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.
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