Re: [math-fun] Cantor set of measure 1?

(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
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>> >>
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