Discussing this subject in previous posts, I came to the following
conclusions:
If S is an uncountable subset of R (the reals), B is the set of
bilateral condensation points of S and U is the set of unilateral
condensation points of S, then:
(1) B is uncountable and U is countable.

(2) B intersect S is uncountable and U intersect S is countable.

(3) B is open and U is closed.

B may not be open. For example let uncountable S = R\Q.

I couldn't give a proof, but it looks to me that, for every
uncountable S, B intersect S is open and U intersect S is closed. Is
this really true? Could anyone please outline a proof or give a
counter example?

The irrationals.

I'm not sure if bilateral and unilateral condensation points are
standard terminology. But they mean exactly what the term suggests. x
is a bilateral condensation point of S if, for every eps>0, both
(x-eps, x) and (x, x+eps) contain uncountably many elements of S; and
x is unilateral if one of these intervals, but not both, contains
uncountably many elements of S.

They'll do nicely for linear ordered spaces.

On Tue, 23 Dec 2003, Amanda wrote:
Discussing this subject in previous posts, I came to the following
conclusions:
If S is an uncountable subset of R (the reals), B is the set of
bilateral condensation points of S and U is the set of unilateral
condensation points of S, then:
(1) B is uncountable and U is countable.

I presented a proof of this in the previous thread which was an
idea taken from another's post of a yet prior thread on this topic.
Did you see my post?
Yes,I did, it was really interesting, thank you. It took me some time

(2) B intersect S is uncountable and U intersect S is countable.

Of course U /\ S is countable as it's subset countable set.
The construction I gave for B was to remove countable many points from S.
Thus B /\ S = B is uncountable.
Sure

(3) B is open and U is closed.

B may not be open. For example let uncountable S = R\Q.
Every point of S is bilateral condensending so B = S.
However the irrationals aren't open, instead they have empty interior.
B is open. I defined B as the set of all condensation points of S,

The construction for U, I gave (in previous thread)
showed it was endpoints of a countable number of disjoint
intervals. Ie, a discrete subset of R, hence closed.
Sure. But, and as for U /\ S. Is it closed too? If S is the set of the

I couldn't give a proof, but it looks to me that, for every
uncountable S, B intersect S is open and U intersect S is closed. Is
this really true? Could anyone please outline a proof or give a
counter example?

The irrationals.

I'm not sure if bilateral and unilateral condensation points are
standard terminology. But they mean exactly what the term suggests. x
is a bilateral condensation point of S if, for every eps>0, both
(x-eps, x) and (x, x+eps) contain uncountably many elements of S; and
x is unilateral if one of these intervals, but not both, contains
uncountably many elements of S.

They'll do nicely for linear ordered spaces.
For other spaces, there's no lateralness.
That`s right.

Interesting problem. An exercise from topology class or text?
No. Actually, I was trying to prove the following proposition a fellow

Interesting problem. An exercise from topology class or text?
No. Actually, I was trying to prove the following proposition a
fellow gave me: "If uncountable S is a subset of R, then S contains
a subset T with the property that, for every distinct elements x and
y in T, there's a z in T between x and y". I noticed this proposition
is just a corollary of "if S is uncountable, then every element of
B/\S is a bilateral condensation point of B/\S", B defined as
before. Anyway, there are different approachs to prove my fellow's
proposition.
Seems you enjoy math.

From: Amanda <sca18@hotmail.com
Interesting problem. An exercise from topology class or text?
No. Actually, I was trying to prove the following proposition a
fellow gave me: "If uncountable S is a subset of R, then S contains
a subset T with the property that, for every distinct elements x and
y in T, there's a z in T between x and y". I noticed this proposition
is just a corollary of "if S is uncountable, then every element of
B/\S is a bilateral condensation point of B/\S", B defined as
before. Anyway, there are different approachs to prove my fellow's
proposition.
Seems you enjoy math.
Really much. I regret I don't have enough time ro study it as I would

This problem is equivalent to showing for uncountable S,
S is somewhere dense,
that is not nowhere dense; int cl S /= nulset

A somewhere dense when A not nowhere dense
iff some open nonnul U with cl U = cl U/\A
iff some open nonnul U with U subset cl A

----

From: Amanda <sca18@hotmail.com

"If uncountable S is a subset of R, then S contains
a subset T with the property that, for every distinct elements
x and y in T, there's a z in T between x and y".
Seems you enjoy math.
Really much. I regret I don't have enough
time ro study it as I would like
Upon retirement you might think so, but as I've discovered

This problem is equivalent to showing for uncountable S,
S is somewhere dense,
that is not nowhere dense; int cl S /= nulset

A somewhere dense when A not nowhere dense
iff some open nonnul U with cl U = cl U/\A
iff some open nonnul U with U subset cl A
I tried to prove this using Baire's Theorem,
but didn't get through. Do u think it's possible?
I'm not finding a topological way either. It likely

From: Amanda <sca18@hotmail.com
Subject: Re: Bilateral condensation points in R

William Elliot <marsh@privacy.net> wrote in message
From: Amanda <sca18@hotmail.com

"If uncountable S is a subset of R, then S contains
a subset T with the property that, for every distinct elements
x and y in T, there's a z in T between x and y".

This problem is equivalent to showing for uncountable S,
S is somewhere dense,
that is not nowhere dense; int cl S /= nulset

See retraction sci.math "nowhere dense real subsets".