I couldn't prove the following propostion, any help is welcome:
If S is an uncountable subset of R, then S contains a subset P such
that, if x and y are distintinct elements of P, then there's a z in P
between x and y.

Hi,
I couldn't prove the following propostion, any help is welcome:
If S is an uncountable subset of R, then S contains a subset P such
that, if x and y are distintinct elements of P, then there's a z in P
between x and y.
I tried to prove the contrapositive of this proposition. If S has no
subset with the given property, then, for every x in S there's a y in
S such that (x,y) doesn't intersect S. This means that every x in S is
squeezed between 2 open intervals in the complement of S. From this I
concluded S had no accumulation points. Since R is separable, every
set that doesn't have accumulation points is countable, which implies
S is countable. But this conclusion is false. But then I noticed I
hadn`t actually proved the contrapositive of the given proposition,
ths thing is more complicated. {0, 1, 1/2,...1/n...} is an example of
a set that doesn`t satisfy the prpoasition and, yet, has olne
accumulation point.
Thank you
Amanda

From: Robert Israel <israel@math.ubc.ca
Subject: Re: A property of uncountable sets in R

Amanda <sca18@hotmail.com> wrote:
I couldn't prove the following propostion, any help is welcome:
If S is an uncountable subset of R, then S contains a subset P such
that, if x and y are distintinct elements of P, then there's a z in
P between x and y.

Question: if S is the Cantor set, is such a P possible?

Hi,
I couldn't prove the following propostion, any help is welcome:
If S is an uncountable subset of R, then S contains a subset P such
that, if x and y are distintinct elements of P, then there's a z in P
between x and y.

I tried to prove the contrapositive of this proposition. If S has no
subset with the given property, then, for every x in S there's a y in
S such that (x,y) doesn't intersect S. This means that every x in S is
squeezed between 2 open intervals in the complement of S. From this I
concluded S had no accumulation points. Since R is separable, every
set that doesn't have accumulation points is countable, which implies
S is countable. But this conclusion is false. But then I noticed I
hadn`t actually proved the contrapositive of the given proposition,
ths thing is more complicated. {0, 1, 1/2,...1/n...} is an example of
a set that doesn`t satisfy the prpoasition and, yet, has olne
accumulation point.

Thank you
Amanda

Hi,
I couldn't prove the following propostion, any help is welcome:
If S is an uncountable subset of R, then S contains a subset P such
that, if x and y are distintinct elements of P, then there's a z in P
between x and y.

Amanda <sca18@hotmail.com> wrote:
I couldn't prove the following propostion, any help is welcome:
If S is an uncountable subset of R, then S contains a subset P such
that, if x and y are distintinct elements of P, then there's a z in
P between x and y.

Hint: remove every point which is in a closed interval of positive
length whose intersection with S is at most countable.
Doubtful. What if S = [0,1] \/ [2,3] \/ Q ?

In article <6f75d9cf.0312161619.439e7ff8@posting.google.com>,
Amanda <sca18@hotmail.com> wrote:

I couldn't prove the following propostion, any help is welcome:
If S is an uncountable subset of R, then S contains a subset P such
that, if x and y are distintinct elements of P, then there's a z in P
between x and y.

Hint: remove every point which is in a closed interval of positive length
whose intersection with S is at most countable.

israel@math.ubc.ca (Robert Israel) wrote in message news:<bro92p$89s$1@nntp.itservices.ubc.ca>...
In article <6f75d9cf.0312161619.439e7ff8@posting.google.com>,
Amanda <sca18@hotmail.com> wrote:

I couldn't prove the following propostion, any help is welcome:
If S is an uncountable subset of R, then S contains a subset P such
that, if x and y are distintinct elements of P, then there's a z in P
between x and y.

Hint: remove every point which is in a closed interval of positive length
whose intersection with S is at most countable.

But this doesn't necessarily lead to a set with the desired property.
For example, if S =[0,1] U [2,3], then those points form an empty set

In article <6f75d9cf.0312161619.439e7ff8@posting.google.com>,
Amanda <sca18@hotmail.com> wrote:
Hi,
I couldn't prove the following propostion, any help is welcome:
If S is an uncountable subset of R, then S contains a subset P such
that, if x and y are distintinct elements of P, then there's a z in P
between x and y.

First prove:

LEMMA. If S is an uncountable subset of R, then there exists an
element x in S such that there are uncountably many elements of S
smaller than x, and uncountably many elements of S larger than x.

Then you can use the Axiom of Choice to pick x_1 with that property;
then pick x_{21} < x_1 < x_{22}, x_{21} with the property that there
are uncountably many points of S smaller than x_{21}, and uncountably
many between x_{21} and x_1; etc. Keep making a choice of point in
between any two, and then take the union. That will be P.

To prove the lemma, consider two sets:

L = {x in S:there are only countably many elements of S smaller than x}
B = {x in S:there are only countably many elements of S larger than x}

Then L is downward closed: if x in S, and y in S satisfies y<x, then y
in L; B is upward closed: if x in B and y in S satisfies x<y, then y
in B.

Show that L is bounded above, and B is bounded below (hint: if L is
not bounded above, then every (-infty,n] intersects s for sufficiently
large integer n; this implies S is countable [prove it]).
If L is not bounded above, then, for every natural n, there is z in L

Now prove that each of L and B are at most countable. E.g. if L is
empty, you are done. If it is not empty, let x_0 be its supremum. Then
for every integer n, (-infty,x_0-1/n) intersects S in at most
countably many elements, and L is either equal to the union of all
these, or is the union of all these plus x_0 (if the sup of L lies in
L). Do something similar with B.
Well, there's nothing left to prove, you did it all.

Therefore, S cannot be equal to the union of L and B. Pick any point
in S not in L union B, and you have the lemma.
Sure.

sca18@hotmail.com (Amanda) wrote in message news:<6f75d9cf.0312170603.717b5063@posting.google.com>...
israel@math.ubc.ca (Robert Israel) wrote in message news:<bro92p$89s$1@nntp.itservices.ubc.ca>...
In article <6f75d9cf.0312161619.439e7ff8@posting.google.com>,
Amanda <sca18@hotmail.com> wrote:

I couldn't prove the following propostion, any help is welcome:
If S is an uncountable subset of R, then S contains a subset P such
that, if x and y are distintinct elements of P, then there's a z in P
between x and y.

Hint: remove every point which is in a closed interval of positive length
whose intersection with S is at most countable.

But this doesn't necessarily lead to a set with the desired property.
For example, if S =[0,1] U [2,3], then those points form an empty set

No: 0, 1, 2 and 3 will be removed (e.g. [1,2] is a closed interval of positive
length whose intersection with S has cardinality 2.
Yes, you're right! (of couirse!)

magidin@math.berkeley.edu (Arturo Magidin) wrote in message news:<brpvpc$1rsa$1@agate.berkeley.edu>...
In article <6f75d9cf.0312161619.439e7ff8@posting.google.com>,
Amanda <sca18@hotmail.com> wrote:
Hi,
I couldn't prove the following propostion, any help is welcome:
If S is an uncountable subset of R, then S contains a subset P such
that, if x and y are distintinct elements of P, then there's a z in P
between x and y.

First prove:

LEMMA. If S is an uncountable subset of R, then there exists an
element x in S such that there are uncountably many elements of S
smaller than x, and uncountably many elements of S larger than x.

Then you can use the Axiom of Choice to pick x_1 with that property;
then pick x_{21} < x_1 < x_{22}, x_{21} with the property that there
are uncountably many points of S smaller than x_{21}, and uncountably
many between x_{21} and x_1; etc. Keep making a choice of point in
between any two, and then take the union. That will be P.

To prove the lemma, consider two sets:

L = {x in S:there are only countably many elements of S smaller than x}
B = {x in S:there are only countably many elements of S larger than x}

Then L is downward closed: if x in S, and y in S satisfies y<x, then y
in L; B is upward closed: if x in B and y in S satisfies x<y, then y
in B.

Show that L is bounded above, and B is bounded below (hint: if L is
not bounded above, then every (-infty,n] intersects s for sufficiently
large integer n; this implies S is countable [prove it]).

If L is not bounded above, then, for every natural n, there is z in L
such that z >n. Since L is downward closed, this means the set I_n =
{x in S : x< n} is countable

If y is in S, let m be the smallest
integer > y. Then I_m = (-inf, m) intersection S is countable, which
shows every element of S belongs to an open interval that contains at
most finitely many elements of S. Therefore, S is countable (it has no
condensations points). I think we can come to this same conclusion if
we observe that S = Union (S /\I_n), n=1,2,3...This equation shows S
is the union of a countable collection of countable sets. Right?

Now prove that each of L and B are at most countable. E.g. if L is
empty, you are done. If it is not empty, let x_0 be its supremum. Then
for every integer n, (-infty,x_0-1/n) intersects S in at most
countably many elements, and L is either equal to the union of all
these, or is the union of all these plus x_0 (if the sup of L lies in
L). Do something similar with B.

Well, there's nothing left to prove, you did it all.

Therefore, S cannot be equal to the union of L and B. Pick any point
in S not in L union B, and you have the lemma.

Sure.

Now, can we use this result to prove that, if S is uncountable, then S
has bilateral condensation points (not necessarily in S) and the set
of the bilateral condensation points of S that are in S is
uncountable? x is a bilateral condensation point of S if, for every
eps>0, the intervals (x-eps, x) and (x, x+eps) intersect S in
uncountably many elements.

magidin@math.berkeley.edu (Arturo Magidin) wrote in message news:<brt3mr$2rdp$1@agate.berkeley.edu>...
In article <6f75d9cf.0312180709.497467ae@posting.google.com>,
Amanda <sca18@hotmail.com> wrote:
magidin@math.berkeley.edu (Arturo Magidin) wrote in message news:<brpvpc$1rsa$1@agate.berkeley.edu>...
In article <6f75d9cf.0312161619.439e7ff8@posting.google.com>,
Amanda <sca18@hotmail.com> wrote:
Hi,
I couldn't prove the following propostion, any help is welcome:
If S is an uncountable subset of R, then S contains a subset P such
that, if x and y are distintinct elements of P, then there's a z in P
between x and y.

[skip...]

Now, can we use this result to prove that, if S is uncountable, then S
has bilateral condensation points (not necessarily in S) and the set
of the bilateral condensation points of S that are in S is
uncountable? x is a bilateral condensation point of S if, for every
eps>0, the intervals (x-eps, x) and (x, x+eps) intersect S in
uncountably many elements.

Hmmm... You can certainly use the Lemma to prove that there exist
bilateral condensation points, but it seems unnecessary: prove that if
S is uncountable, then for every epsilon there exists an x in S,
possibly depending on epsilon, such that (x-eps, x+eps) intersect S is
uncountable.

Now let x_1 be the point you find for eps=1/2.
Then let x_2 be the point you find for eps=1/4, but for the set
(x_1-1/2,x_1+1/2) intersect S. Then let x_3 be the point for eps=1/8,
and the set (x_2-1/4,x_2+1/4) intersect S, etc. Then
x_1,....,x_n,... is a Cauchy sequence, and so converges; the limit
point will be a condensation point.

Sure. But this doesn't prove it's a bilateral, it may be unilatereal, right?

In article <6f75d9cf.0312180709.497467ae@posting.google.com>,
Amanda <sca18@hotmail.com> wrote:
magidin@math.berkeley.edu (Arturo Magidin) wrote in message news:<brpvpc$1rsa$1@agate.berkeley.edu>...
In article <6f75d9cf.0312161619.439e7ff8@posting.google.com>,
Amanda <sca18@hotmail.com> wrote:
Hi,
I couldn't prove the following propostion, any help is welcome:
If S is an uncountable subset of R, then S contains a subset P such
that, if x and y are distintinct elements of P, then there's a z in P
between x and y.

[skip...]

Now, can we use this result to prove that, if S is uncountable, then S
has bilateral condensation points (not necessarily in S) and the set
of the bilateral condensation points of S that are in S is
uncountable? x is a bilateral condensation point of S if, for every
eps>0, the intervals (x-eps, x) and (x, x+eps) intersect S in
uncountably many elements.

Hmmm... You can certainly use the Lemma to prove that there exist
bilateral condensation points, but it seems unnecessary: prove that if
S is uncountable, then for every epsilon there exists an x in S,
possibly depending on epsilon, such that (x-eps, x+eps) intersect S is
uncountable.

Now let x_1 be the point you find for eps=1/2.
Then let x_2 be the point you find for eps=1/4, but for the set
(x_1-1/2,x_1+1/2) intersect S. Then let x_3 be the point for eps=1/8,
and the set (x_2-1/4,x_2+1/4) intersect S, etc. Then
x_1,....,x_n,... is a Cauchy sequence, and so converges; the limit
point will be a condensation point.

In article <6f75d9cf.0312180709.497467ae@posting.google.com>,
Amanda <sca18@hotmail.com> wrote:
magidin@math.berkeley.edu (Arturo Magidin) wrote in message news:<brpvpc$1rsa$1@agate.berkeley.edu>...
In article <6f75d9cf.0312161619.439e7ff8@posting.google.com>,
Amanda <sca18@hotmail.com> wrote:
Hi,
I couldn't prove the following propostion, any help is welcome:
If S is an uncountable subset of R, then S contains a subset P such
that, if x and y are distintinct elements of P, then there's a z in P
between x and y.

[skip...]

Now, can we use this result to prove that, if S is uncountable, then S
has bilateral condensation points (not necessarily in S) and the set
of the bilateral condensation points of S that are in S is
uncountable? x is a bilateral condensation point of S if, for every
eps>0, the intervals (x-eps, x) and (x, x+eps) intersect S in
uncountably many elements.

Hmmm... You can certainly use the Lemma to prove that there exist
bilateral condensation points, but it seems unnecessary: prove that if
S is uncountable, then for every epsilon there exists an x in S,
possibly depending on epsilon, such that (x-eps, x+eps) intersect S is
uncountable.

Now let x_1 be the point you find for eps=1/2.
Then let x_2 be the point you find for eps=1/4, but for the set
(x_1-1/2,x_1+1/2) intersect S. Then let x_3 be the point for eps=1/8,
and the set (x_2-1/4,x_2+1/4) intersect S, etc. Then
x_1,....,x_n,... is a Cauchy sequence, and so converges; the limit
point will be a condensation point.