable base I_0, I_1, ... .
> Associate with each V_alpha, complement of K_alpha, the set
> {k: I_k subset of V_alpha}... and you have an increasing sequence
> of subsets of w; fresh integers appear at every step.
>
> Ilias
>
>
David C. Ullrich
Ilias Kastanas <ika...@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
(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.)
Edward C. Hook
|> In article <8fi4he$js4$1...@sun500.nas.nasa.gov>,
|> ho...@nas.nasa.gov (Edward C. Hook) wrote:
|> > In article <8fhcj9$99t$1...@nnrp1.deja.com>, ull...@math.okstate.edu
<*snip*>
|> > |>
|> > |> The problem is that if you remove the isolated points
|> > |> more pop up (points that were limits of isolated points
|> > |> can become isolated after the original isolalated points
|> > |> are removed.)
|> >
|> > Maybe the following is a reasonable approach:
|> >
|> > Start with [0,1] (say) and an enumeration of
|> > the rationals in there. At the k'th stage,
|> > remove the points in an open interval of length
|> > <whatever>/3^k centered at the k'th rational.
|> > When you're done, you're left with a compact
|> > set containing nothing but irrationals and of
|> > positive measure, hence uncountable. Then appeal
|> > to the Cantor-Bendixson theorem (which is yet
|> > another problem in Rudin -- or, at least, it
|> > was when I had it as a textbook 25+ years ago):
oops: 35+
|> >
|> > Any closed set in a separable metric space can
|> > be written as the disjoint union of a (possibly
|> > empty) perfect set and a set that is at most
|> > countable.
|> >
|> > So just cite that theorem and use the perfect
|> > subset as the example ...
|>
|> Or instead of citing the theorem prove it (there's
|> my bad attitude showing up). What happens if you remove
|> all the isolated points, then remove all the new ones, do
|> that infinitely many times if needed, then do it again...
This seems to have evolved into an interesting exchange
between yourself and Ilias, which is refreshing (given
the recent history of sci.math -- judging from the number
of postings that my killfile processing removes, I'd guess
that Pertti is still at it ?? ) Anyhow, my suggestion
above was an attempt to avoid having to deal with the
problem of isolated points continuing to pop up as one
attempted to get rid of them. Basically, there's _no_
iteration here - you build the first approximation, then
cite the theorem to get the perfect set in one fell swoop.
And it's not clear (to me, at least) that this amounts
to removing isolated points. If you want to investigate
whether that's what is really happening, here's a sketch
of the proof:
Let X be a separable metric space and let A be a closed
subset of X. Say that p is a _condensation_point_ of A
iff every neighborhood of p contains uncountably many
points of A (so these are "industrial strength" limit
points). Since condensation points _are_ limit points,
the closed set A contains any of its condensation points.
Define P = {p in X | p a condensation point of A}
and let C = A \ P. Then (obviously) A is the disjoint
union of P and C - I claim that P is perfect and C is
(at most) countable. Taking the claim about C first,
suppose that q in X is _not_ a condensation point of A -
then q has a *basic* neighborhood V(q) such that V(q) \cap A
is at most countable. Then C is a subset of the union of
the various distinct V(q) \cap A's and there are only
countably many V(q)'s possible -- hence, C is at most
countable. Now for the claim concerning P. First, every
p in P is a limit point of P: if V is a neighborhood of
p, then V \cap A = (V \cap P) \cup (V \cap C) is
uncountable, while V \cap C is at most countable, so
V \cap P is uncountable. (So p is, in fact, a condensation
point of P). Second, P is closed: any q in X \ P has a
neighborhood V such that V \cap A is at most countable
and this V cannot contain _any_ points of P -- so
X \ P is open. That finishes it off ...
|> Is it clear that after some countable-ordinal number
|> of steps there must be no isolated points left?
|> I bet it is. Let's see, suppose not: If K_alpha
|> is what's left after removing all the isolated points
|> alpha times (taking intersections at limits), and if
|> for any countable alpha we can find a x_alpha which is
|> in K_alpha but not in any K_beta for beta > alpha
That is, we're supposing that the process never
stabilizes while we're still in countable-ordinal
range ?
|> then for every alpha there is d_alpha > 0 such that
|> d(x_alpha, x_beta) > d_alpha for all beta > alpha.
Why's that ?
You're clearly better at this than I am -- I'm
not following this argument.
Jeez -- just think what you could do, if only you'd
put in the effort to master Clifford algebras ... :-)
|> Now there's an uncountable set of ordinals S and
|> a number r > 0 such that d_alpha > r for all
|> alpha in S. Hence d(x_alpha, x_beta) > r for all
|> alpha, beta in S, so we have uncountably many disjoint
|> balls of radius r/2, unlikely in a compact metric
|> space. Hah, our proof is saved.
ull...@math.okstate.edu
This is (presumably going to be) more or less the
argument Ilias suggested yesterday, presumably the standard
proof. It took me a minute to see it, but I believe it is
removing the same points as the points I removed one step at
a time: You see by induction on alpha that an isolated point
of one of the K_alpha must be a non-condensation point, ie
must have a neighborhood containing only countably many
points of the original set. (Oh: it's not even induction:
each K_alpha has only countable many isolated points, so
at any stage there are only countably many points that have
been removed so far. An isolated point of K_alpha has a
neighborhood containing no other points of K_alpha, hence
only countably many points of the original set.) Otoh a
non-condensation point eventually disappears from the
sequence K_alpha (by the argument below).
So (I'm pretty sure that) it's the same points being
removed. The other day it seemed pretty clear that the
points I was removing one step at a time had some simple
charcterization, was distracted by the nonsense elsewhere,
settled for the first contradicion that popped up.
Yes - the K_alpha are dereasing, so saying they
never become constant is the same as saying for every
alpha there is x_alpha as above.
> |> then for every alpha there is d_alpha > 0 such that
> |> d(x_alpha, x_beta) > d_alpha for all beta > alpha.
>
> Why's that ?
K_(alpha+1) is a compact set not containing x_alpha,
so there is d_alpha > 0 such that d(x_alpha, x) > d_alpha
for all x in K_(alpha+1). Each K_beta (beta > alpha) is
a subset of K_(alpha+1).
> You're clearly better at this than I am -- I'm
> not following this argument.
Golly, I hadn't realized that was the criterion.
If my ability to write something you can't follow at
first because I left out a lot of details makes me
smart then I'm a bloody genius, no doubt about it.
Keen.
> Jeez -- just think what you could do, if only you'd
> put in the effort to master Clifford algebras ... :-)
>
> |> Now there's an uncountable set of ordinals S and
> |> a number r > 0 such that d_alpha > r for all
> |> alpha in S. Hence d(x_alpha, x_beta) > r for all
> |> alpha, beta in S, so we have uncountably many disjoint
> |> balls of radius r/2, unlikely in a compact metric
> |> space. Hah, our proof is saved.
>
> |> I wish I was a good mathematician like Pertti.
> |> He could probably find a better proof using Clifford
> |> algebras. (There are plenty of people in the parallel
> |> thread on the same question who've given very nice
> |> proofs, but none of them involve Clifford algebras -
> |> bad bad bad.)
>
> --
> Ed Hook | Copula eam, se non posit
> Computer Sciences Corporation | acceptera jocularum.
> NAS, NASA Ames Research Center | All opinions herein
expressed are
> Internet: ho...@nas.nasa.gov | mine alone
>
Dave L. Renfro
[sci.math Sun, 14 May 2000 18:30:17 GMT]
<http://forum.swarthmore.edu/epigone/sci.math/flarzhozal>
wrote (in part)
> It's a more interesting question than I realized at first.
> Since you've appeared I don't have to worry about trying to
> figure it out<g>: I imagine it's easy to construct examples
> of compact subsets of the line such that the K_alpha become
> constant at precisely any given alpha_0?
> Assuming yes, that's sort of interesting - makes the
> countable ordinals a little more "real"...
What you're getting at is something called the Cantor-Bendixson
rank of a closed set. Define the derived set of a set A to be
A' = {x in R: x is a limit point of A} [R = the reals].
Now define by transfinite induction the b'th derived set of A,
denoted here by A^b, for any ordinal b as follows.
Let A^0 = A. If b is a successor ordinal, let A^b = [A^(b-1)]'.
If b is a nonzero limit ordinal, let A^b be the intersection of
all the sets A^c such that c < b.
Note that if A is a closed set, then A^b is a closed set for
each ordinal b. In this case, we get a non-increasing (with respect
to set inclusion) transfinite sequence of closed sets:
A^0 -> A^1 -> A^2 -> ... -> A^w -> A^(w+1) -> ... -> A^(w^w) -> ...
It is easy to see that any non-increasing transfinite sequence of
closed sets, whether of the above form or not, is eventually
stationary. [[ It suffices to prove that if {F_c : c < b} is a
strictly decreasing b-sequence of closed sets, then b < w_1. To
show this, for each c < b choose x_c in F_c and not F_{c+1}.
Since the complement of F_{c+1} is open, there exists an open
interval (r_c, s_c) with rational endpoints containing x_c that
belongs to the complement of F_{c+1}. Because distinct values
of c give rise to distinct rational intervals, the map from
the ordinal b = {c : c < b} to the collection of rational open
intervals in R defined by c |--> (r_c, s_c) is injective. Hence,
b must be countable since there are only countably many rational
open intervals in R. ]] This continues to hold in any second
countable topological space.
Using this fact it is not difficult to prove the Cantor-Bendixson
theorem: Given any closed set F in the reals, F is the union of
a perfect set and a countable set. [This is how Cantor and
Bendixson (independently) originally proved the C-B theorem
in 1883.]
I'm not sure what Bendixson was motivated by, although I do know
that he was closely following Cantor's work and that they exchanged
a number of letters during this time, but Cantor was motivated
by questions involving trigonometric sets of uniqueness. In 1870
Cantor proved that if two convergent trigonometric series are
equal, then their coefficients must be term-wise equal. Two years
later Cantor managed to improve this result by showing that
"equality everywhere" of the series expansions can be weakened
to "equality except for a closed countable set having finite
Cantor-Bendixson rank". Cantor's further researches into the
nature of closed sets (e.g. the C-B theorem) led him into those
areas for which he is better known for (cardinal and ordinal
numbers) and, as a result of the successes he had there, he
never returned to questions involving trigonometric sets of
uniqueness. [At least, if Cantor did return to these questions,
there is no record of any further results by him.]
[[ Call a set of real numbers a U-set if the agreement of any
two convergent trigonometric series on the complement of
this set implies that their coefficients are term-wise equal.
(i.e. An "interpolating" set for trigonometric series, to
use the terminology employed by complex analysts.) W. H. Young
improved Cantor's result in 1909 by showing that any countable
set is a U-set. It was known that any measurable U-set has
measure zero. (There exist U-sets with positive outer measure:
Take any set of positive outer measure not containing a perfect
subset. Then a short argument--due to W. H. Young--on page 344
of Zygmund's treatise TRIGONOMETRIC SERIES shows this set is
a U-set.) It was thought the converse was true, that any set
of measure zero is a U-set. However, the Russian D. E. Men'shov
proved in 1916 that the symmetric Cantor set whose n'th stage
dissection ratio is 1/(n+1), which is easily shown to have
measure zero, is NOT a U-set. At this point it still wasn't
known if a Borel U-set could be uncountable. This was settled
by Nina Bary (Russia, 1922) and A. Rajchman (Poland, 1923),
who showed that uncountable closed U-sets exist. Bary also
improved Young's result by showing that any countable union
of closed U-sets is a U-set. (As far as I know, it is still
an unsolved problem as to whether the union of two Borel, or
even just two G_delta, U-sets is a U-set.) Recently (1987),
G. Debs and J. Saint-Reymond proved that any U-set having the
Baire property is a first category set in the reals. Their
proof was rather involved and used descriptive set-theoretic
notions. A more elegant proof using functional analysis ideas
was given by A. S. Kechris and A. Louveau (see their monograph,
or see Colloq. Math. 59 (1990), pp. 63-79). Very recently,
Miroslav Zeleny ["Sets of extended uniqueness and
sigma-porosity", Comment. Math. Univ. Carolinae 38 (1997),
337-341] has proved there exists a closed U-set that isn't
sigma-porous (a slightly stronger property than being
simultaneously measure zero and first category). ]]
Note that the Cantor-Bendixson theorem implies any closed set
of real numbers is either countable or has cardinality of the
continuum. Thus, the continuum hypothesis is true for closed sets.
This is as far as Cantor got in settling CH. (Cantor had proved
that any nonempty perfect set has cardinality of the continuum.)
Of course, once you know this for closed sets, you automatically
have it for F_sigma sets. However, I don't think these notions
(F_sigma and/or G_delta sets) were employed by Cantor. W. H. Young
extended CH to G_delta sets in 1903, and then Hausdorff proved
that the continuum hypothesis holds for any Borel set in 1916.
This was further extended to any continuous image of a Borel set
(the analytic sets) by Suslin in 1917 (announcement only; I don't
believe a proof was published until Lusin's 1930 monograph on
analytic sets). Further up the projective set hierarchy the
problem becomes very difficult, since it is independent of the
ZFC axioms, and so the results vary depending on which large
cardinal hypotheses (and/or other non-ZFC hypotheses) are assumed.
Back to the Cantor-Bendixson (C-B) rank of a closed set . . .
It is easy to show that for any ordinal b < w_1, there exists a
closed subset of the rational numbers having C-B rank equal to b.
THEOREM: Let (A, <) be an unbounded and dense linearly ordered
set and let (B, <') be a countable linearly ordered set.
Then (B, <') is order-isomorphic to a subset of (A, <).
PROOF: Re-order B as b_1, b_2, ... . We will define an order
isomorphism f from (B, <') to (A, <) by (ordinary)
induction. Let f(b_1) be any element of A, say a_1. Now
consider the <' relation of b_2 to b_1. If b_2 >' b_1,
choose a_2 in A such that a_2 > a_1. If b_2 <' b_1, choose
a_2 in A such that a_2 < a_1. Such an a_2 can be found in
either case because (A, <) is unbounded. Now assume that
a_1, a_2, ..., a_n have been chosen in A having the same
<-order relationship as b_1, b_2, ..., b_n have in (B, <').
If b_{n+1} <' b_n or b_{n+1} >' b_n, choose a_{n+1} < a_n
or a_{n+1} > a_n, respectively, using the fact that (A, <)
is unbounded. If b_{n+1} lies between (in the <' ordering)
two elements of {b_1, b_2, ..., b_n}, choose a_{n+1} to be
an element of A lying between (in the < ordering) the two
corresponding elements of {a_1, a_2, ..., a_n}. This is
possible because (A, <) is dense. Now define f: B --> A
by f(b_k) = a_k. It is straightforward to check that f
is an injective order preserving map from (B, <') onto
a subset of (A, <).
Since the rational numbers are unbounded and dense in their usual
ordering, any countable linear order type can be represented
by some subset of the rationals. In particular, any countable
WELL ordering (i.e. a countable ordinal) can be represented by
a subset of the rationals. To be explicit, if you take a
subset of the rationals having order type w^b + 1 (ordinal
exponentiation), then its b'th derived set is nonempty while its
(b+1)'st derived set is empty. {See page 197 of Hausdorff [1],
5.2(3) on page 78 of Rosenstein [2], or 8.6.6 on page 154 of
Semandeni [3] for this specific construction. Semandeni's book,
by the way, gives some interesting applications in functional
analysis of these ideas in his Chapter II.8 "Compact 0-dimensional
spaces".}
[1] Felix Hausdorff, SET THEORY, 3'rd edition, Chelsea, 1935/1978.
[MR 19 (p. 111)]
[2] Joseph G. Rosenstein, LINEAR ORDERINGS, Pure and Applied
Mathematics 98, Academic Press, 1982.
[MR 84m:06001; Zbl 488.04002]
[3] Zbigniew Semadeni, BANACH SPACES OF CONTINUOUS FUNCTIONS,
Monografie Matematyczne 55, PWN-Polish Scientific Publishers,
1971. [MR 45 #5730; Zbl 225.46030]
The Cantor-Bendixson rank of a closed set is often used in
certain areas. [It plays a nontrivial role in one of my papers,
in fact.] Moreover, the ideas can be generalized to the setting
of a monotone operator defined on the power set of a given set.
This setting, where the monotone operator is transfinitely
iterated until a fixed point is obtained, allows one to
"constructively obtain" things such as: the linear span of
a set of vectors in a vector space, the closure of a set in a
topological space, the sigma-algebra generated by a collection
of subsets of a fixed set, the set of well-formed formulas in
first order logic, etc. In fact, one of the common proofs of
the Schroder-Berstein theorem can even be put into this
context. {See the exercises on page 69 of Harbacek/Jech [4].}
[4] Karel Hrbacek and Thomas Jech, INTRODUCTION TO SET THEORY,
3'rd edition, Pure and Applied Mathematics 220, Marcel
Dekker, 1999.
There are two other methods of proving the C-B theorem that I
know of, and each of these methods has given rise to a number
of papers exploiting their generalizations. [It remains to be
seen, however, whether there is a proof of the C-B theorem using
Clifford algebras.] There is the condensation point proof due to
Lindelof (1903), which has been generalized to the topological
localization of ideals [See Kuratowski's TOPOLOGY treatise,
Henry Blumberg's papers, and the numerous papers by Troy R. Hamlett
and/or Dragan Jankovic (e.g. "New topologies from old via ideals",
Amer. Math. Monthly 97 (1990), 295-310).], and there is a
"maximal dense-in-itself set" proof due to Hausdorff, which
leads to the generalization to hereditarily Lindelof topological
spaces that one can find in some topology texts (e.g. Kelly's
text, page 57; Willard's text, page 114).
Dave L. Renfro
ull...@math.okstate.edu
dlre...@gateway.net (Dave L. Renfro) wrote:
> ullrich <ull...@math.okstate.edu>
> [sci.math Sun, 14 May 2000 18:30:17 GMT]
> <http://forum.swarthmore.edu/epigone/sci.math/flarzhozal>
>
> wrote (in part)
>
> > It's a more interesting question than I realized at first.
> > Since you've appeared I don't have to worry about trying to
> > figure it out<g>: I imagine it's easy to construct examples
> > of compact subsets of the line such that the K_alpha become
> > constant at precisely any given alpha_0?
> > Assuming yes, that's sort of interesting - makes the
> > countable ordinals a little more "real"...
>
> What you're getting at is something called the Cantor-Bendixson
> rank of a closed set. Define the derived set of a set A to be
>
kinda, category, only a small bit of which was stuff I actually
proved the other day...<g>]
> To be explicit, if you take a
> subset of the rationals having order type w^b + 1 (ordinal
> exponentiation), then its b'th derived set is nonempty while its
> (b+1)'st derived set is empty.
I certainly knew you could get any countable ordinal as a subset
of the rationals, and hence it seemed that something like this
was what I wanted (exactly this set was on my list of possibilites,
decided to let someone else worry about it.)
This is the second time very recently that a post of yours
made it all the way from my monitor to the printer down the hall.
There should be a moderated sci.math archive somewhere, with
just the good stuff. (Sugar-honey fruits on Clifford algebras
might even be acceptable, but no posts containing the string
"I am the greatest".) Thanks.
Dave L. Renfro
[sci.math 15 May 00 13:59:58 -0400 (EDT)]
<http://forum.swarthmore.edu/epigone/sci.math/flarzhozal>
wrote (in part):
> In particular, any countable WELL ordering (i.e. a countable
> ordinal) can be represented by a subset of the rationals. To be
> explicit, if you take a subset of the rationals having order type
> w^b + 1 (ordinal exponentiation), then its b'th derived set is
> nonempty while its (b+1)'st derived set is empty.
Of course, I want b < w_1 here.
> first order logic, etc. In fact, one of the common proofs of
> the Schroder-Berstein theorem can even be put into this
And here I should have written: Schroder-Bernstein.
I'll leave to Pertti the task of finding other errors that
might be present.
Dave L. Renfro
Pertti Lounesto
> Dave L. Renfro <dlre...@gateway.net>
>
> > the Schroder-Berstein theorem can even be put into this
>
> And here I should have written: Schroder-Bernstein.
>
> I'll leave to Pertti the task of finding other errors that
> might be present.
I do not scan errors unless they are mathematical and
1. have been proved,
2. have been published in refereed journals,
3. have passed reviews in abstracts without being noticed,
4. reveal genuine misconceptions, whose corrections
might result in cognitive growth in the mistake-maker.
Here in sci.math, I have exposed errors only for the purpose
of demonstrating that also the best mathematicians in this
forum make mistakes, while the best mathematicians were
doubtful about frequency of (their own) errors. See,
http://www.hit.fi/~lounesto/counterexamples.htm