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Set of bilateral condensation points of uncountable subsets of R

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Artur

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Mar 27, 2007, 10:10:01 AM3/27/07

Good morning. I'd like opinions about my proof for the following
statement:

Let S be an uncountable subset of R. Then, the set of bilateral
condensation points of S is uncountable

We say x is a bilateral condensation point of S, if, for every eps>0,
the intervals (x -eps, x) and (x, x+ eps) contain uncountably many
elements of S. If this is true for only one of such intervals, then x
is a unilateral condensation point. According to these definitions,
bilateral condensation points are not unilateral ones.

Let C be the set of all condensation points of S, B be the set of the
bilateral ones and U be the set of the unilateral ones. Then C = B
Union U ,with B and U disjoint. According to the properties of
separable metric spaces (R is one), C is closed and uncountable. So,
it's complement C' is open and, therefore, is given by the union of a
countable collection (a_n , b_n) of pairwise disjoint intervals (we
can have a_n = - oo and/or b_m = oo for some n and m) .

For each n, (a_n, b_n) contains no condensation point of S and,
therefore, (a_n, b_n) Inter S is countable (this is a known fact about
separable metric spaces). Since b_n is a condensation point of S, then
the points of S must condensate to the right of b_n, so that it's a
unilateral condensation point. Similarly, every a_n is a left hand
unilateral condensation point. Therefore the countable set {a_n, b_n |
n=1,2,3...} is a countable subset of U.

Now, suppose x is in C and is not end point of any of the intervals
(a_n , b_n). If, for some eps >0, (x, x + eps) contains no
condensation point of S, then (x , x + eps) is in C' and, therefore,
in one of it's component intervals (a_n , b_n). So, either x is in
(a_n , b_n) - so in C' - or x = a_n. In both cases this contradicts
the choice of x, so that, for every eps >0, (x, x + eps) contains
condensation points of S. Therefore, (x, x+ eps) Inter S is
uncountable for every eps >0. By a similar reasoning, the same is true
of (x - eps , x), and we conclude x is a bilateral condensation point
of S. In addition, this implies {a_n, b_n | n=1,2,3...} is all of U and
U is countable.

Since C = B Union U, C is uncountable and U is countable, it follows B
is uncountable, as desired.

Another conclusion is that the set of bilateral condensation points of
S that are in S is uncountable. This follows from the equality C Inter
S = (B inter S) Union (U inter S). We know C inter S is uncountable
(property of separable metric spaces) and, since U inter S is subset
of the countable U, it's countable. In order for this set equality to
be true, B inter S must be uncountable.

Thank you for any comments. Probably, there's a neater and nicer
proof.

Artur

Dave L. Renfro

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Mar 27, 2007, 3:15:51 PM3/27/07

Artur wrote (in part):

[reposted, because it's been 15 minutes and my post
still hasn't shown up at google]

> Let S be an uncountable subset of R. Then, the set
> of bilateral condensation points of S is uncountable

As I often do, instead of answering your question,
I'll use your post to dive into something of interest
to me that is probably not very well known.

A more precisely stated result, whose proof is essentially
the same, is the following.

THEOREM (YOUNG): Given any subset E of the reals, we can
write E = C union Z, where Z is countable and every point
of C is a bilateral condensation point of C (not just
a bilateral condensation point of E).

The earliest appearance of this result that I know of
is p. 76 of the following paper.

William H. Young, "On the distinction of right and left
at points of discontinuity", Quarterly Journal of Pure
and Applied Mathematics 39 (1908), 67-83. [JFM 38.0420.02]

More generally, let s-I be a sigma-ideal in R containing
all singleton sets (equivalently, containing all countable
subsets). For example, s-I could be the collection of all
Lebesgue measure zero sets or the collection of all first
category sets. [A sigma-ideal is a collection of sets closed
under countable unions and which contains every subset of
each of its elements.]

Nothing significantly different is needed to prove the
following more general result:

THEOREM (HANSON): Given any subset E of the reals, we can
write E = C union Z, where Z belongs to s-I and C is bilaterally
non-(s-I)-dense at each point of C.

"F is bilaterally non-(s-I)-dense at a point x" (my phrase)
means that each unilateral neighborhood of x intersects F
in a set that does not belong to s-I. Note that if s-I is
the collection of countable subsets of R, then we get the
bilateral Cantor-Bendixson theorem due to Young.

This more general version has been rediscovered and published
many times (e.g. Geza Freud's MR 21 #7494 paper)
http://www.springerlink.com/content/w57m4348805u62r6/
The earliest appearance I've uncovered is p. 10
(of Part II, which was never published) of Hanson's
Ph.D. Dissertation under Henry Blumberg:

Eugene H. Hanson, "Various Properties of the \tau - Limit",
Part II of Ph.D. Dissertation under Henry Blumberg, Ohio
State University, 1935, 46 pages.

Note that the previous theorem has the following compromises
built into it. If we desire Z to be smaller, then we must
contend with a weaker density requirement on C. If we
desire C to satisfy a stronger density requirement,
then we must contend with a larger exceptional set Z.

There is a higher dimensional version of the bilateral
Cantor-Bendixson theorem that does not seem to be very
well known. As far as I can determine, it was first
formulated and proved by A. J. Ward in 1933 (J. London
Math. Soc. 8, pp. 109-112). Jack G. Ceder (Ann. Scuola Norm
Sup. Pisa (3) 24, 1970, pp. 53-63), unaware of Ward's paper,
rediscovered this result for R^2 and R^3. Ceder's proofs
were quite a bit more complicated (Ceder's proofs involved
an application of Zorn's Lemma, of all things) and less
general than Ward's proof, which holds for any R^n.

THEOEREM (WARD): Given any subset E of R^n, we can write
write E = C union Z, where Z is countable and every point
of C has a bilateral line of condensation relative to C.

To say that a point x has a bilateral line of condensation
relative to C means there exists a line L through x such
that the interior of every cone with vertex x and axis L
has an uncountable intersection with the set C.

The sigma-ideal version of this theorem also holds
(straightforward modifications of Ward's proof will
take care of it), but I'm not aware that it's ever
been published.

The conclusion of Ward's theorem for R^n, with n > 1, cannot
be strengthened to "C has more than one bilateral condensation
direction at each of its points". A counterexample is trivial:
let E be a line. More generally, any differentiable curve
presents the same difficulty, since the existence of a tangent
line implies a unique condensation direction at each point.
However, differentiable curves are rather small sets
in many situations. For example, they are simultaneously
first category and measure zero. The following question
arises. If we allow ourselves a bit more leeway with the
size of Z, can we establish the existence of more than
one bilateral condensation direction at each point of C?

It turns out that by allowing Z to belong to a certain
sigma-ideal, one that is larger than the collection of
countable sets, but smaller than the collection of sets
simultaneously first category and measure zero (even smaller
that the collection of lower porous sets), then we can
arrange to have, at each point x of C, _every_direction_
be a bilateral condensation direction at x relative to C.

The sigma-ideal that works is the collection of "sparse sets"
(terminology of Blumberg and Ludek Zajicek). [The use of "sparse"
for these sets is due to Blumberg, while the concept is due to
Young. Young used the term "ridé" in Bulletin Sci. Math. (2)
52, 1928, pp. 265-280, but the notion itself seems to date
back to Young/Young (Proc. London Math. Soc. (2) 16, 1917
pp. 337-351).]

These are sets that are countable unions of what one might
call "cone free sets" -- sets which have the property that
at each of their points there exists a cone with vertex
at that point whose interior is disjoint from the set.
Cone free sets are obviously nowhere dense. The Lebesgue
density theorem implies they have measure zero. Indeed,
it's obvious, if you know the definition, that each cone
free set is lower porous.

There is a close connection between sparse sets and Lipschitz
functions. Specifically, a subset E of R^2 is sparse if and
only if E can be covered by countably many sets, each of which
is congruent to the graph of a Lipschitz function from R to R.
[Obviously, we can't have the sparse sets being equal to
the collection of countable unions of sets congruent to
Lipschitz graphs, since the latter are all F_sigma sets
while the former can be non-Borel or worse (any subset
of a line in R^2 is cone free, and hence sparse, for
instance).]

If anyone's interested, I gave a talk on these issues
(with proofs) at an April 2001 MAA Section meeting in
Des Moines (Iowa), titled "What do Lipschitz functions
have to do with the Cantor-Bendixson theorem?", but I
haven't gotten around to writing it up for publication.
However, the LaTex slides I used are reasonably complete,
and I have another much longer manuscript-in-progress
(titled "Cantor-Bendixson Analogs") that includes a number
of things I haven't touched on here. Maybe one day I'll
see about sticking these into the arXiv.org preprint
archive, if nothing else.

Dave L. Renfro

Arturo Magidin

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Mar 27, 2007, 3:26:36 PM3/27/07

In article <1175004601.5...@n76g2000hsh.googlegroups.com>,

Artur <artur_...@yahoo.com> wrote:
>Good morning. I'd like opinions about my proof for the following
>statement:
>
>Let S be an uncountable subset of R. Then, the set of bilateral
>condensation points of S is uncountable
>
>We say x is a bilateral condensation point of S, if, for every eps>0,
>the intervals (x -eps, x) and (x, x+ eps) contain uncountably many
>elements of S. If this is true for only one of such intervals, then x
>is a unilateral condensation point. According to these definitions,
>bilateral condensation points are not unilateral ones.
>
>Let C be the set of all condensation points of S, B be the set of the
>bilateral ones and U be the set of the unilateral ones. Then C = B
>Union U ,with B and U disjoint. According to the properties of
>separable metric spaces (R is one), C is closed and uncountable.

I confess that this strikes me as a very strong assumption to
make. You are saying that you already know that either there are
uncountably many bilateral condensation points, or uncountably many
unilateral condensation points, or both. Are you really allowed to
assume this from the outset?

The rest of your argument looks good to me, except that you use "it's"
where it should be "its":

>Now, suppose x is in C and is not end point of any of the intervals
>(a_n , b_n). If, for some eps >0, (x, x + eps) contains no
>condensation point of S, then (x , x + eps) is in C' and, therefore,
>in one of it's component intervals (a_n , b_n).

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

Artur

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Mar 27, 2007, 5:11:35 PM3/27/07

On 27 mar, 16:26, magi...@math.berkeley.edu (Arturo Magidin) wrote:
> In article <1175004601.595177.294...@n76g2000hsh.googlegroups.com>,

>
>
>
>
>
> Artur <artur_stei...@yahoo.com> wrote:
> >Good morning. I'd like opinions about my proof for the following
> >statement:
>
> >Let S be an uncountable subset of R. Then, the set of bilateral
> >condensation points of S is uncountable
>
> >We say x is a bilateral condensation point of S, if, for every eps>0,
> >the intervals (x -eps, x) and (x, x+ eps) contain uncountably many
> >elements of S. If this is true for only one of such intervals, then x
> >is a unilateral condensation point. According to these definitions,
> >bilateral condensation points are not unilateral ones.
>
> >Let C be the set of all condensation points of S, B be the set of the
> >bilateral ones and U be the set of the unilateral ones. Then C = B
> >Union U ,with B and U disjoint. According to the properties of
> >separable metric spaces (R is one), C is closed and uncountable.
>
> I confess that this strikes me as a very strong assumption to
> make. You are saying that you already know that either there are
> uncountably many bilateral condensation points, or uncountably many
> unilateral condensation points, or both. Are you really allowed to
> assume this from the outset?
>
> The rest of your argument looks good to me, except that you use "it's"
> where it should be "its":

Ok, sorry for the wrong it's, I'll pay more attention.

Actually, My proof is based on the following statement:

If S is an uncountable subset of a separable metric space X, then the
set C, formed by the condensation points of S, is closed and
uncountable.

My proof:

Since X is separable metric space, it has a countable topological
basis {B_n}. This implies C', the complement of C, equals the set W =
(Union B_n such that B_n Inter S is countable). Actually, if x is in
C', then x is not a condensation point of S and, therefore, x has a
basic neighborhood B_n such that B_n Inter S is countable. This
implies x is W and C' is subset of W. If, OTH, y is in W, then y
belongs to some basic B_n that intersects S according to only
countably many elements. Therefore, y is not a condensation point of
S, so that y is C' and W is subset of C'. We conclude C' = W. And
since W is open (union of open sets), C' os open and C is closed
(actually, C is closed in any topological space).

Now, we see that C' Inter S = C' inter W = Union (B_n Inter S), where
n runs over the basic neighborhoods such that B_n Inter S is
countable. This implies C' Inter S is countable, for it is a countable
union of countable sets.

We have S = (S Inter C) Union (S Inter C'). Since S, by assumption, is
uncountable and S Inter C' is countable, it follows S inter C is
uncountable, that is: the set of the condensation points of S that are
in S is uncountable. It follows immediately that C is uncountable.

And there's another interesting conclusion: If x is in C then, for
every neighborhood V of x, it's true that S Inter V = (S inter V Inter
C) Union (S Inter V Inter C'). Since S Inter V is uncountable and S
Inter V Inter C' is countable (subset of the countable S Inter C'), it
follows S inter V Inter C is uncountable. This shows every
condensation point of S is condensation point of S Inter C.

Actually, at the very beginning of my proof for bilateral condensation
points of uncountable subsets of R, I already knew that the set of all
condensation points was closed and uncountable

Do you agree with these points?

Thank you very much.
Artur

Arturo Magidin

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Mar 28, 2007, 9:39:17 AM3/28/07

In article <1175029895....@o5g2000hsb.googlegroups.com>,

Artur <artur_...@yahoo.com> wrote:
>On 27 mar, 16:26, magi...@math.berkeley.edu (Arturo Magidin) wrote:
>> In article <1175004601.595177.294...@n76g2000hsh.googlegroups.com>,
>>
>>
>>
>>
>>
>> Artur <artur_stei...@yahoo.com> wrote:
>> >Good morning. I'd like opinions about my proof for the following
>> >statement:
>>
>> >Let S be an uncountable subset of R. Then, the set of bilateral
>> >condensation points of S is uncountable
>>
>> >We say x is a bilateral condensation point of S, if, for every eps>0,
>> >the intervals (x -eps, x) and (x, x+ eps) contain uncountably many
>> >elements of S. If this is true for only one of such intervals, then x
>> >is a unilateral condensation point. According to these definitions,
>> >bilateral condensation points are not unilateral ones.
>>
>> >Let C be the set of all condensation points of S, B be the set of the
>> >bilateral ones and U be the set of the unilateral ones. Then C = B
>> >Union U ,with B and U disjoint. According to the properties of
>> >separable metric spaces (R is one), C is closed and uncountable.
>>
>> I confess that this strikes me as a very strong assumption to
>> make. You are saying that you already know that either there are
>> uncountably many bilateral condensation points, or uncountably many
>> unilateral condensation points, or both. Are you really allowed to
>> assume this from the outset?

[...]

>Actually, My proof is based on the following statement:
>
>If S is an uncountable subset of a separable metric space X, then the
>set C, formed by the condensation points of S, is closed and
>uncountable.
>
>My proof:

[.snip what looks like a correct proof.]

I did not mean to imply it was false, but it did seem like something a
little too big (in the context of the statement you were proving to
begin with) to simply toss in as an assumption. If you had already
proven it, of course, then it is fine to assume it.

Artur

unread,

Mar 28, 2007, 10:08:38 AM3/28/07

On 28 mar, 10:39, magi...@math.berkeley.edu (Arturo Magidin) wrote:
> In article <1175029895.082396.22...@o5g2000hsb.googlegroups.com>,

>entre aspas -

OK, thank you. Maybe there's a simpler proof. I came across this
problem involving condensation points when I was trying to prove a
statement posted in another thread about a month ago: If S isn an
uncountable subset of R, then S has a subset T (with at least 2
elements, to avoid trivialities) with the in-between property: for
every distict elements x and y in T, there's a z in T between x and y.

If B is the set of bilateral condensation points of S, then T = B
Inter S satisfies the desired condition, because every element of T is
bilateral condensation point of T.

In that thread I posted this proof , but I made a mistake. I concluded
B needed to be open, but actually it does'nt.

Best regards

Artur

Dave L. Renfro

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Mar 28, 2007, 5:33:08 PM3/28/07

Dave L. Renfro wrote (in part):

http://groups.google.com/group/sci.math/msg/adad4f80670bd5d0

> It turns out that by allowing Z to belong to a certain
> sigma-ideal, one that is larger than the collection of
> countable sets, but smaller than the collection of sets
> simultaneously first category and measure zero (even smaller
> that the collection of lower porous sets), then we can
> arrange to have, at each point x of C, _every_direction_
> be a bilateral condensation direction at x relative to C.

When I got home yesterday I looked over my notes on
this topic and saw that I said more than I had actually
proved. The result I have, which I might post a proof of
tomorrow, is that at each point x of C, every direction
is a bilateral condensation direction at x relative to E.
I suspect we can replace "E" with "C" at the end of the
previous sentence, but I wasn't able to prove or disprove
this back in 1997 (when I first wrote this stuff up) or
in April 2001 (when I gave a talk on it), and I haven't
worked on it since then.

The other statements in my yesterday's post seem fine.
In particular, in all the other versions of the
Cantor-Bendixson theorem I gave, the density conditions
are relative to C (and not just to E).

Dave L. Renfro

b vcb

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Apr 2, 2007, 3:51:52 AM4/2/07

vcv

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Apr 2, 2007, 11:51:38 PM4/2/07

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