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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