unique limit point in infinite bounded set
frank
I was thinking about the following exercise: "Suppose that S is a
bounded, infinite subset of R having exactly one limit point x0. Prove
that, for any e>0, the neighborhood N'(x0,e) contains all but finitely
many of the points of S."
It would appear that the proof is rather straightforward, by invoking
the Bolzano-Weierstrass theorem on the set S\N(x0,e). If this set were
to contain infinitely many points, the Bolzano-Weierstrass theorem
would predict that it contains a limit point. This limit point would
also be a limit point for S, since S\N(x0,e) is contained in S. This
would violate the assumption that S has exactly one limit point.
I have extreme difficulty, however, imagining a set that would be a
bounded, infinite subset of R having exactly one limit point x0. Could
someone give an example of such a set?
Sorry if this is a double posting: I posted the question yesterday,
but could not find it back.
Thanks,
Frank
Arturo Magidin
frank <frank.d...@gmail.com> wrote:
>Hello
>
>I was thinking about the following exercise: "Suppose that S is a
>bounded, infinite subset of R having exactly one limit point x0. Prove
>that, for any e>0, the neighborhood N'(x0,e) contains all but finitely
>many of the points of S."
[...]
>I have extreme difficulty, however, imagining a set that would be a
>bounded, infinite subset of R having exactly one limit point x0. Could
>someone give an example of such a set?
{1/n: n a positive integer}
--
======================================================================
"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
frank
> In article <1181236929.182886.171...@w5g2000hsg.googlegroups.com>,
Thanks very much. This an instructive example indeed: I can now see
that a limit point needs only to be arbitrarily close to some other
point of the set, not "immediately adjacent" (i.e. without a gap) to
it -that was the mistake which led to my posting.
Frank
Dave L. Renfro
> Thanks very much. This an instructive example indeed:
> I can now see that a limit point needs only to be
> arbitrarily close to some other point of the set,
> not "immediately adjacent" (i.e. without a gap) to
> it -that was the mistake which led to my posting.
It turns out that the collection of closed sets is
equal to the collection of sets that are a "limit
point set" of some set. That is, the set of limit
points of any subset of the reals is a closed set
of the reals, and any closed set of the reals is
the set of limit points for some set. [The last set
mentioned is usually, but not always, different from
the second to last set mentioned.]
Dave L. Renfro
Faton Berisha
> I have extreme difficulty, however, imagining a set that would be a
> bounded, infinite subset of R having exactly one limit point x0. Could
> someone give an example of such a set?
>
The set of values { a_n : n\in N } for any convergent sequence (a_n)
satisfying the following condition
(\forall n)(\exists m)(m > n \land a_m \ne a_n)
will do. (The condition could be understood as not converging "constantly".)
>
> Frank
>
Faton Berisha
frank
> frank wrote:
> > Thanks very much. This an instructive example indeed:
> > arbitrarily close to some otherpointof the set,
> > not "immediately adjacent" (i.e. without a gap) to
> > it -that was the mistake which led to my posting.
>
> It turns out that the collection of closed sets is
> points of any subset of the reals is a closed set
> of the reals, and any closed set of the reals is
> mentioned is usually, but not always, different from
> the second to last set mentioned.]
>
> Dave L. Renfro
Thanks. I have been thinking on this. To go from a set S of reals to
the corresponding set L of limit points, I would think the recipe is
to add all the boundary points of S and to discard all the isolated
points of S. Then L would be uniquely defined by S.
To go from a given set L to a set S for which L is a limit point set,
you can add as many isolated points as you would care for, and you can
discard as many boundary points from L as you would care for. So this
means that if L is different from R, infinitely many sets S would be
possible.
Right?
Frank
frank
>
> It turns out that the collection of closed sets is
> equal to the collection of sets that are a "limitpointset" of some set. That is, the set oflimit
> points of any subset of the reals is a closed set
> of the reals, and any closed set of the reals is
> the set oflimitpoints for some set. [The last set
> mentioned is usually, but not always, different from
> the second to last set mentioned.]
>
> Dave L. Renfro
Dave,
Could you have a look at this comment, just to see if I got it right?
I posted it earlier but got no response - perhaps it was not clear
that is was meant as a response on your post.
Here it goes:
"Thanks. I have been thinking on this. To go from a set S of reals to
the corresponding set L of limit points, I would think the recipe is
to add all the boundary points of S and to discard all the isolated
points of S. Then L would be uniquely defined by S.
To go from a given set L to a set S for which L is a limit point set,
you can add as many isolated points as you would care for, and you
can
discard as many boundary points from L as you would care for. So this
means that if L is different from R, infinitely many sets S would be
possible."
Right? Thanks again,
Frank
Dave L. Renfro
>> It turns out that the collection of closed sets
>> is equal to the collection of sets that are a
>> "limitpointset" of some set. That is, the set of limit
>> points of any subset of the reals is a closed set
>> of the reals, and any closed set of the reals is
>> the set oflimitpoints for some set. [The last set
>> mentioned is usually, but not always, different from
>> the second to last set mentioned.]
frank wrote:
> Dave,
> Could you have a look at this comment, just to see
> if I got it right? I posted it earlier but got no
> response - perhaps it was not clear that is was meant
> as a response on your post. Here it goes:
> "Thanks. I have been thinking on this. To go from
> a set S of reals to the corresponding set L of limit
> points, I would think the recipe is to add all the
> boundary points of S and to discard all the isolated
> points of S. Then L would be uniquely defined by S.
> To go from a given set L to a set S for which L is
> a limit point set, you can add as many isolated points
> as you would care for, and you can discard as many
> boundary points from L as you would care for. So this
> means that if L is different from R, infinitely many
> sets S would be possible."
> Right? Thanks again,
Sorry. I haven't had internet access at home for the past
several weeks because of a problem with the computer I
use for the internet, something I've been rather lazy
about looking into (i.e. taking the computer to the
shop I bought it from 1.5 years ago and having them look
at it). I saw your post when I came into work Sat. for
a few minutes to post something I had typed up
at home (the mixed partial derivatives history post),
but I didn't stay to answer your question becasue I had
something else I wanted to get done this past weekend
(a LaTex'ed and translation of Suslin's 1917 paper
on Borel and analytic sets), and I figured I'd take
care of your question Monday, only I was busy with
other things and completely forgot.
I'm not quite sure what your construction involves,
but maybe I'm misreading it -- I only have a few moments
now (it seems a bit "hand-wavey'ish"). Here's one way.
First, the set of limit points of any set is a closed
set. This can be seen by showing that the set of limit
points contains all of *it's* limit points, and one way
to show this is to show that a "limit of limits is a limit".
If each y_1, y_2, ... is a limit point of a set S and the
sequence {y_n} converges to a point (we'll call) y, then
we need to show that y is also a limit point of S. O-K,
since each y_n is a limit point of S, then we know for
each n there is a sequence {s_n_k} in S that converges
to y_n. See if you can show the "diagonal sequence"
s_1_1, s_2_2, s_3_3, ... converges to y. It may help
to draw a schematic diagram of a special case, such as
S = rationals, y_n = 1 / sqrt(n), and s_n_k = y_n + 1/k.
Now for the other half, where for each closed set C
I claim there is a set S such that the collection
of limit points of S is equal to C. Write C as C* union C#,
where each point in C* is a limit point of C and each
point in C# is not a limit point of C (i.e. C# is the
collection of points isolated from C). For each point
x in C#, let S_x be a sequence converging to x, all
of whose terms are within a distance of r_x/2 from
each point of C besides x, where r_x is such that
the ball B(x, r_x) contains only the point x of C.
[I'm using r_x/2 because there might be a problem
with the sequences, for different points x, having
initial terms that cluster to points not in C.]
Then I believe C* union union{S_x: x in C#}, that is,
union{C* union S_x: x in C#}, is a set whose collection
of limit points is exactly the set C.
There might be an easier way this other half, by the way.
Dave L. Renfro
Dave L. Renfro
> It may help to draw a schematic diagram of
> a special case, such as S = rationals,
> y_n = 1 / sqrt(n), and s_n_k = y_n + 1/k.
It helps even more if you get it correct!
(Which I didn't.) The points s_n_k have
to be points in S (the rationals), and they
come from sequences of rationals that
you pick converging to the points y_n.
Dave L. Renfro
frank
>
> Sorry. I haven't had internet access at home for the past
> several weeks because of a problem with the computer I
> use for the internet, something I've been rather lazy
> about looking into (i.e. taking the computer to the
> shop I bought it from 1.5 years ago and having them look
> at it). I saw your post when I came into work Sat. for
> a few minutes to post something I had typed up
> at home (the mixed partial derivatives history post),
> but I didn't stay to answer your question becasue I had
> something else I wanted to get done this past weekend
> (a LaTex'ed and translation of Suslin's 1917 paper
> on Borel and analytic sets), and I figured I'd take
> care of your question Monday, only I was busy with
> other things and completely forgot.
Sorry, I did not mean to push you that hard; there was no urgency
involved. Actually, I am a nuclear physician with a special interest
in mathematics. My busy job implies that I have only limited time to
study. After work, I am mostly too tired to embark on difficult
reasonings - today is no exception.So that means my efforts, if any,
are mostly confined to weekends. Anyhow, I am much obliged that you
took the time to respond to my musings; especially now that I can see
that you, as I expected, have more serious business to attend to..
>
> I'm not quite sure what your construction involves,
> but maybe I'm misreading it -- I only have a few moments
> now (it seems a bit "hand-wavey'ish"). Here's one way.
My construction was not meant as a proof, but rather as an intuitive
reasoning. As a result of that reasoning, I came to the conclusion
that each set S in R would have a UNIIQUE corresponding set of its
limit points, whereas there are INFINITELY many sets which have S as
the set of their limit points.My question was whether you would
approve this conclusion.
>
> First, the set oflimitpoints of any set is a closed
> set. This can be seen by showing that the set oflimit
> points contains all of *it's*limitpoints, and one way
> to show this is to show that a "limitof limits is alimit".
>
> If each y_1, y_2, ... is alimitpointof a set S and the
> sequence {y_n} converges to apoint(we'll call) y, then
> we need to show that y is also alimitpointof S. O-K,
> since each y_n is alimitpointof S, then we know for
> each n there is a sequence {s_n_k} in S that converges
> to y_n. See if you can show the "diagonal sequence"
> s_1_1, s_2_2, s_3_3, ... converges to y. It may help
> to draw a schematic diagram of a special case, such as
> S = rationals, y_n = 1 / sqrt(n), and s_n_k = y_n + 1/k.
>
> Now for the other half, where for each closed set C
> I claim there is a set S such that the collection
> oflimitpoints of S is equal to C. Write C as C* union C#,
> where eachpointin C* is alimitpointof C and eachpointin C# is not alimitpointof C (i.e. C# is the
> collection of points isolated from C). For eachpoint
> x in C#, let S_x be a sequence converging to x, all
> of whose terms are within a distance of r_x/2 from
> eachpointof C besides x, where r_x is such that
> the ball B(x, r_x) contains only thepointx of C.
> [I'm using r_x/2 because there might be a problem
> with the sequences, for different points x, having
> initial terms that cluster to points not in C.]
> Then I believe C* union union{S_x: x in C#}, that is,
> union{C* union S_x: x in C#}, is a set whose collection
> oflimitpoints is exactly the set C.
As to part one, I have an inkling on what it is about, but part two is
definitely weekend stuff - my mind is too obtunded now. I'll get in
touch in some days. Anyhow, I can see you use different (but, I
presume, equivalent) definitions of limit points - the one in my book
(SA Douglass) is any point the deleted neigbourhood of which contains
elements of the set. Anyhow, for the time being, many thanks.
Dave L. Renfro
> My construction was not meant as a proof, but rather
> as an intuitive reasoning. As a result of that
> reasoning, I came to the conclusion that each set S
> in R would have a UNIIQUE corresponding set of its
> limit points, whereas there are INFINITELY many
> sets which have S as the set of their limit points.
> My question was whether you would approve this
> conclusion.
This conclusion seems correct to me. If the closed set
is the empty set, then any subset of the positive
integers will work. If the closed set contains a point,
say x, then the union of any *infinite* subset of
{x + 1/n : n = 1, 2, 3, ...},
along with a fixed example of a set whose collection
of limit points is the given closed set, will work.
To be a little more explicit with the second part,
let C be the given closed set, let x be a point of C,
and let E be a set such that the collection of limit
points of E is equal to C. Then the collection of
limit points of 'E union D', where D is any infinite
subset of {x + 1/n : n = 1, 2, 3, ...}, is equal to C.
One way to prove this is to use the fact that
L(A union B) = L(A) union L(B), where 'L' is the
operation whose input is a set and whose output is
the collection of all limit points of the input set.
(This operation is often called the "derived set
operator").
Remark for others who might be reading: It seems to
me that the derived set operator is continuum-to-one
at each countable closed set and 2^continuum-to-one
at each uncountable closed set.
> As to part one, I have an inkling on what it is about,
> but part two is definitely weekend stuff - my mind
> is too obtunded now. I'll get in touch in some days.
> Anyhow, I can see you use different (but, I presume,
> equivalent) definitions of limit points - the one
> in my book (SA Douglass) is any point the deleted
> neigbourhood of which contains elements of the set.
> Anyhow, for the time being, many thanks.
For the real numbers (even R^n; even metric spaces),
it is not difficult to see that your use of "x is a
limit point of E" is equivalent to "there exists a
sequence in E - {x} that converges to x". [Note the
small correction I'm making. I noticed, while proving
this, that I need to exclude the case where x is
in E and we use a sequence such that all of its terms
are equal to x after some sequence-index value.]
For the ==> half, suppose E has the property you gave
with respect to x. Pick x_1 in E - {x} and in the interior
of the ball of radius 1 about x. Pick x_2 in E - {x}
and in the interior of the ball of radius |x - x_1|
about x. Pick x_3 in E - {x} and in the interior of
the ball of radius |x - x_2| about x. Keep going in
this way. Then {x_n} is a sequence in E that converges
to x.
For the <== half, suppose x is the limit of some sequence
of points in E - {x}, say x is the limit of {x_n}. Let
B(x,r) be any ball about x. Because x_n --> x, there must
be an index k such that x_k is within a distance of r
from x. [This follows from the much stronger result we
know to be true, namely that there must be an index k
such that *every* term of the sequence, whose index is
k or greater than k, belongs to B(x,r).] Since {x_n}
is a sequence in E - {x}, x_k is not equal to x. Hence,
we have a point of E, namely x_k, that belongs to
the deleated r-neighborhood of x.
Dave L. Renfro
Dave L. Renfro
>> My construction was not meant as a proof, but rather
>> as an intuitive reasoning. As a result of that
>> reasoning, I came to the conclusion that each set S
>> in R would have a UNIIQUE corresponding set of its
>> limit points, whereas there are INFINITELY many
>> sets which have S as the set of their limit points.
>> My question was whether you would approve this
>> conclusion.
Dave L. Renfro wrote (in part):
> This conclusion seems correct to me. If the closed set
> is the empty set, then any subset of the positive
> integers will work. If the closed set contains a point,
> say x, then the union of any *infinite* subset of
>
> {x + 1/n : n = 1, 2, 3, ...},
>
> along with a fixed example of a set whose collection
> of limit points is the given closed set, will work.
> To be a little more explicit with the second part,
> let C be the given closed set, let x be a point of C,
> and let E be a set such that the collection of limit
> points of E is equal to C. Then the collection of
> limit points of 'E union D', where D is any infinite
> subset of {x + 1/n : n = 1, 2, 3, ...}, is equal to C.
> One way to prove this is to use the fact that
> L(A union B) = L(A) union L(B), where 'L' is the
> operation whose input is a set and whose output is
> the collection of all limit points of the input set.
> (This operation is often called the "derived set
> operator").
Curiously, while working on something else I just happened
to come across a passage in a paper by Cantor, who originated
the notion of the derived set of a set (= the set of limit
points of a set), where he makes the observation that each
closed set is equal to the derived set of infinitely many
other sets. The comment can be found on p. 470 of the following
paper, a paper that is on the internet at the URL I've given.
I think the relevant passage is one of the two italicized
sentences just below the halfway position from the top of
the page.
Georg Cantor, "Ueber unendliche, lineare Punktmannichfaltigkeiten",
Mathematische Annalen 23 (1884), 453-488.
http://dz1.gdz-cms.de/no_cache/dms/load/img/?IDDOC=26526
Dave L. Renfro