[math-fun] Cantor set of measure 1?
Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3, ...) as the set obtained from S_{n-1} by replacing each of the 2^n closed intervals [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half, duplicate the midpoint, shift the left half to the left by 1/3^n, and shift the right half to the right by 1/3^n). It seems to me that at each stage we get 2^n disjoint closed intervals of length 1/2^n separated from one another by 1/3^n, and that every point (including the duplicate points created along the way) moves a finite distance and therefore has a well-defined location "at time infinity". So we seem to get a well-defined set "at time infinity". Am I right? And do we obtain a set of measure 1 that's homeomorphic to the Cantor set? I've seen constructions of "fat Cantor sets" on the web, where one modifies the culling procedure, but I've never seen one that uses sliding and preserves the measure throughout the procedure. So I'm worried that I'm overlooking something. Jim Propp
This is just the IFS { x -> (1/2) x - 1/3, x -> (1/2) x + 5/6 }. It satisfies the open set condition, so yes, it is homeomorphic to a Cantor set and has Hausdorff and similarity dimension = 1. -Thomas C On Tue, Sep 8, 2015 at 9:42 AM, James Propp <jamespropp@gmail.com> wrote:
Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3, ...) as the set obtained from S_{n-1} by replacing each of the 2^n closed intervals [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half, duplicate the midpoint, shift the left half to the left by 1/3^n, and shift the right half to the right by 1/3^n).
It seems to me that at each stage we get 2^n disjoint closed intervals of length 1/2^n separated from one another by 1/3^n, and that every point (including the duplicate points created along the way) moves a finite distance and therefore has a well-defined location "at time infinity". So we seem to get a well-defined set "at time infinity".
Am I right? And do we obtain a set of measure 1 that's homeomorphic to the Cantor set?
I've seen constructions of "fat Cantor sets" on the web, where one modifies the culling procedure, but I've never seen one that uses sliding and preserves the measure throughout the procedure. So I'm worried that I'm overlooking something.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
----- "It seems to me that at each stage we get 2^n disjoint closed intervals of length 1/2^n separated from one another by 1/3^n, and that every point (including the duplicate points created along the way) moves a finite distance and therefore has a well-defined location 'at time infinity'. So we seem to get a well-defined set 'at time infinity'." ----- If I understand the construction correctly, it can just as well go on within the limiting convex hull of all the points created,* namely [0-K, 1+K] where K = 1/3 + 1/3^3 + 1/3^3 +... = 1/2, i.e., [-1/2, 3/2]. * It can just as well go on in [-1/2,3/2] because we can always dilate each stage about its midpoint by the appropriate factor so it exactly fits into [-1/2,3/2]. If that and the quoted paragraph above are correct, then IF this dilatation of each stage into [-1/2,3/2] is done, then the nth stage will consist of 2^n closed intervals of equal length L_n, separated by 2^n - 1 open intervals of length M_n = (2/3)^n * L_n, such that 2 (= 3/2 - (-1/2)) = 2^n * L_n + (2^n - 1) * (2/3)^n * L_n. (And so if we care to find L_n, it must be L_n = 2 / [2^n + (2^n - 1)*(2/3)^n].) THEN it appears to me that, however the limiting set X is defined, there will (by symmetry) be a countable dense set of c in [0,2] for which the translations x |-> x + c will take a nonempty interval's worth of X to a different equal-size interval's worth of X. This implies that X contains points dense in some interval. But since the Cantor set is closed in R no matter how it is embedded, this implies it contains an entire interval, which is impossible for a Cantor set. Meanwhile, Thomas Colthurst's post (Hi, Tom!) has arrived, completely negating what I have hypothesized here. —Dan
On Sep 8, 2015, at 6:42 AM, James Propp <jamespropp@gmail.com> wrote:
Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3, ...) as the set obtained from S_{n-1} by replacing each of the 2^n closed intervals [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half, duplicate the midpoint, shift the left half to the left by 1/3^n, and shift the right half to the right by 1/3^n).
It seems to me that at each stage we get 2^n disjoint closed intervals of length 1/2^n separated from one another by 1/3^n, and that every point (including the duplicate points created along the way) moves a finite distance and therefore has a well-defined location "at time infinity". So we seem to get a well-defined set "at time infinity".
Am I right? And do we obtain a set of measure 1 that's homeomorphic to the Cantor set?
I've seen constructions of "fat Cantor sets" on the web, where one modifies the culling procedure, but I've never seen one that uses sliding and preserves the measure throughout the procedure. So I'm worried that I'm overlooking something.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Hi, Dan! For what it is worth, I get a different convex hull: [-2/3, 5/3]. The left end point is the fixed point of x -> (1/2) x - 1/3 and the right end point is the fixed point of x -> (1/2) x + 5/6. And actually, you are right about this not being a Cantor set; I was wrong about it satisfying the open set condition. Specifically, the first function maps [-2/3, 5/3] to [-2/3, 1/2] and the second function maps [-2/3, 5/3] to [1/2, 5/3]. So the final attractor is just the line segment [-2/3, 5/3]. Oh well, at least I was right about it having measure 1. :) -Thomas C On Tue, Sep 8, 2015 at 10:39 AM, Dan Asimov <asimov@msri.org> wrote:
----- "It seems to me that at each stage we get 2^n disjoint closed intervals of length 1/2^n separated from one another by 1/3^n, and that every point (including the duplicate points created along the way) moves a finite distance and therefore has a well-defined location 'at time infinity'. So we seem to get a well-defined set 'at time infinity'." -----
If I understand the construction correctly, it can just as well go on within the limiting convex hull of all the points created,* namely [0-K, 1+K] where
K = 1/3 + 1/3^3 + 1/3^3 +... = 1/2,
i.e., [-1/2, 3/2].
* It can just as well go on in [-1/2,3/2] because we can always dilate each stage about its midpoint by the appropriate factor so it exactly fits into [-1/2,3/2].
If that and the quoted paragraph above are correct, then IF this dilatation of each stage into [-1/2,3/2] is done, then the nth stage will consist of 2^n closed intervals of equal length L_n, separated by 2^n - 1 open intervals of length
M_n = (2/3)^n * L_n,
such that
2 (= 3/2 - (-1/2)) = 2^n * L_n + (2^n - 1) * (2/3)^n * L_n.
(And so if we care to find L_n, it must be L_n = 2 / [2^n + (2^n - 1)*(2/3)^n].)
THEN it appears to me that, however the limiting set X is defined, there will (by symmetry) be a countable dense set of c in [0,2] for which the translations
x |-> x + c
will take a nonempty interval's worth of X to a different equal-size interval's worth of X.
This implies that X contains points dense in some interval. But since the Cantor set is closed in R no matter how it is embedded, this implies it contains an entire interval, which is impossible for a Cantor set.
Meanwhile, Thomas Colthurst's post (Hi, Tom!) has arrived, completely negating what I have hypothesized here.
—Dan
On Sep 8, 2015, at 6:42 AM, James Propp <jamespropp@gmail.com> wrote:
Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3, ...) as the set obtained from S_{n-1} by replacing each of the 2^n closed intervals [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half, duplicate the midpoint, shift the left half to the left by 1/3^n, and shift the right half to the right by 1/3^n).
It seems to me that at each stage we get 2^n disjoint closed intervals of length 1/2^n separated from one another by 1/3^n, and that every point (including the duplicate points created along the way) moves a finite distance and therefore has a well-defined location "at time infinity". So we seem to get a well-defined set "at time infinity".
Am I right? And do we obtain a set of measure 1 that's homeomorphic to the Cantor set?
I've seen constructions of "fat Cantor sets" on the web, where one modifies the culling procedure, but I've never seen one that uses sliding and preserves the measure throughout the procedure. So I'm worried that I'm overlooking something.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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On Sep 8, 2015, at 7:51 AM, Thomas Colthurst <thomaswc@gmail.com> wrote:
Hi, Dan!
For what it is worth, I get a different convex hull: [-2/3, 5/3]. The left end point is the fixed point of x -> (1/2) x - 1/3 and the right end point is the fixed point of x -> (1/2) x + 5/6.
And actually, you are right about this not being a Cantor set; I was wrong about it satisfying the open set condition. Specifically, the first function maps [-2/3, 5/3] to [-2/3, 1/2] and the second function maps [-2/3, 5/3] to [1/2, 5/3]. So the final attractor is just the line segment [-2/3, 5/3].
Oh well, at least I was right about it having measure 1. :)
Okay. (I also agreed with that.) —Dan
On Tue, Sep 8, 2015 at 10:39 AM, Dan Asimov <asimov@msri.org> wrote:
----- "It seems to me that at each stage we get 2^n disjoint closed intervals of length 1/2^n separated from one another by 1/3^n, and that every point (including the duplicate points created along the way) moves a finite distance and therefore has a well-defined location 'at time infinity'. So we seem to get a well-defined set 'at time infinity'." -----
If I understand the construction correctly, it can just as well go on within the limiting convex hull of all the points created,* namely [0-K, 1+K] where
K = 1/3 + 1/3^3 + 1/3^3 +... = 1/2,
i.e., [-1/2, 3/2].
* It can just as well go on in [-1/2,3/2] because we can always dilate each stage about its midpoint by the appropriate factor so it exactly fits into [-1/2,3/2].
If that and the quoted paragraph above are correct, then IF this dilatation of each stage into [-1/2,3/2] is done, then the nth stage will consist of 2^n closed intervals of equal length L_n, separated by 2^n - 1 open intervals of length
M_n = (2/3)^n * L_n,
such that
2 (= 3/2 - (-1/2)) = 2^n * L_n + (2^n - 1) * (2/3)^n * L_n.
(And so if we care to find L_n, it must be L_n = 2 / [2^n + (2^n - 1)*(2/3)^n].)
THEN it appears to me that, however the limiting set X is defined, there will (by symmetry) be a countable dense set of c in [0,2] for which the translations
x |-> x + c
will take a nonempty interval's worth of X to a different equal-size interval's worth of X.
This implies that X contains points dense in some interval. But since the Cantor set is closed in R no matter how it is embedded, this implies it contains an entire interval, which is impossible for a Cantor set.
Meanwhile, Thomas Colthurst's post (Hi, Tom!) has arrived, completely negating what I have hypothesized here.
—Dan
On Sep 8, 2015, at 6:42 AM, James Propp <jamespropp@gmail.com> wrote:
Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3, ...) as the set obtained from S_{n-1} by replacing each of the 2^n closed intervals [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half, duplicate the midpoint, shift the left half to the left by 1/3^n, and shift the right half to the right by 1/3^n).
It seems to me that at each stage we get 2^n disjoint closed intervals of length 1/2^n separated from one another by 1/3^n, and that every point (including the duplicate points created along the way) moves a finite distance and therefore has a well-defined location "at time infinity". So we seem to get a well-defined set "at time infinity".
Am I right? And do we obtain a set of measure 1 that's homeomorphic to the Cantor set?
I've seen constructions of "fat Cantor sets" on the web, where one modifies the culling procedure, but I've never seen one that uses sliding and preserves the measure throughout the procedure. So I'm worried that I'm overlooking something.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I still don't see why the bijection between my original set and the product set {0,1} x {0,1} x ... fails to be a homeomorphism. Can someone explain that to me? Jim On Tue, Sep 8, 2015 at 10:58 AM, Dan Asimov <asimov@msri.org> wrote:
On Sep 8, 2015, at 7:51 AM, Thomas Colthurst <thomaswc@gmail.com> wrote:
Hi, Dan!
For what it is worth, I get a different convex hull: [-2/3, 5/3]. The left end point is the fixed point of x -> (1/2) x - 1/3 and the right end point is the fixed point of x -> (1/2) x + 5/6.
And actually, you are right about this not being a Cantor set; I was wrong about it satisfying the open set condition. Specifically, the first function maps [-2/3, 5/3] to [-2/3, 1/2] and the second function maps [-2/3, 5/3] to [1/2, 5/3]. So the final attractor is just the line segment [-2/3, 5/3].
Oh well, at least I was right about it having measure 1. :)
Okay. (I also agreed with that.)
—Dan
On Tue, Sep 8, 2015 at 10:39 AM, Dan Asimov <asimov@msri.org> wrote:
----- "It seems to me that at each stage we get 2^n disjoint closed intervals of length 1/2^n separated from one another by 1/3^n, and that every point (including the duplicate points created along the way) moves a finite distance and therefore has a well-defined location 'at time infinity'. So we seem to get a well-defined set 'at time infinity'." -----
If I understand the construction correctly, it can just as well go on within the limiting convex hull of all the points created,* namely [0-K, 1+K] where
K = 1/3 + 1/3^3 + 1/3^3 +... = 1/2,
i.e., [-1/2, 3/2].
* It can just as well go on in [-1/2,3/2] because we can always dilate each stage about its midpoint by the appropriate factor so it exactly fits into [-1/2,3/2].
If that and the quoted paragraph above are correct, then IF this dilatation of each stage into [-1/2,3/2] is done, then the nth stage will consist of 2^n closed intervals of equal length L_n, separated by 2^n - 1 open intervals of length
M_n = (2/3)^n * L_n,
such that
2 (= 3/2 - (-1/2)) = 2^n * L_n + (2^n - 1) * (2/3)^n * L_n.
(And so if we care to find L_n, it must be L_n = 2 / [2^n + (2^n - 1)*(2/3)^n].)
THEN it appears to me that, however the limiting set X is defined, there will (by symmetry) be a countable dense set of c in [0,2] for which the translations
x |-> x + c
will take a nonempty interval's worth of X to a different equal-size interval's worth of X.
This implies that X contains points dense in some interval. But since the Cantor set is closed in R no matter how it is embedded, this implies it contains an entire interval, which is impossible for a Cantor set.
Meanwhile, Thomas Colthurst's post (Hi, Tom!) has arrived, completely negating what I have hypothesized here.
—Dan
On Sep 8, 2015, at 6:42 AM, James Propp <jamespropp@gmail.com> wrote:
Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3, ...) as the set obtained from S_{n-1} by replacing each of the 2^n closed intervals [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half, duplicate the midpoint, shift the left half to the left by 1/3^n, and shift the right half to the right by 1/3^n).
It seems to me that at each stage we get 2^n disjoint closed intervals of length 1/2^n separated from one another by 1/3^n, and that every point (including the duplicate points created along the way) moves a finite distance and therefore has a well-defined location "at time infinity". So we seem to get a well-defined set "at time infinity".
Am I right? And do we obtain a set of measure 1 that's homeomorphic to the Cantor set?
I've seen constructions of "fat Cantor sets" on the web, where one modifies the culling procedure, but I've never seen one that uses sliding and preserves the measure throughout the procedure. So I'm worried that I'm overlooking something.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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(Or am I confused about the ordinary Cantor set being homeomorphic to {0,1}
x {0,1} x ...?)
Jim
On Tue, Sep 8, 2015 at 12:20 PM, James Propp <jamespropp@gmail.com> wrote:
> I still don't see why the bijection between my original set and the
> product set {0,1} x {0,1} x ... fails to be a homeomorphism. Can someone
> explain that to me?
>
> Jim
>
> On Tue, Sep 8, 2015 at 10:58 AM, Dan Asimov <asimov@msri.org> wrote:
>
>> > On Sep 8, 2015, at 7:51 AM, Thomas Colthurst <thomaswc@gmail.com>
>> wrote:
>> >
>> > Hi, Dan!
>> >
>> > For what it is worth, I get a different convex hull: [-2/3, 5/3]. The
>> > left end point is the fixed point of x -> (1/2) x - 1/3 and the right
>> end
>> > point is the fixed point of x -> (1/2) x + 5/6.
>> >
>> > And actually, you are right about this not being a Cantor set; I was
>> wrong
>> > about it satisfying the open set condition. Specifically, the first
>> > function maps [-2/3, 5/3] to [-2/3, 1/2] and the second function maps
>> > [-2/3, 5/3] to [1/2, 5/3]. So the final attractor is just the line
>> segment
>> > [-2/3, 5/3].
>> >
>> > Oh well, at least I was right about it having measure 1. :)
>>
>> Okay. (I also agreed with that.)
>>
>> —Dan
>>
>>
>> > On Tue, Sep 8, 2015 at 10:39 AM, Dan Asimov <asimov@msri.org> wrote:
>> >
>> >> -----
>> >> "It seems to me that at each stage we get 2^n disjoint closed
>> intervals of
>> >> length 1/2^n separated from one another by 1/3^n, and that every point
>> >> (including the duplicate points created along the way) moves a finite
>> >> distance and therefore has a well-defined location 'at time infinity'.
>> So
>> >> we seem to get a well-defined set 'at time infinity'."
>> >> -----
>> >>
>> >> If I understand the construction correctly, it can just as well go on
>> >> within the limiting convex hull of all the points created,* namely
>> [0-K,
>> >> 1+K] where
>> >>
>> >> K = 1/3 + 1/3^3 + 1/3^3 +... = 1/2,
>> >>
>> >> i.e., [-1/2, 3/2].
>> >>
>> >> * It can just as well go on in [-1/2,3/2] because we can always dilate
>> >> each stage about its midpoint by the appropriate factor so it exactly
>> fits
>> >> into [-1/2,3/2].
>> >>
>> >> If that and the quoted paragraph above are correct, then IF this
>> >> dilatation of each stage into [-1/2,3/2] is done, then the nth stage
>> will
>> >> consist of 2^n closed intervals of equal length L_n, separated by 2^n
>> - 1
>> >> open intervals of length
>> >>
>> >> M_n = (2/3)^n * L_n,
>> >>
>> >> such that
>> >>
>> >> 2 (= 3/2 - (-1/2)) = 2^n * L_n + (2^n - 1) * (2/3)^n * L_n.
>> >>
>> >> (And so if we care to find L_n, it must be L_n = 2 / [2^n + (2^n -
>> >> 1)*(2/3)^n].)
>> >>
>> >> THEN it appears to me that, however the limiting set X is defined,
>> there
>> >> will (by symmetry) be a countable dense set of c in [0,2] for which the
>> >> translations
>> >>
>> >> x |-> x + c
>> >>
>> >> will take a nonempty interval's worth of X to a different equal-size
>> >> interval's worth of X.
>> >>
>> >> This implies that X contains points dense in some interval. But since
>> the
>> >> Cantor set is closed in R no matter how it is embedded, this implies it
>> >> contains an entire interval, which is impossible for a Cantor set.
>> >>
>> >> Meanwhile, Thomas Colthurst's post (Hi, Tom!) has arrived, completely
>> >> negating what I have hypothesized here.
>> >>
>> >> —Dan
>> >>
>> >>
>> >>> On Sep 8, 2015, at 6:42 AM, James Propp <jamespropp@gmail.com> wrote:
>> >>>
>> >>> Starting from S_0 = [0,1], iteratively define S_n (for n = 1, 2, 3,
>> ...)
>> >> as
>> >>> the set obtained from S_{n-1} by replacing each of the 2^n closed
>> >> intervals
>> >>> [a,b] by the two closed intervals [a-1/3^n,(a+b)/2-1/3^n] and
>> >>> [(a+b)/2+1/3^n, b+1/3^n] (that is, we break the interval in half,
>> >> duplicate
>> >>> the midpoint, shift the left half to the left by 1/3^n, and shift the
>> >> right
>> >>> half to the right by 1/3^n).
>> >>
>> >>> It seems to me that at each stage we get 2^n disjoint closed
>> intervals of
>> >>> length 1/2^n separated from one another by 1/3^n, and that every point
>> >>> (including the duplicate points created along the way) moves a finite
>> >>> distance and therefore has a well-defined location "at time
>> infinity". So
>> >>> we seem to get a well-defined set "at time infinity".
>> >>
>> >>> Am I right? And do we obtain a set of measure 1 that's homeomorphic to
>> >> the
>> >>> Cantor set?
>> >>>
>> >>> I've seen constructions of "fat Cantor sets" on the web, where one
>> >> modifies
>> >>> the culling procedure, but I've never seen one that uses sliding and
>> >>> preserves the measure throughout the procedure. So I'm worried that
>> I'm
>> >>> overlooking something.
>> >>>
>> >>> Jim Propp
>> >>> _______________________________________________
>> >>> math-fun mailing list
>> >>> math-fun@mailman.xmission.com
>> >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
>> >>
>> >>
>> >> _______________________________________________
>> >> math-fun mailing list
>> >> math-fun@mailman.xmission.com
>> >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
>> >>
>> > _______________________________________________
>> > math-fun mailing list
>> > math-fun@mailman.xmission.com
>> > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
>>
>>
>> _______________________________________________
>> math-fun mailing list
>> math-fun@mailman.xmission.com
>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
>>
>
>
To evaluate whether the bijection is a homeomorphism, it would be immensely helpful if you defined it. —Dan
On Sep 8, 2015, at 9:20 AM, James Propp <jamespropp@gmail.com> wrote:
I still don't see why the bijection between my original set and the product set {0,1} x {0,1} x ... fails to be a homeomorphism. Can someone explain that to me?
Good point, Dan! For all but countably many points (namely the dyadic rationals), we use the binary representation of the original point in [0,1]. For the dyadic rationals, we need to know which of the two "clones" we're using. E.g., when we split [0,1] into two pieces and give each piece an endpoint (creating a new point out of thin air), we give the right endpoint of the left piece the label .0111... and the left endpoint of the right piece the label .1000... In this way, we get a labelling of the points of my set using infinite strings of bits, where each string corresponds to a unique point in the set, and vice versa. Hopefully that's clear. Jim On Tue, Sep 8, 2015 at 12:22 PM, Dan Asimov <asimov@msri.org> wrote:
To evaluate whether the bijection is a homeomorphism, it would be immensely helpful if you defined it.
—Dan
On Sep 8, 2015, at 9:20 AM, James Propp <jamespropp@gmail.com> wrote:
I still don't see why the bijection between my original set and the product set {0,1} x {0,1} x ... fails to be a homeomorphism. Can someone explain that to me?
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Sep 8, 2015, at 9:27 AM, James Propp <jamespropp@gmail.com> wrote:
Good point, Dan!
For all but countably many points (namely the dyadic rationals), we use the binary representation of the original point in [0,1].
For the dyadic rationals, we need to know which of the two "clones" we're using.
E.g., when we split [0,1] into two pieces and give each piece an endpoint (creating a new point out of thin air), we give the right endpoint of the left piece the label .0111... and the left endpoint of the right piece the label .1000...
In this way, we get a labelling of the points of my set using infinite strings of bits, where each string corresponds to a unique point in the set, and vice versa.
Hopefully that's clear.
Not to me it isn't. I don't even know how you are defining your set, since you never stated that. (Yes, you stated what the stages are, but not how the final set is defined in terms of the stages. And you don't even seem to be defining your bijection above in terms of the final set, but only in terms of the stages. —Dan
Jim
On Tue, Sep 8, 2015 at 12:22 PM, Dan Asimov <asimov@msri.org <mailto:asimov@msri.org>> wrote:
To evaluate whether the bijection is a homeomorphism, it would be immensely helpful if you defined it.
Part of what I had in mind is that points in the final set arise from points in the original interval [0,1] through a process of repeated sliding (and, in the case of dyadic rationals, "cloning"). If we look at the sequence of slide operation (Left or Right) that a point undergoes, e.g., Left, Right, Right, Left, Right, Right, ... and replace each Left by a 0 and each Right by a 1, we get an infinite string of bits; this is the intended bijection between my set of points and the product set {0,1} x {0,1} x {0,1} x ... Clear now? (Apologies in advance if still murky.) Jim On Tue, Sep 8, 2015 at 12:35 PM, Dan Asimov <dasimov@earthlink.net> wrote:
On Sep 8, 2015, at 9:27 AM, James Propp <jamespropp@gmail.com> wrote:
Good point, Dan!
For all but countably many points (namely the dyadic rationals), we use the binary representation of the original point in [0,1].
For the dyadic rationals, we need to know which of the two "clones" we're using.
E.g., when we split [0,1] into two pieces and give each piece an endpoint (creating a new point out of thin air), we give the right endpoint of the left piece the label .0111... and the left endpoint of the right piece the label .1000...
In this way, we get a labelling of the points of my set using infinite strings of bits, where each string corresponds to a unique point in the set, and vice versa.
Hopefully that's clear.
Not to me it isn't.
I don't even know how you are defining your set, since you never stated that. (Yes, you stated what the stages are, but not how the final set is defined in terms of the stages.
And you don't even seem to be defining your bijection above in terms of the final set, but only in terms of the stages.
—Dan
Jim
On Tue, Sep 8, 2015 at 12:22 PM, Dan Asimov <asimov@msri.org <mailto: asimov@msri.org>> wrote:
To evaluate whether the bijection is a homeomorphism, it would be immensely helpful if you defined it.
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Possibly dumb question: Do you know that the various intervals created (or at least defined) never overlap? 'Cause if they did, then some incarnations of a point x could at a later stage be asked to slide by one amount, and other incarnations by a different amount (or direction). —Dan
On Sep 8, 2015, at 12:32 PM, James Propp <jamespropp@gmail.com> wrote:
Part of what I had in mind is that points in the final set arise from points in the original interval [0,1] through a process of repeated sliding (and, in the case of dyadic rationals, "cloning"). If we look at the sequence of slide operation (Left or Right) that a point undergoes, e.g., Left, Right, Right, Left, Right, Right, ... and replace each Left by a 0 and each Right by a 1, we get an infinite string of bits; this is the intended bijection between my set of points and the product set {0,1} x {0,1} x {0,1} x ...
Clear now? (Apologies in advance if still murky.)
Jim
On Tue, Sep 8, 2015 at 12:35 PM, Dan Asimov <dasimov@earthlink.net> wrote:
On Sep 8, 2015, at 9:27 AM, James Propp <jamespropp@gmail.com> wrote:
Good point, Dan!
For all but countably many points (namely the dyadic rationals), we use the binary representation of the original point in [0,1].
For the dyadic rationals, we need to know which of the two "clones" we're using.
E.g., when we split [0,1] into two pieces and give each piece an endpoint (creating a new point out of thin air), we give the right endpoint of the left piece the label .0111... and the left endpoint of the right piece the label .1000...
In this way, we get a labelling of the points of my set using infinite strings of bits, where each string corresponds to a unique point in the set, and vice versa.
Hopefully that's clear.
Not to me it isn't.
I don't even know how you are defining your set, since you never stated that. (Yes, you stated what the stages are, but not how the final set is defined in terms of the stages.
And you don't even seem to be defining your bijection above in terms of the final set, but only in terms of the stages.
—Dan
Jim
On Tue, Sep 8, 2015 at 12:22 PM, Dan Asimov <asimov@msri.org <mailto: asimov@msri.org>> wrote:
To evaluate whether the bijection is a homeomorphism, it would be immensely helpful if you defined it.
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Assign the original set S_0 = [0,1] the empty codename. Assign the two new intervals of length 1/2 in S_1 (obtained by applying the split-and-slide operation to S_0) the codenames 0 and 1, respectively (from left to right). Assign the four new intervals of length 1/4 in S_2 (obtained by applying the split-and-slide operation to the two intervals in S_1) the codenames 00, 01, 10, and 11, respectively (from left to right). And so on like that. Claim: The 2^k intervals of length 1/2^k that constitute S_k are ordered left-to-right in accordance with their k-bit binary codenames, and the gaps between the intervals are all of size > 1/3^k. Proof: By straightforward induction (unless I'm missing something). Jim On Tue, Sep 8, 2015 at 8:14 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Possibly dumb question: Do you know that the various intervals created (or at least defined) never overlap? 'Cause if they did, then some incarnations of a point x could at a later stage be asked to slide by one amount, and other incarnations by a different amount (or direction).
—Dan
On Sep 8, 2015, at 12:32 PM, James Propp <jamespropp@gmail.com> wrote:
Part of what I had in mind is that points in the final set arise from points in the original interval [0,1] through a process of repeated sliding (and, in the case of dyadic rationals, "cloning"). If we look at the sequence of slide operation (Left or Right) that a point undergoes, e.g., Left, Right, Right, Left, Right, Right, ... and replace each Left by a 0 and each Right by a 1, we get an infinite string of bits; this is the intended bijection between my set of points and the product set {0,1} x {0,1} x {0,1} x ...
Clear now? (Apologies in advance if still murky.)
Jim
On Tue, Sep 8, 2015 at 12:35 PM, Dan Asimov <dasimov@earthlink.net> wrote:
On Sep 8, 2015, at 9:27 AM, James Propp <jamespropp@gmail.com> wrote:
Good point, Dan!
For all but countably many points (namely the dyadic rationals), we use the binary representation of the original point in [0,1].
For the dyadic rationals, we need to know which of the two "clones"
we're
using.
E.g., when we split [0,1] into two pieces and give each piece an endpoint (creating a new point out of thin air), we give the right endpoint of the left piece the label .0111... and the left endpoint of the right piece the label .1000...
In this way, we get a labelling of the points of my set using infinite strings of bits, where each string corresponds to a unique point in the set, and vice versa.
Hopefully that's clear.
Not to me it isn't.
I don't even know how you are defining your set, since you never stated that. (Yes, you stated what the stages are, but not how the final set is defined in terms of the stages.
And you don't even seem to be defining your bijection above in terms of the final set, but only in terms of the stages.
—Dan
Jim
On Tue, Sep 8, 2015 at 12:22 PM, Dan Asimov <asimov@msri.org <mailto: asimov@msri.org>> wrote:
To evaluate whether the bijection is a homeomorphism, it would be immensely helpful if you defined it.
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First, disregard everything I said above -- this isn't an iterated function system. I think I can prove it is a Cantor set, though. The idea is that step n (n
0) introduces 2^(n-1) gaps of size 2 / 3^n . These gaps narrow over subsequent steps, but do not vanish. Rather, in the final set they have size 2 / 3^n - 2 (sum_{m > n} 1 / 3^m ) = 2 / 3^n - 2 ( 1 / 2 ) ( 1 / 3^n ) = 1 / 3^n.
This makes sense, because if you add up the size of the gaps, you get sum_{n > 0} 2^(n-1) / 3^n = (1/3) sum_{n >= 0) (2/3)^n = 1. So over the final convex hull [-1/2, 3/2], half is gaps and half is non-gaps. -Thomas C On Tue, Sep 8, 2015 at 10:24 PM, James Propp <jamespropp@gmail.com> wrote:
Assign the original set S_0 = [0,1] the empty codename. Assign the two new intervals of length 1/2 in S_1 (obtained by applying the split-and-slide operation to S_0) the codenames 0 and 1, respectively (from left to right). Assign the four new intervals of length 1/4 in S_2 (obtained by applying the split-and-slide operation to the two intervals in S_1) the codenames 00, 01, 10, and 11, respectively (from left to right). And so on like that.
Claim: The 2^k intervals of length 1/2^k that constitute S_k are ordered left-to-right in accordance with their k-bit binary codenames, and the gaps between the intervals are all of size > 1/3^k.
Proof: By straightforward induction (unless I'm missing something).
Jim
On Tue, Sep 8, 2015 at 8:14 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Possibly dumb question: Do you know that the various intervals created (or at least defined) never overlap? 'Cause if they did, then some incarnations of a point x could at a later stage be asked to slide by one amount, and other incarnations by a different amount (or direction).
—Dan
On Sep 8, 2015, at 12:32 PM, James Propp <jamespropp@gmail.com> wrote:
Part of what I had in mind is that points in the final set arise from points in the original interval [0,1] through a process of repeated sliding (and, in the case of dyadic rationals, "cloning"). If we look at the sequence of slide operation (Left or Right) that a point undergoes, e.g., Left, Right, Right, Left, Right, Right, ... and replace each Left by a 0 and each Right by a 1, we get an infinite string of bits; this is the intended bijection between my set of points and the product set {0,1} x {0,1} x {0,1} x ...
Clear now? (Apologies in advance if still murky.)
Jim
On Tue, Sep 8, 2015 at 12:35 PM, Dan Asimov <dasimov@earthlink.net> wrote:
On Sep 8, 2015, at 9:27 AM, James Propp <jamespropp@gmail.com>
wrote:
Good point, Dan!
For all but countably many points (namely the dyadic rationals), we
use the
binary representation of the original point in [0,1].
For the dyadic rationals, we need to know which of the two "clones" we're using.
E.g., when we split [0,1] into two pieces and give each piece an endpoint (creating a new point out of thin air), we give the right endpoint of the left piece the label .0111... and the left endpoint of the right piece the label .1000...
In this way, we get a labelling of the points of my set using infinite strings of bits, where each string corresponds to a unique point in the set, and vice versa.
Hopefully that's clear.
Not to me it isn't.
I don't even know how you are defining your set, since you never stated that. (Yes, you stated what the stages are, but not how the final set is defined in terms of the stages.
And you don't even seem to be defining your bijection above in terms of the final set, but only in terms of the stages.
—Dan
Jim
On Tue, Sep 8, 2015 at 12:22 PM, Dan Asimov <asimov@msri.org <mailto: asimov@msri.org>> wrote:
To evaluate whether the bijection is a homeomorphism, it would be immensely helpful if you defined it.
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Perhaps it's worth mentioning that my original project was finding a way to embed {0,1}x{0,1}x{0,1}x... in the real line that respects lexicographic order and the natural product measure on the set. So then I got curious as to whether the embedding I found respects the topology as well (in both directions). I suspect that with the right description of the map and its inverse, it's easy to prove or disprove that it's a homeomorphism. I've just been too busy with other things to think about it. (Actually, the main "other thing" that's been occupying me is my upcoming blog post on .999..., which actually inspired the question. I got to thinking, What's the nicest way to picture the product set S=DxDxDx... with D={0,1,2,...,8,9} as a subset of the reals so that lexicographic ordering of S is a restriction of the usual ordering of the reals? Eventually I decided that this was too peripheral, but by then I'd gotten a bit hooked on the cut-and-slide construction.) Jim On Friday, September 11, 2015, Thomas Colthurst <thomaswc@gmail.com> wrote:
First, disregard everything I said above -- this isn't an iterated function system.
I think I can prove it is a Cantor set, though. The idea is that step n (n
0) introduces 2^(n-1) gaps of size 2 / 3^n . These gaps narrow over subsequent steps, but do not vanish. Rather, in the final set they have size 2 / 3^n - 2 (sum_{m > n} 1 / 3^m ) = 2 / 3^n - 2 ( 1 / 2 ) ( 1 / 3^n ) = 1 / 3^n.
This makes sense, because if you add up the size of the gaps, you get sum_{n > 0} 2^(n-1) / 3^n = (1/3) sum_{n >= 0) (2/3)^n = 1. So over the final convex hull [-1/2, 3/2], half is gaps and half is non-gaps.
-Thomas C
On Tue, Sep 8, 2015 at 10:24 PM, James Propp <jamespropp@gmail.com <javascript:;>> wrote:
Assign the original set S_0 = [0,1] the empty codename. Assign the two new intervals of length 1/2 in S_1 (obtained by applying the split-and-slide operation to S_0) the codenames 0 and 1, respectively (from left to right). Assign the four new intervals of length 1/4 in S_2 (obtained by applying the split-and-slide operation to the two intervals in S_1) the codenames 00, 01, 10, and 11, respectively (from left to right). And so on like that.
Claim: The 2^k intervals of length 1/2^k that constitute S_k are ordered left-to-right in accordance with their k-bit binary codenames, and the gaps between the intervals are all of size > 1/3^k.
Proof: By straightforward induction (unless I'm missing something).
Jim
On Tue, Sep 8, 2015 at 8:14 PM, Dan Asimov <dasimov@earthlink.net <javascript:;>> wrote:
Possibly dumb question: Do you know that the various intervals created (or at least defined) never overlap? 'Cause if they did, then some incarnations of a point x could at a later stage be asked to slide by one amount, and other incarnations by a different amount (or direction).
—Dan
On Sep 8, 2015, at 12:32 PM, James Propp <jamespropp@gmail.com <javascript:;>> wrote:
Part of what I had in mind is that points in the final set arise from points in the original interval [0,1] through a process of repeated sliding (and, in the case of dyadic rationals, "cloning"). If we look at the sequence of slide operation (Left or Right) that a point undergoes, e.g., Left, Right, Right, Left, Right, Right, ... and replace each Left by a 0 and each Right by a 1, we get an infinite string of bits; this is the intended bijection between my set of points and the product set {0,1} x {0,1} x {0,1} x ...
Clear now? (Apologies in advance if still murky.)
Jim
On Tue, Sep 8, 2015 at 12:35 PM, Dan Asimov <dasimov@earthlink.net <javascript:;>> wrote:
On Sep 8, 2015, at 9:27 AM, James Propp <jamespropp@gmail.com
<javascript:;>> wrote:
Good point, Dan!
For all but countably many points (namely the dyadic rationals), we
use the
binary representation of the original point in [0,1].
For the dyadic rationals, we need to know which of the two "clones" we're using.
E.g., when we split [0,1] into two pieces and give each piece an endpoint (creating a new point out of thin air), we give the right endpoint of the left piece the label .0111... and the left endpoint of the right piece the label .1000...
In this way, we get a labelling of the points of my set using infinite strings of bits, where each string corresponds to a unique point in the set, and vice versa.
Hopefully that's clear.
Not to me it isn't.
I don't even know how you are defining your set, since you never stated that. (Yes, you stated what the stages are, but not how the final set is defined in terms of the stages.
And you don't even seem to be defining your bijection above in terms of the final set, but only in terms of the stages.
—Dan
Jim
On Tue, Sep 8, 2015 at 12:22 PM, Dan Asimov <asimov@msri.org <javascript:;> <mailto: asimov@msri.org <javascript:;>>> wrote:
> To evaluate whether the bijection is a homeomorphism, it would be > immensely helpful if you defined it.
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What's the nicest way to picture the product set S=DxDxDx... with D={0,1,2,...,8,9} as a subset of the reals so that lexicographic ordering of S is a restriction of the usual ordering of the reals?
f : (D x D x D x ...) --> R (d_0, d_1, d_2, ...) |--> sum(d_n / 11^n : n in {0, 1, 2, ...}) has the property you describe, unless I'm missing something. Sincerely, Adam P. Goucher
I meant to add "and preserves the natural product measure on the set". Jim Propp On Saturday, September 12, 2015, Adam P. Goucher <apgoucher@gmx.com> wrote:
What's the nicest way to picture the product set S=DxDxDx... with D={0,1,2,...,8,9} as a subset of the reals so that lexicographic ordering of S is a restriction of the usual ordering of the reals?
f : (D x D x D x ...) --> R (d_0, d_1, d_2, ...) |--> sum(d_n / 11^n : n in {0, 1, 2, ...})
has the property you describe, unless I'm missing something.
Sincerely,
Adam P. Goucher
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Oh, in that case replace 11^n with: 10^n * (p_0 * p_1 * ... * p_n) / Q where each p_i is greater than 1 but the infinite product: p_0 * p_1 * p_2 * p_3 * ... converges to the finite value Q.
Sent: Saturday, September 12, 2015 at 12:24 PM From: "James Propp" <jamespropp@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Cantor set of measure 1?
I meant to add "and preserves the natural product measure on the set".
Jim Propp
On Saturday, September 12, 2015, Adam P. Goucher <apgoucher@gmx.com> wrote:
What's the nicest way to picture the product set S=DxDxDx... with D={0,1,2,...,8,9} as a subset of the reals so that lexicographic ordering of S is a restriction of the usual ordering of the reals?
f : (D x D x D x ...) --> R (d_0, d_1, d_2, ...) |--> sum(d_n / 11^n : n in {0, 1, 2, ...})
has the property you describe, unless I'm missing something.
Sincerely,
Adam P. Goucher
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Here's a definition of Jim's set. For simplicity, I'll define a slightly different set, that when it splits an interval in half, just moves the right half a bit to the right, rather than both doing this and moving the left half to the left. For any subset S of Z+ (think of S as "the set of places that are 1 in the binary representation of the number before we start to move it"), sum_{s \in S} (1/2^k + 1/3^k) To get Jim's original set, add sum{s not in S} -1/3^k to this. You can get the K'th "stage" of Jim's construction by adding 1/2^k + 1/3^k for k<=K, and 1/2^k for k > K. I'm pretty sure the set is a Cantor set, but I'm still confused as to its measure. But having a precise definition of the set, without defining it as a limit, should make it easier to figure this out. Andy On Tue, Sep 8, 2015 at 12:35 PM, Dan Asimov <dasimov@earthlink.net> wrote:
On Sep 8, 2015, at 9:27 AM, James Propp <jamespropp@gmail.com> wrote:
Good point, Dan!
For all but countably many points (namely the dyadic rationals), we use the binary representation of the original point in [0,1].
For the dyadic rationals, we need to know which of the two "clones" we're using.
E.g., when we split [0,1] into two pieces and give each piece an endpoint (creating a new point out of thin air), we give the right endpoint of the left piece the label .0111... and the left endpoint of the right piece the label .1000...
In this way, we get a labelling of the points of my set using infinite strings of bits, where each string corresponds to a unique point in the set, and vice versa.
Hopefully that's clear.
Not to me it isn't.
I don't even know how you are defining your set, since you never stated that. (Yes, you stated what the stages are, but not how the final set is defined in terms of the stages.
And you don't even seem to be defining your bijection above in terms of the final set, but only in terms of the stages.
—Dan
Jim
On Tue, Sep 8, 2015 at 12:22 PM, Dan Asimov <asimov@msri.org <mailto:asimov@msri.org>> wrote:
To evaluate whether the bijection is a homeomorphism, it would be immensely helpful if you defined it.
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participants (6)
-
Adam P. Goucher -
Andy Latto -
Dan Asimov -
Dan Asimov -
James Propp -
Thomas Colthurst