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Science Forum Index » Mathematics Forum » condensation points in R
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| Amanda |
 Posted: Fri Dec 19, 2003 11:38 am |
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Some days ago I posted a message about uncountable sets in R. I am
still confused about some points and would like some ideas.
Let S be an uncountable set in R. I want to prove:
(1) If P is the set of bilateral condensation points of S, then P
intersection S is uncountable
(2) The set of condensation points of S that don't belong to S is
countable.
To prove (1), I tried to prove that the set of unilateral
condensation points of S is countable. Since it's known the set of all
condensation points of S is uncountable, this gives the conclusion.
But I could`t get through yet.
PS.: x is a bilateral condensation point of S if the points of S
condensate to the left and to the right of x, that is, for every
eps>0, the intervals (x-eps, x) and (x, x+eps) contain uncountably
many elements of S.
Amanda |
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| The World Wide Wade |
| Posted: Fri Dec 19, 2003 3:18 pm |
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In article <6f75d9cf.0312190838.414e8579@posting.google.com>,
sca18@hotmail.com (Amanda) wrote:
Quote: Let S be an uncountable set in R. I want to prove:
(1) If P is the set of bilateral condensation points of S, then P
intersection S is uncountable
(2) The set of condensation points of S that don't belong to S is
countable.
To prove (1), I tried to prove that the set of unilateral
condensation points of S is countable. Since it's known the set of all
condensation points of S is uncountable, this gives the conclusion.
But I could`t get through yet.
PS.: x is a bilateral condensation point of S if the points of S
condensate to the left and to the right of x, that is, for every
eps>0, the intervals (x-eps, x) and (x, x+eps) contain uncountably
many elements of S.
For (1), let V be the union of all open intervals whose intersection with S
is countable. Then V intersect S is countable. Write V as the pairwise
disjoint union of open intervals (an,bn). If you put V* = U [an,bn], then
V* intersect S is still countable and it looks to me like every point of S
\ V* is a bilateral condensation point of S. |
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| Amanda |
| Posted: Fri Dec 19, 2003 11:01 pm |
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The World Wide Wade <waderameyxiii@comcast.remove13.net> wrote in message news:<waderameyxiii-C49B13.12184519122003@news.supernews.com>...
Quote: In article <6f75d9cf.0312190838.414e8579@posting.google.com>,
sca18@hotmail.com (Amanda) wrote:
Let S be an uncountable set in R. I want to prove:
(1) If P is the set of bilateral condensation points of S, then P
intersection S is uncountable
(2) The set of condensation points of S that don't belong to S is
countable.
To prove (1), I tried to prove that the set of unilateral
condensation points of S is countable. Since it's known the set of all
condensation points of S is uncountable, this gives the conclusion.
But I could`t get through yet.
PS.: x is a bilateral condensation point of S if the points of S
condensate to the left and to the right of x, that is, for every
eps>0, the intervals (x-eps, x) and (x, x+eps) contain uncountably
many elements of S.
For (1), let V be the union of all open intervals whose intersection with S
is countable. Then V intersect S is countable.
Here, V must be the union of a countable collection of open intervals
that form a _countable_ topological basis for R, right? For example,
the collection of all open intervals centered at rational elements and
with rational radius. This is essential to ensure V intersect S is
given by a countable union of countable sets, so that V is countable.
Right?
Write V as the pairwise
Quote: disjoint union of open intervals (an,bn). If you put V* = U [an,bn], then
V* intersect S is still countable and it looks to me like every point of S
\ V* is a bilateral condensation point of S.
It's clear U (a_n, b_n) is the set of all real numbers that are not
condensation points of S. Therefore, all of the a_n's and the b_n's
are condensation points. But for every eps satisfying, 0< eps <b_n -
a_n, the open interval (a_n, a_n+eps) contains, by construction, only
a countable number of elements of S. This shows the elements of S
don't condensate to the right of a_n and, therefore, a_n is
unilateral. Similarly, all of the bn's are unilateral and the elements
of S condensate to the right. We conclude U [an,bn] contains at least
some of the unilateral condensation points of S and contains all the
elements of R that are not condensation points of any kind. all.
To complete the proof, it remains to show - if it's true - that,
except for the a_n's and b_n's, S doesn't have any other unilateral
condensation point. And that's where I got confused.....
Anyway, it looks to me you're conclusion is true. Supposing V
originates from a countable collections of open intervals that forms a
basis for R.
Thank you.
Amanda |
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| William Elliot |
| Posted: Sat Dec 20, 2003 5:46 am |
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Guest
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From: Amanda <sca18@hotmail.com>
Subject: condensation points in R
Quote: Let S be an uncountable set in R. I want to prove:
(1) If P is the set of bilateral condensation points of S,
then P intersection S is uncountable
(2) The set of condensation points of S that
don't belong to S is countable.
Remove from S all maximal closed intervals
[a,b], a < b for which [a,b] /\ S is countable
/\ intersect
[a,b] is maximal when for all (c,d) containing [a,b]
(c,d) is uncountable.
There are at most countable many of these pairwise disjoint closed
intervals. Thus removing all of them from S, leaves S unchanged
cardinally.
The points in C, the remainder of S, are bi-sided condensending.
If x in C wasn't, then for some a, either [x,a] or [a,x] would be
countable which contradicts the construction of C.
For every so constructed maximal [a,b],
the endpts a,b, are mono-sided condensending
and the interior points at best, are mundane accumulators.
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| Amanda |
| Posted: Sat Dec 20, 2003 10:41 am |
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Guest
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The World Wide Wade <waderameyxiii@comcast.remove13.net> wrote in message news:<waderameyxiii-C49B13.12184519122003@news.supernews.com>...
Quote: In article <6f75d9cf.0312190838.414e8579@posting.google.com>,
sca18@hotmail.com (Amanda) wrote:
Let S be an uncountable set in R. I want to prove:
(1) If P is the set of bilateral condensation points of S, then P
intersection S is uncountable
(2) The set of condensation points of S that don't belong to S is
countable.
To prove (1), I tried to prove that the set of unilateral
condensation points of S is countable. Since it's known the set of all
condensation points of S is uncountable, this gives the conclusion.
But I could`t get through yet.
PS.: x is a bilateral condensation point of S if the points of S
condensate to the left and to the right of x, that is, for every
eps>0, the intervals (x-eps, x) and (x, x+eps) contain uncountably
many elements of S.
For (1), let V be the union of all open intervals whose intersection with S
is countable. Then V intersect S is countable. Write V as the pairwise
disjoint union of open intervals (an,bn). If you put V* = U [an,bn], then
V* intersect S is still countable and it looks to me like every point of S
\ V* is a bilateral condensation point of S.
Yes, I think you're right. The numbers a_n a and b_n are unilateral
condensation points of S. (the elements of S condensate to the left of
the a_n's and the the right of the b_n's. If P is the set of all
condensation points of S, then S is a union of pairwise disjoint
closed intervals. If x is in P, then there is a natural n such that x
is in [b_n, a_n+1], which is an interval whose elements are
condensation popints of S. (b_n may be -inf and a_n+1 may be inf.)
Suppose b_n < x < a_n+1. If the elements of S don't condensate to the
right of x, then there's eps>0 such that (x, x+eps) is in (x, a_n+1)
and intersects S in only countably many elements. Therefore, no
element of (x, x+eps) is a condensation point of S, which is a
contadiction. Similarly, we see the elements of S also condensate to
the left of x, and so we concluse x is a bilateral condensation point.
With this we have proved that:
(a) The set B of all bilateral condensation points of S is open and,
so,uncountable.
(b) the set U of the unilateral condensation points of s is {a1, b1,
a2,b2.....}, therefore countable.
(c) Since the set P of all condensation points of S is closed, it
follows U is closed, because U = P / B
(d)The set U intersect S is clearly countable.
(e)We have P intersect S = (B intersect S) Union (U intersect S).
Since P intersect S is uncountable, it follows from (d) that B
intersect S is uncountable.
Now it looks to me that B intersect S is open and U intersect S is
closed, but I'm not sure.
Amanda |
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