Real Dense Problem
William Elliot
Gee, I haven't the foggiest how,
tho I surmise it requires axiom of choice.
Anybody have any ideas?
Jim Ferry
Let S be the union of the rationals on [0,1] and the irrationals
elsewhere . . .
| Jim Ferry | Center for Simulation |
+------------------------------------+ of Advanced Rockets |
| http://www.uiuc.edu/ph/www/jferry/ +------------------------+
| jferry@[delete_this]uiuc.edu | University of Illinois |
Zdislav V. Kovarik
Hint: Play with Cantor set (uncountable, closed, nowhere dense).
Cheers, ZVK(Slavek).
Denis Feldmann
Yes; no choice needed. Stay in base 10. A is the set of numbers n (in Z)
+0.a_1a_2a_3... such that exists n_0 such that for k>n_0, a_k=2 or 3. B is
the complementary set. It is easy to check that A and B are dense and
uncountable (in fact, with the cardinality of IR)
Adrien
B=((IR\Q inter IR+) U (Q-))\{0}
IR+ is the set of positive reals (>=0)
Q+ is the set of positive rationals
IR- is the set of negative reals (<=0)
Q- is the set of negative rationals
does it work ?
Mike Oliver
> Divide the real line into two uncountable disjoint dense subsets.
>
> Gee, I haven't the foggiest how,
> tho I surmise it requires axiom of choice.
Comeager and conull both imply dense, and there are lots of
ways to partition the reals into a comeager set and a conull set.
For example, the set of reals not normal to any base is null
but comeager.
Dave L. Renfro
[sci.math Jan 20 2003 10:30:29:000AM]
http://mathforum.org/discuss/sci.math/m/475274/475274
wrote
Aww come on, let's be really greedy -->
Partition the reals into c = 2^(aleph_0) pairwise disjoint sets
{P_i: i < c} such that for each P_i and for each open interval
(a,b), the outer Lebesgue measure of 'P_i intersect (a,b)' is b-a.
Dave L. Renfro
William Elliot
> William Elliot wrote:
> > Divide the real line into two uncountable disjoint dense subsets.
> >
> > Anybody have any ideas?
>
> Let S be the union of the rationals on [0,1] and the irrationals
> elsewhere . . .
>
Oh no, it was trivial problem.
William Elliot
K = Negative irrationals and positive rationals.
L = Positive irrationals and negative rationals.
put 0 in either K or L. Ok, it works.
To make it a real problem, let the sets be nowhere countable.
A set S is nowhere countable if for all open U, S intersect U uncountable.
William Elliot
> William Elliot <ma...@xx.com>
> wrote
>
> > Divide the real line into two uncountable disjoint dense subsets.
> > Anybody have any ideas?
>
> Aww come on, let's be really greedy -->
>
Why? Am I an Enron CEO. ;-)
> Partition the reals into c = 2^(aleph_0) pairwise disjoint sets
> {P_i: i < c} such that for each P_i and for each open interval
> (a,b), the outer Lebesgue measure of 'P_i intersect (a,b)' is b-a.
>
The cofactors of R/Q ? But they're countable, hence measure 0?
Gosh, this is looking like a real problem, maybe even a hard problem.
What if I used a representative set of the cofactors? They're
uncountable. How many are there that are mutually disjoint?
As each cofactor is countable, I'd surmise there's at most countably
many mutually disjoint representative sets.
Ok, tho it's not really as greedy as a corporate monarch, isn't it as
greedy as a board member?
Oh oh, aren't my sets unmeasurable?
A set S is nowhere countable when for all open U, S /\ U is uncountable.
Do your sets have the additional property that they are nowhere
countable?
I don't know if mine do. This seems related to a the theorem that an
uncountable subset of a 2nd countable has uncountably many condensation
points. Proof of this I'm posting sci.math, 'infinite sets in compact
spaces, summary'
Virgil
William Elliot <ma...@xx.com> wrote:
Take the reals, R, as a vector space over the rationals, Q.
The "Hamel" basis of this space is uncountable, and can, in
theory, be partitioned into disjoint uncountable subsets,
say, U and V.
Let one set, S, in R be the set of all finite Q_linear
combinations of elements in U, and the other set T be R\S.
I think that both sets, S and T, are both dense and
uncountable in every open subset of R. The proofs should be
easy enough.
Zdislav V. Kovarik
On Mon, 20 Jan 2003, William Elliot wrote:
[on splitting R^1 into disjoint dense uncountable sets]
[...]
:
: To make it a real problem, let the sets be nowhere countable.
: A set S is nowhere countable if for all open U, S intersect U
:uncountable.
That's what I hinted at elsewhere. OK, so I'll serve the whole
deal and then some.
There is a set S, homeomorphic to the Cantor set, with the extra
property that its intersection with every open set in R1 is
either empty or of positive Lebesgue measure. (Just keep
removing open intervals less greedily than Cantor did.)
- (that's "uncountable and then some")
Take the set T = S+Q , the union of all rational shifts of S.
T is dense and locally of positive Lebesgue measure; now by
Baire's Category Theorem (with some extra lines of proving), its
complement is also dense and locally uncountable (of cardinality
of continuum). And I did not check the details, but I suspect
that this complement is also locally of positive Lebesgue
measure.
Cheers, ZVK(Slavek).
Dave L. Renfro
[sci.math Jan 21 2003 1:30:34:000AM]
http://mathforum.org/discuss/sci.math/m/475274/475519
wrote (in part, responding to my earlier post):
>> Partition the reals into c = 2^(aleph_0) pairwise disjoint sets
>> {P_i: i < c} such that for each P_i and for each open interval
>> (a,b), the outer Lebesgue measure of 'P_i intersect (a,b)' is b-a.
>
> A set S is nowhere countable when for all open U, S /\ U is
> uncountable. Do your sets have the additional property that
> they are nowhere countable?
<< I'm posting this in response to my earlier post because
William Elliot's post is not yet at google and I don't
feel like waiting until this afternoon (or later) for
it to show up before I can post so as to have my reply
show up as a reply to his post in google's tree chart
for this thread. (Whew, that's a mouthful!) >>
These sets will be a lot more than just nowhere countable, since
countable sets have outer measure zero. If P intersect (a,b)
were countable, then the outer measure of P intersect (a,b)
would be 0, which is quite a bit less than b-a. By the way, the
sets have to be nonmeasurable, since even just two disjoint
measurable sets in (a,b) can't both have outer measure greater
than (b-a)/2 -- let alone both having outer measure equal to b-a,
let alone countably many having outer measure equal to b-a, let
alone uncountably many having outer measure equal to b-a, let alone
c-many of them having outer measure equal to b-a. [Depending on
your model of ZFC, and yes I'm throwing all caution about using
AC to the wind, "uncountable" could be "c". But by specifying
c-many, I'm asking for an example whose cardinality is c in every
model of ZFC.]
If we only stick with measurable sets, then all but countably many
of them will have measure zero if the sets are pairwise disjoint,
regardless of any other specification we might impose on the sets.
I'll leave this as an exercise for sci.math readers. [Hint: If A
is any uncountable set of positive real numbers, then
sup{sum(F): F is a finite subset of A} = oo,
where sum(F) is the sum of the numbers belonging to F. You
can use this fact to show that all but countably many of them
will have a measure zero intersection with every interval of
the form [-n,n].]
With this in mind, here's a collection of c-many pairwise disjoint
sets with cardinality c in every open interval. In fact, each of
these sets will have a positive Hausdorff dimension intersection
with every open interval and they are not very complicated from
a descriptive set theoretic point of view (they're all F_sigma_delta
sets, I believe).
For each real number r between 0 and 1 (inclusive), let P_r be
the set of irrational numbers whose decimal expansions have r
for their limiting proportion of 5's. That is, if N(x,5,n) is the
number of appearances of the digit 5 to the right of the decimal
point in the first n digits of the decimal expansion of a real
number x, then P_r is the set of real numbers x such that the limit
as n --> oo of N(x,5,n)/n exists and is equal to r.
Then it is known that each P_r is a Borel set with positive
Hausdorff dimension in every open interval (see the sci.math
posts below), which is quite a bit more than simply saying that
each open interval has c-many numbers belonging to each of the
P_r sets.
http://mathforum.org/discuss/sci.math/m/342661/342664
http://mathforum.org/discuss/sci.math/a/t/422475
On the other hand, these sets are small in a couple of other
ways. All of them except for P_(1/10) have measure zero, and
all of them, including P_(1/10), are first category sets. The
P_r sets don't include all the real numbers. If you let Q be
the remaining real numbers, then Q is a measure zero set with
a first category complement. Hence, Q is co-meager in every
open interval, and so it too will have c-many elements in every
open interval.
Dave L. Renfro
Mike Oliver
> To make it a real problem, let the sets be nowhere countable.
> A set S is nowhere countable if for all open U, S intersect U uncountable.
My answer works for that too. I'll reproduce it:
Comeager and conull both imply dense, and there are lots of
ways to partition the reals into a comeager set and a conull set.
For example, the set of reals not normal to any base is null
but comeager.
Comeager and conull both imply nowhere countable as well (in fact
a bit more: They imply that the cardinality of the set on any
interval is 2^aleph_0). Note you don't need any choice for this.
David C. Ullrich
wrote:
>William Elliot <ma...@xx.com>
>[sci.math Jan 21 2003 1:30:34:000AM]
>http://mathforum.org/discuss/sci.math/m/475274/475519
>
>wrote (in part, responding to my earlier post):
>
>>> Partition the reals into c = 2^(aleph_0) pairwise disjoint sets
>>> {P_i: i < c} such that for each P_i and for each open interval
>>> (a,b), the outer Lebesgue measure of 'P_i intersect (a,b)' is b-a.
>>
[...]
>
>With this in mind, here's a collection of c-many pairwise disjoint
>sets with cardinality c in every open interval.
For a second I thought you were going to give us a collection of
sets as above. Imagine my disappointment... (is doing the above
_much_ harder, or can you do it by the sort of thing you do below
with some modifications?)
> In fact, each of
>these sets will have a positive Hausdorff dimension intersection
>with every open interval and they are not very complicated from
>a descriptive set theoretic point of view (they're all F_sigma_delta
>sets, I believe).
>
>For each real number r between 0 and 1 (inclusive), let P_r be
>the set of irrational numbers whose decimal expansions have r
>for their limiting proportion of 5's.
Very cute. It's clear that this does that, but not the other thing...
David C. Ullrich
Mike Oliver
> On 21 Jan 2003 05:55:31 -0800, renf...@cmich.edu (Dave L. Renfro)
> wrote:
>>>> Partition the reals into c = 2^(aleph_0) pairwise disjoint sets
>>>> {P_i: i < c} such that for each P_i and for each open interval
>>>> (a,b), the outer Lebesgue measure of 'P_i intersect (a,b)' is b-a.
>>>
>[...]
>>
>> With this in mind, here's a collection of c-many pairwise disjoint
>> sets with cardinality c in every open interval.
>
> For a second I thought you were going to give us a collection of
> sets as above. Imagine my disappointment... (is doing the above
> _much_ harder, or can you do it by the sort of thing you do below
> with some modifications?)
To do the full-outer-measure-in-every-interval thing clearly
requires using nonmeasurable sets, and (just from ZFC) we know
that no such set can be Sigma^1_2 or simpler. (Using large
cardinals we can go much further). So no modification of
the density-of-5s thing can work.
Presumably to get the harder result, you wellorder open sets
having less than full measure in an interval, in order-type 2^aleph_0,
and go through them one at a time, killing the possibility of
that open set covering any of the sets in your partition.
Mike Oliver
> To do the full-outer-measure-in-every-interval thing clearly
> requires using nonmeasurable sets, and (just from ZFC) we know
> that no such set can be Sigma^1_2 or simpler.
Sorry, that's wrong. No such set can be Sigma^1_1 or simpler.
There's a Delta^1_2 set that's not measurable in L.
A single measurable gets you Sigma^1_2. After that you need
Woodin cardinals: n Woodin cardinals with a measurable above
get you that every Sigma^1_{n+2} set is measurable.
Robert Israel
OK, now can you take U and V in such a way that this solves
Dave Renfro's version of the problem, i.e. S intersect (a,b)
and T intersect (a,b) have outer measure b-a for every open
interval (a,b)?
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Robert Israel
Zdislav V. Kovarik <kov...@mcmail.cis.mcmaster.ca> wrote:
>There is a set S, homeomorphic to the Cantor set, with the extra
>property that its intersection with every open set in R1 is
>either empty or of positive Lebesgue measure. (Just keep
>removing open intervals less greedily than Cantor did.)
>- (that's "uncountable and then some")
>Take the set T = S+Q , the union of all rational shifts of S.
>T is dense and locally of positive Lebesgue measure; now by
>Baire's Category Theorem (with some extra lines of proving), its
>complement is also dense and locally uncountable (of cardinality
>of continuum). And I did not check the details, but I suspect
>that this complement is also locally of positive Lebesgue
>measure.
Actually in this case the complement will have measure 0. Consider
this: by the Lebesgue Density Theorem, for any epsilon > 0 there is
an interval (a,b) such that m(S intersect (a,b)) > (1-epsilon)(b-a)
and b-a < epsilon.
Moreover we can take a and b to be rational; say b = a+r. Thus
m(S intersect (a,a+r)) > (1-epsilon) r, and for any integer j
m(T intersect (a+jr, a+(j+1)r)) > (1-epsilon) r.
But then m(T intersect (a+jr, a+kr)) > (1-epsilon)(kr - jr).
For any open interval (A,B) with B>A, you have
A <= a+jr < a+kr < B with B - A <= 2 epsilon + kr - jr,
so m(T intersect (A,B) > (1-epsilon)(B - A - 2 epsilon)
and taking epsilon -> 0 we have m(T intersect (A,B)) = B-A.
What you want to do is to take a union, not of translates of S
but of scaled-down copies of S, say one for each interval (a,b).
If done carefully, you can ensure that you don't cover all
the measure of any one of these intervals.
Dave L. Renfro
http://mathforum.org/discuss/sci.math/m/475274/475519
wrote (in part):
> This seems related to a the theorem that an uncountable subset
> of a 2nd countable has uncountably many condensation points.
> Proof of this I'm posting sci.math, 'infinite sets in compact
> spaces, summary'
I glanced over it -->
http://mathforum.org/discuss/sci.math/m/475237/475541
> does every condensation point of A have a dense nhood?
No. Every point of the Cantor set is a condensation point
of the Cantor set.
Also, if it'll help, what you're calling "saturation point" is
often called a "complete accumulation point".
http://www.google.com/search?q=+%22complete+accumulation+point%22&filter=0
Here are some Cantor-Bendixson decomposition results that
you might be interested in -->
1. Given any closed set in a 2'nd countable topological space
X (no separation axioms needed, I believe), we can write
F = P union N, where
(a) P is perfect
(b) N is countable.
If X is a complete metric space, then we additionally have
P = {x in F: x is a condensation point of F}.
If X is a complete separable metric space, then this
representation is unique.
2. Given any subset A of a topological space X (no separation
or other axioms needed), we can write
A = P union N, where
(a) P is dense in itself (hence, P is perfect if A is closed)
(b) P is closed in A
(c) N is scattered
(d) P and N are disjoint.
Moreover, the decomposition of A is unique with respect
to (a) through (d).
If X is a hereditarily Lindelöf space (every open cover has
a countable subcover; no separation axioms needed for this
result), then (i) each condensation point of A that belongs
to A will also belong to P, and (ii) N is countable.
NOTE: The ordinal space omega_1 + 1 is Lindelöf but not
Hereditarily Lindelöf. [omega_1 is a non-Lindelöf
subspace of omega_1 + 1.]
2'nd countable implies both hereditarily Lindelöf and
hereditarily separable. [The latter two notions are
distinct properties whose relationship is quite
nontrivial. See Stevo Todorcevic, "Partition Problems
in Topology", Contemporary Mathematics (Amer. Math.
Soc.) Volume 84, 1989.] The Sorgenfrey line (the reals
with the basis generated by sets of the form [a,b)) is
both hereditarily Lindelöf and hereditarily separable,
but not 2'nd countable.
A regular topological space X is hereditarily Lindelöf if
and only if each uncountable subset E of X contains a
condensation point of E. [Problem 3.12.7(d) on p. 285
of Engelking's "General Topology" (1977 version).]
3. Given any subset of a topological space X, we define A_j
for any ordinal j as follows:
A_0 = A
A_(j+1) = A_j intersect (A_j)'
A_j = intersection(i < j) of A_i, if j is a limit ordinal,
where (A_j)' denotes the limit points of A_j.
Let P and N be the decomposition of A given in #2 above,
and assume that the cardinality of X is the cardinal aleph_b
(i.e. aleph_b is the b'th cardinal). Then
(1) P = A_(omega_(b+1))
(2) N = union(i < omega_(b+1)) of the set differences
A_i - A_(i+1).
Moreover, for each fixed set A an ordinal smaller than
omega_(b+1) will suffice, but in general no smaller ordinal
will work uniformly for all sets A.
If X is a T_1 space, then we can replace omega_(b+1) with
omega_(d+1), where aleph_d = min{card(X), w(X)} and w(X) is
the weight of X (the smallest cardinal of a basis for X).
[In general, neither card(X) nor w(X) dominates the other,
but w(X) never exceeds 2^card(X) and in T_0 spaces card(X)
never exceeds 2^w(X).]
4. The following posts of mine discuss matters that
pertain to the Cantor-Bendixson decomposition -->
http://mathforum.org/discuss/sci.math/m/270118/270130
http://mathforum.org/epigone/math-teach/fraxvimpcrim
Dave L. Renfro (who at one time was very interested in topology,
but later found real analysis pathology to be
more interesting than topological classification)
Dave L. Renfro
I happened to notice that I defined "Lindelöf"
when I meant to define "hereditarily Lindelöf". >>
William Elliot <ma...@xx.com>
[sci.math Jan 21 2003 1:30:34:000AM]
http://mathforum.org/discuss/sci.math/m/475274/475519
wrote (in part):
> This seems related to a the theorem that an uncountable subset
> of a 2nd countable has uncountably many condensation points.
> Proof of this I'm posting sci.math, 'infinite sets in compact
> spaces, summary'
I glanced over it -->
http://www.google.com/search?q=+%22complete+accumulation+point%22&filter=0
If X is a hereditarily Lindelöf space (every relatively
open cover of every subset of X has a countable subcover;
dull...@sprynet.com
wrote:
>Mike Oliver wrote:
>
>> To do the full-outer-measure-in-every-interval thing clearly
>> requires using nonmeasurable sets, and (just from ZFC) we know
>> that no such set can be Sigma^1_2 or simpler.
>
>Sorry, that's wrong. No such set can be Sigma^1_1 or simpler.
>There's a Delta^1_2 set that's not measurable in L.
Details... regardless of the details, it's clear I hadn't thought
about the question carefully before asking. We "know" that
we can't show a non-measurable set exists without AC,
(to be more accurate, we know we've been told that),
so it's clear the construction has to be somewhat
more sophisticated than the density-of-5's thing above.
>A single measurable gets you Sigma^1_2. After that you need
>Woodin cardinals: n Woodin cardinals with a measurable above
>get you that every Sigma^1_{n+2} set is measurable.
David C. Ullrich
William Elliot
I learned what I can from your posts yet some are hard to keep up
with because when it comes to modern analysis, I don't measure up.
Well I do suppose countable sets have measure zero.
Pandora's Paradox: math is infinite, mind finite.
David Libert
[...]
> Presumably to get the harder result, you wellorder open sets
> having less than full measure in an interval, in order-type 2^aleph_0,
> and go through them one at a time, killing the possibility of
> that open set covering any of the sets in your partition.
Yes, this works. Suppose we are at a stage handling interval
(a,b), and handling O an open subset of (a,b) of measure
< b-a, and needing to arrange that O is not a superset of one
specific constructed partition element.
We seek to do this by throwing a previously uncommitted point in
(a,b) - O into the constructed partition element, thereby ruining
O as a superset of that constructed set.
We need to see that we can find such a point. (a,b) - O is a
closed set of positive measure, so it must be uncountable (if
countable then measure 0, contracting O not full measure in (a,b)),
so since it is closed an uncountable it must contain a perfect set,
and hence be not merely uncountable but actually cardinality
continuum.
There are c (= continuum) many partitition elements to handle,
and for each of these c many (a,b) intervals, and for each of
these c many open subsets O of non-full measure, so the cases are
naturally indexed by c^3, which we re-order as a well-ordering of c.
So at a stage needing an unused point in the c sized (a,b) - O,
there were < c previous stages, each only using one point, so we
still have room to find a new point to commit to the current partition
element with no conflict with previous decisions.
--
David Libert ah...@FreeNet.Carleton.CA
Fred Galvin
To put it more concisely: Let (P_i, t_i), i < c, be a listing of all
pairs (P, t) where P is a perfect set and t is a real number. By
transfinite induction, choose distinct points x_i in P_i. For each
real t, let X_t = {x_i: t_i = t}. Then the sets X_t are pairwise
disjoint, and each X_t meets every perfect set, and every uncountable
analytic set. I think this sort of construction is associated with the
name Bernstein.
--
It takes steel balls to play pinball.
Dave L. Renfro
>>> Partition the reals into c = 2^(aleph_0) pairwise disjoint sets
>>> {P_i: i < c} such that for each P_i and for each open interval
>>> (a,b), the outer Lebesgue measure of 'P_i intersect (a,b)' is b-a.
Dave L. Renfro then followed with ...
>> If we only stick with measurable sets, then all but countably many
>> of them will have measure zero if the sets are pairwise disjoint,
[snip]
>> With this in mind, here's a collection of c-many pairwise disjoint
>> sets with cardinality c in every open interval.
David C. Ullrich then wrote ...
> For a second I thought you were going to give us a collection of
> sets as above. Imagine my disappointment...
No one seems to have eased your disappointment yet, so I'll post
a detailed solution to this tomorrow. Those with really good
memories might recall a post that I made on Oct. 24, 2000 in a
sci.math thread titled "What is the point of rigor?", in which I
listed a lot of things that (as I put it) "... I doubt would have
been discovered without the application of a fair amount of rigor".
Here's the URL for those interested -->
http://mathforum.org/discuss/sci.math/m/299086/299105
One of the assertions in this list was that
(rationals)^2 union (irrationals)^2
is a pathwise connected subset of R^2. I gave a detailed proof of
this result on May 15, 2002 in the sci.math thread "Topological
Sandpaper" -->
http://mathforum.org/discuss/sci.math/m/411108/411120
Another assertion in this list was that there exists a collection
c-many pairwise disjoint subsets of [0,1] all of which have outer
Lebesgue measure one. Since I posted a proof of the other result,
I suppose it's time for me to post a proof of this one as well!
But, unlike the other result, this one is by no means original with
me. I know of three marginally different ways to do it, which I'll
discuss and provide references for when I write it up.
I guess now is a good time to mention a sort of intermediate
result. Earlier in this thread I gave an example of c-many pairwise
disjoint subsets of the reals such that each has cardinality c in
every open interval (in fact, they all had positive Hausdorff
dimension in every open interval), and I mentioned that you have
to resort to non-measurable sets if you want more than countably
many of them to have nonzero outer measure. Tomorrow I'll take care
of the c-many case, which will involve sets that are non-measurable
to a fairly high degree. However, if we back off on the cardinality
requirement and just ask for countably many pairwise disjoint sets,
then we can do a lot better than positive Hausdorff dimension in
every open interval.
There exists a countable collection of pairwise disjoint measurable
subsets of the reals each of whose intersection with every open
interval has positive measure and is non-meager. [Note that I said
"non-meager", and not "co-meager".] This is proved in
Frank S. Cater, "A partition of the unit interval", Amer. Math.
Monthly 91 (1984), 564-566.
Technically speaking, Cater only does this for [0,1]. Take appropriate
unions of the analogous sets in every interval [n, n+1] to get the
collection I claimed exists.
The sets are measurable, but not much more. Cater gives a short
argument at the end of his paper showing that no such collection
is possible using Borel sets. However, if we toss out the locally
non-meager requirement, I believe we can satisfy the locally positive
measure requirement by using F_sigma sets.
Dave L. Renfro