Discrete uncountable set?
Joona I Palaste
example of a discrete uncountable set in R^n with the usual topology.
With other topologies it is easy. For example with the discrete
topology, each point is its own neighbourhood, so any uncountable set
is discrete. But how can I do it with the usual topology? Or can it be
done at all?
--
/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
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Robin Chapman
> This has me stumped. I can't for the life of me come up with an
> example of a discrete uncountable set in R^n with the usual topology.
I'd only worry about that if someone else could come up with
such a set.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"His mind has been corrupted by colours, sounds and shapes."
The League of Gentlemen
Lee Rudolph
>Joona I Palaste wrote:
>
>> This has me stumped. I can't for the life of me come up with an
>> example of a discrete uncountable set in R^n with the usual topology.
>
>I'd only worry about that if someone else could come up with
>such a set.
Depends on the value of n, surely?
Q: what's the smallest n that works? (Is this a question for
our logicians? I have to rush off, thank goodness, so I can't
even pretend to think about it.)
Lee Rudolph
Terry E
"Joona I Palaste" <pal...@cc.helsinki.fi> wrote in message
news:b8omqr$bvr$1...@oravannahka.helsinki.fi...
> This has me stumped. I can't for the life of me come up with an
> example of a discrete uncountable set in R^n with the usual topology.
> With other topologies it is easy. For example with the discrete
> topology, each point is its own neighbourhood, so any uncountable set
> is discrete. But how can I do it with the usual topology? Or can it be
> done at all?
Let S be a set of discrete points in R^n. Now 'chop' R^n up into 'boxes'
with side of length 1. That is if n=2 you would make the plane into a big
checker board.
1. Let B be one of the boxes. What is the cardinality of B intersect S ?
2. How many boxes do you have?
Remember to be discrete about it.
Terry
Joona I Palaste
> "Joona I Palaste" <pal...@cc.helsinki.fi> wrote in message
> news:b8omqr$bvr$1...@oravannahka.helsinki.fi...
>> This has me stumped. I can't for the life of me come up with an
>> example of a discrete uncountable set in R^n with the usual topology.
>> With other topologies it is easy. For example with the discrete
>> topology, each point is its own neighbourhood, so any uncountable set
>> is discrete. But how can I do it with the usual topology? Or can it be
>> done at all?
> Let S be a set of discrete points in R^n. Now 'chop' R^n up into 'boxes'
> with side of length 1. That is if n=2 you would make the plane into a big
> checker board.
> 1. Let B be one of the boxes. What is the cardinality of B intersect S ?
> 2. How many boxes do you have?
> Remember to be discrete about it.
The answer to 2 is of course "countably infinitely many", and it
appears that the answer to 1 should also be "countably many" but I can't
think of how to prove it.
--
/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
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"We sorcerers don't like to eat our words, so to say."
- Sparrowhawk
Robin Chapman
> Robin Chapman <r...@ivorynospamtower.freeserve.co.uk> writes:
>
>>Joona I Palaste wrote:
>>
>>> This has me stumped. I can't for the life of me come up with an
>>> example of a discrete uncountable set in R^n with the usual topology.
>>
>>I'd only worry about that if someone else could come up with
>>such a set.
>
> Depends on the value of n, surely?
n = aleph_0 ?
> Q: what's the smallest n that works? (Is this a question for
> our logicians? I have to rush off, thank goodness, so I can't
> even pretend to think about it.)
Might depend on whether you accept AC :-)
ArtflDodgr
Joona I Palaste <pal...@cc.helsinki.fi> wrote:
> This has me stumped. I can't for the life of me come up with an
> example of a discrete uncountable set in R^n with the usual topology.
Do you mean by "discrete" that each point of the set is isolated, in
that it has a neighborhod devoid of other points of the set?
--
A.
Rob Johnson
>example of a discrete uncountable set in R^n with the usual topology.
>With other topologies it is easy. For example with the discrete
>topology, each point is its own neighbourhood, so any uncountable set
>is discrete. But how can I do it with the usual topology? Or can it be
>done at all?
Let S be an uncountable set in R^n.
Divide R^n into unit cubes on the integer lattice. There are a
countable number of these cubes. If each cube contained only a
countable number of elements of S, then S would be countable. Thus,
some unit cube in R^n must contain an uncountable number of elements
of S. Under the normal topology, R^n is locally compact, so there must
be an accumulation point of S in that unit cube. Thus, S cannot be
discrete.
Rob Johnson
r...@whim.org
Joona I Palaste
Yes. Rob Johnson already gave a proof that under the usual topology in
R^n, uncountable discrete sets are impossible. Thanks to Rob.
--
/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
\----------------------------------------- Finland rules! ------------/
Randy Poe
> This has me stumped. I can't for the life of me come up with an
> example of a discrete uncountable set in R^n with the usual topology.
> With other topologies it is easy. For example with the discrete
> topology, each point is its own neighbourhood, so any uncountable set
> is discrete. But how can I do it with the usual topology? Or can it be
> done at all?
Maybe I'm missing something elementary since I don't know
what "the usual topology" is, but isn't the Cantor set
(the one formed by recursively removing the middle
thirds of intervals) a discrete uncountable set?
- Randy
Joona I Palaste
The usual topology is the one defined by the Euclidean metric. A
discrete set is one whose every point is isolated, i.e. has at least
one neighbourhood not containing any other points of the set. If the
Cantor set is indeed discrete under this topology then it goes against
the proof given by Rob Johnson.
--
/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
\----------------------------------------- Finland rules! ------------/
"'I' is the most beautiful word in the world."
- John Nordberg
The World Wide Wade
rpo...@yahoo.com (Randy Poe) wrote:
> Maybe I'm missing something elementary since I don't know
> what "the usual topology" is, but isn't the Cantor set
> (the one formed by recursively removing the middle
> thirds of intervals) a discrete uncountable set?
No, in fact it's a perfect set: No point in the Cantor set is isolated.
fourierr
> message news:<b8omqr$bvr$1...@oravannahka.helsinki.fi
>> example of a discrete uncountable set in R^n with the
>> usual topology.
>> With other topologies it is easy. For example with the discrete
>> topology, each point is its own neighbourhood, so any
>> uncountable set
>> is discrete. But how can I do it with the usual topology?
>> Or can it be
>> done at all?
>
> Maybe I'm missing something elementary since I don't know
> what "the usual topology" is, but isn't the Cantor set
> (the one formed by recursively removing the middle
> thirds of intervals) a discrete uncountable set?
>
> - Randy
The cantor set is totally disconnected-that is probably what
you are thinking of. Totally disconnected sets need not be
discrete.
fourierr
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Eckertson,Fred
This doesn't quite work. Although S must have an accumulation point in
R^n
what you need is S has an accumulation point in S.
{1/n : n in N} has an accumulation point in R but it is still discrete.
The underlying idea can be fixed, but it relies on local caompactness
So it does not work for R^w.
I gave a sketch in another one of Joona's threads
that works in more general situations.
Basically speaking, what makes S discrete?
How large is the smallest pool from which the witnesses can come from?
Eckertson,Fred
> -----Original Message-----
> From: rpo...@yahoo.com (Randy Poe) [mailto:rpo...@yahoo.com]
> Posted At: Wednesday, April 30, 2003 1:26 PM
> Posted To: math
> Conversation: Discrete uncountable set?
> Subject: Re: Discrete uncountable set?
>
>
No. Cantor is not discrete.
Discrete means every point is isolated (in the subspace topology),
and Cantor has no isolated points.
Alternately, compact discrete spaces are finite.
Leonard Blackburn
> example of a discrete uncountable set in R^n with the usual topology.
That's because there is no such example.
> With other topologies it is easy. For example with the discrete
> topology, each point is its own neighbourhood, so any uncountable set
> is discrete. But how can I do it with the usual topology? Or can it be
> done at all?
n=1 case:
Suppose S is an uncountable, discrete subset of R.
For each x in S, let I_x be an open interval centered
at x that contains no other members of S. Now let
J_x be (1/2)I_x, i.e. the open interval centered at x
that has half the length of I_x. It is clear that
{J_x : x in S} is an uncountable collection of pairwise
disjoint open intervals (verify this). For each
x in S, choose a rational number q_x in J_x. Then
{q_x : x in S} is an uncountable collection of rationals,
a contradiction.
Therefore, there is no uncountable, discrete
subset of R with the usual topology. //
The above easily generalizes to R^n for any n.
-Leonard
The World Wide Wade
> Divide R^n into unit cubes on the integer lattice. There are a
> countable number of these cubes. If each cube contained only a
> countable number of elements of S, then S would be countable. Thus,
> some unit cube in R^n must contain an uncountable number of elements
> of S. Under the normal topology, R^n is locally compact, so there must
> be an accumulation point of S in that unit cube. Thus, S cannot be
> discrete.
But a set with an accumulation point can be discrete, so I'm not sure how
the conclusion follows from the above.
Here's another proof: Suppose S is a discrete subset of R^n. For each p in
S we can choose a neighborhood V_p of p such that V_p intersect S = {p}.
Let V = U V_p. Because V is Lindelof, we can choose V_p(1), V_p(2), ...
such that V = U V_p(n). But then S = S intersect [U V_p(n)] = {p1,p2,...},
which shows S is countable.
Related result: If S is an uncountable subset of R^n, then only countably
many points of S fail to be condensation points of S. (A condensation point
of S is a point p in S such that every neighborhood of p contains
uncountably many points of S.) The proof is similar.
Joona I Palaste
> n=1 case:
This is the nicest proof of the non-existence of uncountable discrete
sets in R^n with the Euclidean topology I have seen in the whole
thread. Thanks very much, Leonard.
--
/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
\----------------------------------------- Finland rules! ------------/
"How come even in my fantasies everyone is a jerk?"
- Daria Morgendorfer
Terry E
> Terry E <te...@nwlink.com> scribbled the following:
> > "Joona I Palaste" <pal...@cc.helsinki.fi> wrote in message
> > news:b8omqr$bvr$1...@oravannahka.helsinki.fi...
> >> This has me stumped. I can't for the life of me come up with an
> >> example of a discrete uncountable set in R^n with the usual topology.
> >> With other topologies it is easy. For example with the discrete
> >> topology, each point is its own neighbourhood, so any uncountable set
> >> is discrete. But how can I do it with the usual topology? Or can it be
> >> done at all?
>
> > Let S be a set of discrete points in R^n. Now 'chop' R^n up into
'boxes'
> > with side of length 1. That is if n=2 you would make the plane into a
big
> > checker board.
>
> > 1. Let B be one of the boxes. What is the cardinality of B intersect S
?
> > 2. How many boxes do you have?
>
> > Remember to be discrete about it.
>
> The answer to 2 is of course "countably infinitely many", and it
> appears that the answer to 1 should also be "countably many" but I can't
> think of how to prove it.
How about finite - think compact sets.
David C. Ullrich
<r...@ivorynospamtower.freeserve.co.uk> wrote:
>Lee Rudolph wrote:
>
>> Robin Chapman <r...@ivorynospamtower.freeserve.co.uk> writes:
>>
>>>Joona I Palaste wrote:
>>>
>>>> This has me stumped. I can't for the life of me come up with an
>>>> example of a discrete uncountable set in R^n with the usual topology.
>>>
>>>I'd only worry about that if someone else could come up with
>>>such a set.
>>
>> Depends on the value of n, surely?
>
>n = aleph_0 ?
If we're talking about the product topology then it seems
to me that R^aleph_0 is separable - the points with rational
coordinates, all but finitely many of which are zero, form
a countable dense set.
It clear that aleph_1 works (for example points with
one coordinate equal to 1 and all the rest 0 would be
an uncountable discrete set.)
>> Q: what's the smallest n that works? (Is this a question for
>> our logicians? I have to rush off, thank goodness, so I can't
>> even pretend to think about it.)
>
>Might depend on whether you accept AC :-)
I don't think so. (No sarcasm intended, it's very
easy to be using AC without realizing it, but
I don't think I did above.)
******************
David C. Ullrich
David C. Ullrich
<ull...@math.okstate.edu> wrote:
>On Wed, 30 Apr 2003 18:41:12 +0100, Robin Chapman
><r...@ivorynospamtower.freeserve.co.uk> wrote:
>
>>Lee Rudolph wrote:
>>
>>> Robin Chapman <r...@ivorynospamtower.freeserve.co.uk> writes:
>>>
>>>>Joona I Palaste wrote:
>>>>
>>>>> This has me stumped. I can't for the life of me come up with an
>>>>> example of a discrete uncountable set in R^n with the usual topology.
>>>>
>>>>I'd only worry about that if someone else could come up with
>>>>such a set.
>>>
>>> Depends on the value of n, surely?
>>
>>n = aleph_0 ?
>
>If we're talking about the product topology then it seems
>to me that R^aleph_0 is separable - the points with rational
>coordinates, all but finitely many of which are zero, form
>a countable dense set.
Don't ask me why I thought that showed aleph_0 doesn't work.
But it doesn't. For example, there is a countable basis for the
topology of R^aleph_0. Hence (as Wade pointed out for
R^n, n finite) if {V_alpha} is any collection of open sets,
there is some countable subcollection with the same
union, and it follows that any discrete set is countable.
>It clear that aleph_1 works (for example points with
>one coordinate equal to 1 and all the rest 0 would be
>an uncountable discrete set.)
>
>>> Q: what's the smallest n that works? (Is this a question for
>>> our logicians? I have to rush off, thank goodness, so I can't
>>> even pretend to think about it.)
>>
>>Might depend on whether you accept AC :-)
>
>I don't think so. (No sarcasm intended, it's very
>easy to be using AC without realizing it, but
>I don't think I did above.)
>
>******************
>
>David C. Ullrich
******************
David C. Ullrich
Rob Johnson
Yes, I went awry at the end of the proof.
Once you have an uncountable number of points in a unit cube, assume
they are isolated. Cover them with a disjoint set of open balls. Each
ball has positive volume, and since they are disjoint the sum of their
volumes must be less than the volume of the unit cube, 1.
However, an uncountable sum of positive numbers cannot be finite. For
each positive integer n, let C_n be the set of balls with volume > 1/n.
Each ball must be in some C_n and C_n cannot have any more than n balls
in it. Thus, there are a countable number of balls.
Rob Johnson
r...@whim.org
Terry E
"> > The answer to 2 is of course "countably infinitely many", and it
> > appears that the answer to 1 should also be "countably many" but I can't
> > think of how to prove it.
>
> How about finite - think compact sets.
Make that countable. Reason is in Rob Johnson's correction to his proof.
Dustin Clausen
> example of a discrete uncountable set in R^n with the usual topology.
> With other topologies it is easy. For example with the discrete
> topology, each point is its own neighbourhood, so any uncountable set
> is discrete. But how can I do it with the usual topology? Or can it be
> done at all?
Can't... and the other proofs I've seen seem unnecessarily
complicated. By assumption, around each point of S there is a
neighborhood containing no other points of S, and such a neighborhood
must also contain a point of Q^n (use rational approximations). We
can set a rule for choosing this point because Q^n is countable. Then
we get an injection S --> Q^n, which shows that S is countable.
Dustin Clausen
Crap. I need to stop using google groups to post. It takes too long to show up.
Robin Chapman
No.
Randy Poe
> Randy Poe wrote:
>
> > Joona I Palaste <pal...@cc.helsinki.fi> wrote in message
> > news:<b8omqr$bvr$1...@oravannahka.helsinki.fi>...
> >> This has me stumped. I can't for the life of me come up with an
> >> example of a discrete uncountable set in R^n with the usual topology.
> >> With other topologies it is easy. For example with the discrete
> >> topology, each point is its own neighbourhood, so any uncountable set
> >> is discrete. But how can I do it with the usual topology? Or can it be
> >> done at all?
> >
> > Maybe I'm missing something elementary since I don't know
> > what "the usual topology" is, but isn't the Cantor set
> > (the one formed by recursively removing the middle
> > thirds of intervals) a discrete uncountable set?
>
> No.
Well, that was informative. What am I missing?
From Mathworld:
"A general Cantor set is a closed set consisting entirely of boundary
points. Such sets are uncountable and may have 0 or positive Lebesgue
measure. The Cantor set is the only totally disconnected..."
The Cantor set is uncountable.
From the above, isn't the Cantor set discrete?
- Randy
Eckertson,Fred
> -----Original Message-----
> From: r...@whim.org (Rob Johnson) [mailto:r...@whim.org]
> Posted At: Wednesday, April 30, 2003 5:47 PM
> Posted To: math
> Conversation: Discrete uncountable set?
> Subject: Re: Discrete uncountable set?
>
> Yes, I went awry at the end of the proof.
>
> Once you have an uncountable number of points in a unit cube, assume
> they are isolated. Cover them with a disjoint set of open
> balls. Each
> ball has positive volume, and since they are disjoint the sum of their
> volumes must be less than the volume of the unit cube, 1.
You need to speak more carefully. The unit cube does not contain any
isolated points.
I assume then what you mean is the uncountable set is discrete. But this
only means that each point is contained in a ball that doesn't contain
any other members of the set. It does not mean that there is a disjoint
family
of open sets each memeber of which meets the discrete set in at most one
point.
>
> However, an uncountable sum of positive numbers cannot be finite. For
> each positive integer n, let C_n be the set of balls with
> volume > 1/n.
> Each ball must be in some C_n and C_n cannot have any more
> than n balls
> in it. Thus, there are a countable number of balls.
>
> Rob Johnson
> r...@whim.org
>
Note: Measure theory should be avoided. This is strictly a topological
question.
The first thing somebody will (should) ask is "What's so special about R
here?"
"Is there a general result for all ___ spaces?"
Robert Kolker
Joona I Palaste wrote:
> This has me stumped. I can't for the life of me come up with an
> example of a discrete uncountable set in R^n with the usual topology.
If by discrete you mean non-dense, try the Cantor set which is uncountable.
Bob Kolker
Herman Rubin
>In article <b8omqr$bvr$1...@oravannahka.helsinki.fi>,
>Joona I Palaste <pal...@cc.helsinki.fi> wrote:
>>This has me stumped. I can't for the life of me come up with an
>>example of a discrete uncountable set in R^n with the usual topology.
>>With other topologies it is easy. For example with the discrete
>>topology, each point is its own neighbourhood, so any uncountable set
>>is discrete. But how can I do it with the usual topology? Or can it be
>>done at all?
>Let S be an uncountable set in R^n.
The theorem is true in any metric space with a countable
dense set, and is much more understandable that way.
So let S be a discrete set, and R a countable dense
subset of the metric space.
If x \in S, there is a "sphere" of radius d_x which
contains no other points of S; to avoid any
unnecessary choices, choose d_x to be the first
positive rational distance with this property.
If we then take e_x = d_x/2, the e_x neighborhoods
of the corresponding points are disjoint.
Now let r_i be the i-th point in R. It may or may
not be an element of e_x for some x; if it is, set
y_i = x, unless x = y_j for some j < i. This way,
each x in S corresponds to one integer, and one can
get an explicit enumeration from this.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Deptartment of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Leonard Blackburn
You have to be a little more careful than this. You haven't explained
how to get an injection (you might be choosing the same point in Q^n
for two different points of S--see my earlier proof for the extra step.)
-Leonard
Eckertson,Fred
second countable spaces cannot contain an uncountable discrete set.
Let BB be a base. Let S be discrete. For each s in S fix B_s in BB such
that
B_s /\ S = {s}. This defines an injection from S into BB. So |S| < |BB|
= w.
With a little care AC can be avoided. Hint: fix an enumeration of BB and
use it
to define B_s.
The problem with basing the proof on density is that it just doesn't
generalize.
You end up relying entirely on the metric when you don't need to.
FYI, the following simple result is very useful.
It swats the question dead with hereditarily separable, but the key is
second countable.
In the class of metric spaces the following are equivalent
X is second countable (a hereditary property)
X is separable
X is hereditarily separable
X is Lindel"of
As an aside, compact metric ==> Lindel"of metric.
So compact metric spaces must be hereditarily separable.
Jon Cast
> Well, that was informative. What am I missing?
>
> From Mathworld:
> "A general Cantor set is a closed set consisting entirely of
> boundary points. Such sets are uncountable
Did they mean to say that any set consisting entirely of boundary
points is a general Cantor set? If so, a general Cantor set is /not/
necessarily uncountable; in fact, all countable sets consist entirely
of boundary points...
> and may have 0 or positive Lebesgue measure. The Cantor set is the
> only totally disconnected..."
> The Cantor set is uncountable. From the above, isn't the Cantor set
> discrete?
No. Disconnected (in R) === there is no open interval entirely
contained in E. Discrete === there is no open interval with a
non-empty, non-singleton intersection with E. Discrete is a much
stronger condition than disconnected.
Jon Cast
Mehdi Tibouchi
> >> Q: what's the smallest n that works? (Is this a question for
> >> our logicians? I have to rush off, thank goodness, so I can't
> >> even pretend to think about it.)
> >
> >Might depend on whether you accept AC :-)
>
> I don't think so. (No sarcasm intended, it's very
> easy to be using AC without realizing it, but
> I don't think I did above.)
Isn't the acceptance of AC implied by the mere mention of the "smallest
n that works"? I mean, do cardinals form a well-ordered class without
AC? If not, what's the definition of aleph_1 without AC? I guess
"smallest uncountable cardinals" might fail to work, and that it might
even depend on the definition you give of a countable set, but then I'm
quite thoroughly clueless about set theory, especially without choice.
--
M. Tibouchi <med...@alussinan.org>
> Life's but a walking Shadow, a poore Player
> That struts and frets his houre vpon the Stage
> And then is heard no more [...] Macbeth V, 5.
The World Wide Wade
rpo...@yahoo.com (Randy Poe) wrote:
> From Mathworld:
>
> "A general Cantor set is a closed set consisting entirely of boundary
> points. Such sets are uncountable and may have 0 or positive Lebesgue
> measure. The Cantor set is the only totally disconnected..."
>
> The Cantor set is uncountable.
> From the above, isn't the Cantor set discrete?
No, a topological space is discrete if every point in the space is
isolated. The Cantor K set is the opposite in this respect: Each point in K
is a limit point of K.
Herman Rubin
>> >> Q: what's the smallest n that works? (Is this a question for
>> >> our logicians? I have to rush off, thank goodness, so I can't
>> >> even pretend to think about it.)
>> >Might depend on whether you accept AC :-)
>> I don't think so. (No sarcasm intended, it's very
>> easy to be using AC without realizing it, but
>> I don't think I did above.)
>Isn't the acceptance of AC implied by the mere mention of the "smallest
>n that works"? I mean, do cardinals form a well-ordered class without
>AC? If not, what's the definition of aleph_1 without AC? I guess
>"smallest uncountable cardinals" might fail to work, and that it might
>even depend on the definition you give of a countable set, but then I'm
>quite thoroughly clueless about set theory, especially without choice.
Well-ordered sets exist without AC, and were reasonably
well known. There was no canonical version of ordinal
numbers before von Neumann came up with one, but this makes
very little difference. Cantor, Zermelo, Hartogs, and
others did quite well without the canonical representation,
and even without the representation of a relation as a set
of ordered pairs.
When it comes to "smallest cardinals", there is a major
problem, but not ordinals.
Robin Chapman
No.
Robin Chapman
But what if by discrete you mean discrete?
Rob Johnson
"Eckertson,Fred" <FECKE...@cerner.com> wrote:
>> From: r...@whim.org (Rob Johnson) [mailto:r...@whim.org]
>> Posted At: Wednesday, April 30, 2003 5:47 PM
>> Posted To: math
>> Conversation: Discrete uncountable set?
>> Subject: Re: Discrete uncountable set?
>>
>> Yes, I went awry at the end of the proof.
>>
>> Once you have an uncountable number of points in a unit cube, assume
>> they are isolated. Cover them with a disjoint set of open
>> balls. Each
>> ball has positive volume, and since they are disjoint the sum of their
>> volumes must be less than the volume of the unit cube, 1.
>
>
>You need to speak more carefully. The unit cube does not contain any
>isolated points.
>I assume then what you mean is the uncountable set is discrete. But this
>
>only means that each point is contained in a ball that doesn't contain
>any other members of the set. It does not mean that there is a disjoint
>family
>of open sets each memeber of which meets the discrete set in at most one
>point.
Yes, I meant discrete, since that is what we were trying to contradict.
However, for a set {x_n} in a metric space, if each B(x_n,r_n) does not
contain any other points in {x_n}, then the balls B(x_n,r_n/2) are
disjoint.
>>
>> However, an uncountable sum of positive numbers cannot be finite. For
>> each positive integer n, let C_n be the set of balls with
>> volume > 1/n.
>> Each ball must be in some C_n and C_n cannot have any more
>> than n balls
>> in it. Thus, there are a countable number of balls.
>>
>> Rob Johnson
>> r...@whim.org
>>
>
>
>Note: Measure theory should be avoided. This is strictly a topological
>question.
>The first thing somebody will (should) ask is "What's so special about R
>here?"
>"Is there a general result for all ___ spaces?"
I know that this proof uses a lot of machinery that is not necessary for
the ultimate result. A streamlined proof can expose basic principles
that can be generalized to other situations (like second countability).
However, sometimes using ideas that do not lead to generalities in one
problem expose those ideas for use in other problems (such as a finite
sum can have only countably many non-zero terms). Someone else had
already used a countable dense set argument in another thread of the
original article, so I decided to use another approach in completing
my incomplete proof.
Rob Johnson
r...@whim.org
Robert Kolker
Robin Chapman wrote:
> But what if by discrete you mean discrete?
What is the definition of discrete?
Bob Kolker
Robin Chapman
>
>
> Robin Chapman wrote:
>> But what if by discrete you mean discrete?
>
> What is the definition of discrete?
A discrete topological space is one where all subsets are open.
Robert Kolker
Robert Kolker wrote:
>
>
> Robin Chapman wrote:
>
>> But what if by discrete you mean discrete?
>
>
> What is the definition of discrete?
If you mean you could wrap each element in an open interval which
contains no other memebers of the set, then clearly such a set could be
at most countable.
Bob Kolker
David C. Ullrich
Tibouchi) wrote:
>David C. Ullrich <ull...@math.okstate.edu> wrote:
>
>> >> Q: what's the smallest n that works? (Is this a question for
>> >> our logicians? I have to rush off, thank goodness, so I can't
>> >> even pretend to think about it.)
>> >
>> >Might depend on whether you accept AC :-)
>>
>> I don't think so. (No sarcasm intended, it's very
>> easy to be using AC without realizing it, but
>> I don't think I did above.)
>
>Isn't the acceptance of AC implied by the mere mention of the "smallest
>n that works"? I mean, do cardinals form a well-ordered class without
>AC? If not, what's the definition of aleph_1 without AC? I guess
>"smallest uncountable cardinals" might fail to work, and that it might
>even depend on the definition you give of a countable set, but then I'm
>quite thoroughly clueless about set theory, especially without choice.
Exactly what a cardinal is without AC is a little fuzzy, but there's
no problem talking about _ordinals_ - they work the same as with
AC (and you can still talk about the smallest ordinal with a certain
cardinality - what goes wrong is showing that every set is
equinumerous with some ordinal, hence you can't use
certain ordinals as cardinals.)
In any case, this doesn't have much to do with the question
of what the smallest n that works is. A countable n doesn't
work, and any umcountable n does.
******************
David C. Ullrich
David C. Ullrich
<FECKE...@cerner.com> wrote:
>No one seems to have the real proof:
>second countable spaces cannot contain an uncountable discrete set.
You should read everything everyone has said before making
such claims. Here's a quote from my second post:
"But it doesn't. For example, there is a countable basis for the
topology of R^aleph_0. Hence (as Wade pointed out for
R^n, n finite) if {V_alpha} is any collection of open sets,
there is some countable subcollection with the same
union, and it follows that any discrete set is countable."
Wade said more or less the same thing in different words,
for R^n (n finite): he used the fact that R^n is a Lindelof
space.
>Let BB be a base. Let S be discrete. For each s in S fix B_s in BB such
>that
>B_s /\ S = {s}. This defines an injection from S into BB. So |S| < |BB|
>= w.
>
>With a little care AC can be avoided. Hint: fix an enumeration of BB and
>use it
>to define B_s.
>
>The problem with basing the proof on density is that it just doesn't
>generalize.
>You end up relying entirely on the metric when you don't need to.
>
>
>FYI, the following simple result is very useful.
>It swats the question dead with hereditarily separable, but the key is
>second countable.
>
>In the class of metric spaces the following are equivalent
>X is second countable (a hereditary property)
>X is separable
>X is hereditarily separable
>X is Lindel"of
>
>As an aside, compact metric ==> Lindel"of metric.
>So compact metric spaces must be hereditarily separable.
>
>
>
>
******************
David C. Ullrich
Joona I Palaste
> This has me stumped. I can't for the life of me come up with an
> example of a discrete uncountable set in R^n with the usual topology.
> topology, each point is its own neighbourhood, so any uncountable set
> is discrete. But how can I do it with the usual topology? Or can it be
> done at all?
It has been proven here quite many times that such a set is impossible
under the normal topology. I think a corollary here is that it is also
impossible in every other topology which does not have an uncountable
collection of open singletons.
--
/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
\----------------------------------------- Finland rules! ------------/
"How come even in my fantasies everyone is a jerk?"
- Daria Morgendorfer
Eckertson,Fred
> -----Original Message-----
> From: David C. Ullrich [mailto:ull...@math.okstate.edu]
> Posted At: Thursday, May 01, 2003 6:35 PM
> Posted To: math
> Conversation: Discrete uncountable set?
> Subject: Re: Discrete uncountable set?
>
>
> On Thu, 1 May 2003 10:22:56 -0500 , "Eckertson,Fred"
> <FECKE...@cerner.com> wrote:
>
> >No one seems to have the real proof:
> >second countable spaces cannot contain an uncountable discrete set.
>
> You should read everything everyone has said before making
> such claims. Here's a quote from my second post:
>
> "But it doesn't. For example, there is a countable basis for the
> topology of R^aleph_0. Hence (as Wade pointed out for
> R^n, n finite) if {V_alpha} is any collection of open sets,
> there is some countable subcollection with the same
> union, and it follows that any discrete set is countable."
Sorry David I totally looked passed this.
It certainly does use the jist of the matter:
s(X) <= hd(X) <= w(X)
Wade's proof using is equally good.
s(X) <= hL(X) <= w(X)
Eckertson,Fred
> -----Original Message-----
> From: Joona I Palaste [mailto:pal...@cc.helsinki.fi]
> Posted At: Friday, May 02, 2003 1:15 AM
> Posted To: math
> Conversation: Discrete uncountable set?
> Subject: Re: Discrete uncountable set?
>
>
> Joona I Palaste <pal...@cc.helsinki.fi> scribbled the following:
> > This has me stumped. I can't for the life of me come up with an
> > example of a discrete uncountable set in R^n with the usual
> topology.
> > With other topologies it is easy. For example with the discrete
> > topology, each point is its own neighbourhood, so any
> uncountable set
> > is discrete. But how can I do it with the usual topology?
> Or can it be
> > done at all?
>
> It has been proven here quite many times that such a set is impossible
> under the normal topology. I think a corollary here is that it is also
> impossible in every other topology which does not have an uncountable
> collection of open singletons.
>
No.
Any free sum of uncountably spaces will have an uncountable discrete
set.
If none of the summand spaces has an isolated point than neither will
the sum.
Alternately, David pointed out how that R^(w_1) has an uncountable
discrete set and
no power of R has an isolated point (R^kappa is always infinite and
connected).
Randy Poe
> rpo...@yahoo.com (Randy Poe) writes:
>
> > Well, that was informative. What am I missing?
> >
> > From Mathworld:
>
> > "A general Cantor set is a closed set consisting entirely of
> > boundary points. Such sets are uncountable
>
> Did they mean to say that any set consisting entirely of boundary
> points is a general Cantor set?
I took it out of context. They started with the middle-thirds
construction. I think this refers to generalizing that construction
in some way. I've seen the statement before that Cantor sets
can be generalized, and that some have (Lebesgue?) measure 0
and some have finite measure.
> If so, a general Cantor set is /not/
> necessarily uncountable; in fact, all countable sets consist entirely
> of boundary points...
As I read the above, it seems to say "closed + all elements are
boundary points => uncountable" but that's not true, since
a singleton set is closed and a boundary point. Perhaps they
are saying that however you generalize the Cantor construction,
you get an uncountable set.
> > and may have 0 or positive Lebesgue measure. The Cantor set is the
> > only totally disconnected..."
>
> > The Cantor set is uncountable. From the above, isn't the Cantor set
> > discrete?
>
> No. Disconnected (in R) === there is no open interval entirely
> contained in E. Discrete === there is no open interval with a
> non-empty, non-singleton intersection with E. Discrete is a much
> stronger condition than disconnected.
I had to ponder this a bit to see the distinction. I'd phrase
it this way. Discrete: For each x in E, there exists
an open set containing x and no other element of E. Disconnected:
for each x in E, every neighborhood of x contains external
points.
That seems to be what you're saying with your definitions,
but I'm still confused by a couple of things. First, doesn't
that make the rationals (and for that matter, the
irrationals) totally disconnected? I'm certain I'm still
misunderstanding something.
Second, I believe the Cantor set is nowhere dense. How does
the argument for that go?
- Randy
The World Wide Wade
> I had to ponder this a bit to see the distinction. I'd phrase
> it this way. Discrete: For each x in E, there exists
> an open set containing x and no other element of E. Disconnected:
> for each x in E, every neighborhood of x contains external
> points.
I think you want "totally disconnected" rather than "disconnected" above.
> That seems to be what you're saying with your definitions,
> but I'm still confused by a couple of things. First, doesn't
> that make the rationals (and for that matter, the
> irrationals) totally disconnected? I'm certain I'm still
> misunderstanding something.
Yes, both sets are totally disconnected (TD). A topological space with more
than one point is TD if its only connected subsets are singletons. In R,
the only connected subsets are intervals. So a subset of R with more than
one point without an interval inside it is TD. Therefore the irrationals
and the rationals are both TD subsets of R.
> Second, I believe the Cantor set is nowhere dense. How does
> the argument for that go?
One way to think of it: From [0,1] remove an open interval containing 1/2.
You have two closed intervals left. In each of those, remove an open
interval containing the midpoint. Now you have four closed intervals left.
Remove four open intervals containing the midpoints. Continue ... The union
U of open intervals so removed has to be dense in [0,1], because no open
interval in [0,1] could miss all of U. The complement of any open dense set
is closed and nowhere dense. The Cantor set is a particular case of this
scenario.
Oscar Lanzi III
neither? Continuous is not obvious because you don't have any full line
segments; discrete is not obvious because, except at points
corresponding to terminating base-3 representations, you don't have any
solid gaps.
--OL
Robin Chapman
> How is the Cantor Set rendered here? Discrete or continuous? Or
> neither?
what's a "continuous set"?
Shmuel (Seymour J.) Metz
at 06:04 PM, r...@whim.org (Rob Johnson) said:
>Divide R^n into unit cubes on the integer lattice. There are a
>countable number of these cubes.
Only if n is finite.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Any unsolicited bulk E-mail will be subject to legal action. I reserve the
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Shmuel (Seymour J.) Metz
at 04:56 PM, David C. Ullrich <ull...@math.okstate.edu> said:
>If we're talking about the product topology then it seems to me that
>R^aleph_0 is separable - the points with rational coordinates, all
>but finitely many of which are zero, form a countable dense set.
No. You're thinking of the wrong space. There is no convergence
condition on the elements of R^n.
>It clear that aleph_1 works (for example points with
>one coordinate equal to 1 and all the rest 0 would be
>an uncountable discrete set.)
Also no. "Box" and "Slab" are not interchangable, except for finite n.
Shmuel (Seymour J.) Metz
at 04:41 PM, dtcl...@hotmail.com (Dustin Clausen) said:
>Can't... and the other proofs I've seen seem unnecessarily
>complicated. By assumption, around each point of S there is a
>neighborhood containing no other points of S, and such a neighborhood
>must also contain a point of Q^n (use rational approximations). We
>can set a rule for choosing this point because Q^n is countable.
>Then we get an injection S --> Q^n, which shows that S is
>countable.
Only if n is finite.
--
Shmuel (Seymour J.) Metz
at 06:30 PM, Joona I Palaste <pal...@cc.helsinki.fi> said:
>The usual topology is the one defined by the Euclidean metric.
What euclidean metric? That term is only well defined if you restrict
n to be finite. You can, of course, extend the Euclidena metric to a
subset or R^n for n countable, but you still have problems if n is
uncountable.
OTOH, if by "the usual topology" you mean the product topology, then
there is no discrete uncountable set even if n is uncountable.
Unfortunately, there is no extension of the Euclidean Metric
compatible with that topology.
Shmuel (Seymour J.) Metz
at 06:15 AM, Joona I Palaste <pal...@cc.helsinki.fi> said:
>It has been proven here quite many times that such a set is
>impossible under the normal topology.
No, since nobody has given an unambiguosu definition of "the normal
topology".
Joona I Palaste
> In <b8t2d4$50m$2...@oravannahka.helsinki.fi>, on 05/02/2003
> at 06:15 AM, Joona I Palaste <pal...@cc.helsinki.fi> said:
>>It has been proven here quite many times that such a set is
>>impossible under the normal topology.
> No, since nobody has given an unambiguosu definition of "the normal
> topology".
Here is one: The normal topology of R^n, where n is a natural number, is
the one given by the Euclidean metric, i.e.
n
d(x, y) = sqrt( \sum (x_i - y_i)^2 ).
i=1
--
/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
\----------------------------------------- Finland rules! ------------/
"This is a personnel commuter."
- Train driver in Scientific American
Herman Rubin
Shmuel (Seymour J.) Metz <spam...@library.lspace.org.invalid> wrote:
>In <3ch0bv4t5sojqq6eo...@4ax.com>, on 04/30/2003
> at 04:56 PM, David C. Ullrich <ull...@math.okstate.edu> said:
>>If we're talking about the product topology then it seems to me that
>>R^aleph_0 is separable - the points with rational coordinates, all
>>but finitely many of which are zero, form a countable dense set.
>No. You're thinking of the wrong space. There is no convergence
>condition on the elements of R^n.
For the product topology, this is correct. The space is
homeomorphic to the product of (-1/n^2, 1/n^2) with any
reasonable distance function, and here we see easily
that the above set of points, which can be carried over
by the mapping x_n -> x_n/(n^2*|x_n| + 1).
>>It clear that aleph_1 works (for example points with
>>one coordinate equal to 1 and all the rest 0 would be
>>an uncountable discrete set.)
>Also no. "Box" and "Slab" are not interchangable, except for finite n.
It certainly is discrete. Each point has a neighborhood
containing no other points of the set.
Eckertson,Fred
> -----Original Message-----
> From: The World Wide Wade [mailto:wadera...@attbi.remove13.com]
> Posted At: Friday, May 02, 2003 3:17 PM
> Posted To: math
> Conversation: Discrete uncountable set?
> Subject: Re: Discrete uncountable set?
>
>
> In article <585ab5d8.03050...@posting.google.com>,
> rpo...@yahoo.com (Randy Poe) wrote:
>
> > I had to ponder this a bit to see the distinction. I'd phrase
> > it this way. Discrete: For each x in E, there exists
> > an open set containing x and no other element of E. Disconnected:
> > for each x in E, every neighborhood of x contains external
> > points.
>
> I think you want "totally disconnected" rather than
> "disconnected" above.
>
> > That seems to be what you're saying with your definitions,
> > but I'm still confused by a couple of things. First, doesn't
> > that make the rationals (and for that matter, the
> > irrationals) totally disconnected? I'm certain I'm still
> > misunderstanding something.
>
> Yes, both sets are totally disconnected (TD). A topological
> space with more
> than one point is TD if its only connected subsets are
> singletons.
The contemporary definition of Totally Disconnected is a little stronger
than this:
points can be separated by clopen sets
<->
every quasi-component is a singleton
(a quasi-component is the intersection of all clopen sets containing a
fixed point)
Totally disconnected is slightly stronger than hereditarily disconnected
i.e., every component is a singleton
Shmuel (Seymour J.) Metz
at 09:46 AM, Joona I Palaste <pal...@cc.helsinki.fi> said:
>Here is one: The normal topology of R^n, where n is a natural number,
And where n is *not* a natural number? Nothing was said about the
cardinality being small.