Re: [math-fun] Antoine's necklace
DanA>I'm not sure what "at the limit the areas disappear and everything is boundary" means exactly. rwg>We have a continuous map of a circle onto the closed region depicted in the link: http://www.tweedledum.com/rwg/rad2Image4.gif . The areas in http://www.tweedledum.com/rwg/rad2fill.htm are outlined by connect-the-dots images of (the vertices of) a regular polygon inscribed in the circular domain. Doubling the number of vertices (sample points) corresponds to one iteration of the construction suggested by the upper figure. As with most such closed-loop spacefills (i.e., a sausage encasing a tree), the largest inscribable circle shrinks to a point as the sampling frequency → ∞. Thus the space is filled twice. I.e., every point has at least two inverse images, uncountably many have four, and ℵ₀ of them have six. The polygonal boundaries of the regions are admittedly artifacts of the sampling, but they illustrate the underlying spacefilling function as they crinkle up due to increasing sampling frequency. As they crinkle, the interior area remains constant, then suddenly vanishes at the limit, while the area of the closure is that same constant until it suddenly doubles. Actually, the "constant" may increase slightly to account for the shrinkage of the hexagon in the center of the lower figure. DanA>But if each of the six regions is in the limit space-filling the very same space, Terminology. Regions already fill space. At the limit, their *boundaries* fill the area near the center four times over: twice via the closed loop property, and twice because any extended patch of the lower diagram has two colors. DanA> then isn't this a lot like the sets A_n and B_n in the reals?: Where A_n = union of [k/2^n,(k+1)/2^n] for odd integers k, B_n = union of [k/2^n,(k+1)/2^n] for even integers k. both sets having the common boundary {k/2^n | k in Z}, but for any real x as n -> oo, dist(x,A_n) -> 0 and dist(x,B_n) -> 0. ----- DanA>I agree the illustrations with the Wikipedia article are unhelpful. Here's a picture of the second stage: 4 cyclically linked solid tori going around one, and in each of those 4 there are four smaller cyclically linked tori going around *it*: < http://onionesquereality.files.wordpress.com/2011/07/antoines-necklace.jpg >. (There are a few extraneous lines, which should be ignored.) Yes, three should work. In fact any sequence of numbers of tori at various stages would work -- as long as the diameters of the tori in stage n approach 0, as n -> oo. (I wonder what happens in the case where there is only one torus in each stage, pulled longitudinally around the previous torus and made to link with itself.) --Dan Ick! I think the nth one winds and unwinds 2^n times around inside the outermost one. But we lose the shrinking to a point effect. --rwg RWG wrote: << Robert M. wrote: << . . . . . . But the necklace one bothers me. That transition from linked set of tori to unlinked set of points seems impossible. A similar thing happens with increasingly fine polygonal approximations to spacefilling functions. Inhttp://www.tweedledum.com/rwg/rad2fill.htm the upper figure defines the iteration. For any finite order, six of these equipartition the area near the center of the lower figure. But at the limit, the areas disappear and everything is boundary. I don't get the many-torus illustrations inhttp://en.wikipedia.org/wiki/Antoine%27s_necklace . Doesn't the text say four tori? Shouldn't three work? (Bogus Borromean rings.)
On 12/8/12, Bill Gosper <billgosper@gmail.com> wrote:
rwg>We have a continuous map of a circle onto the closed region depicted in the link: http://www.tweedledum.com/rwg/rad2Image4.gif .
Two anomalous short segments appear in the latter graphic: one green, one cyan. WFL
I had a suspicion I knew that space [the intersection of nested tori, each pulled around the previous one so as to link itself], from a previous encounter with it. Sure enough, it turns out that it's the *complement* of the amazing "Whitehead contractible manifold" W in the 3-sphere. ----- A space is contractible if it can be shrunk to a point within itself, like an n-disk or Euclidean space R^n. W was discovered in 1935 by J.C. Whitehead, after he erroneously thought he'd proved that every contractible 3-manifold (without boundary) is topologically equivalent to 3-space R^3. One way to define it is to create the infinite intersection J of tori as below inside the 3-sphere S^3; then the complement S^3 - J is the Whitehead contractible manifold. The reason it's not topologically R^3 is that W is not "simply-connected at infinity" < http://en.wikipedia.org/wiki/Simply_connected_at_infinity >). ----- --Dan RWG wrote: << I wrote: << (I wonder what happens in the case where there is only one torus in each stage, pulled longitudinally around the previous torus and made to link with itself.)
Ick! I think the nth one winds and unwinds 2^n times around inside the outermost one. But we lose the shrinking to a point effect.
On Sat, Dec 8, 2012 at 4:23 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I had a suspicion I knew that space [the intersection of nested tori, each pulled around the previous one so as to link itself], from a previous encounter with it.
Sure enough, it turns out that it's the *complement* of the amazing "Whitehead contractible manifold" W in the 3-sphere.
When you nest 4 linked tori inside each torus at each step, you've proved that the complement of the intersection is not simply connected, hence not contractible. Why doesn't the same proof work in the case where you place a single self-linked torus inside the torus each time? Andy
----- A space is contractible if it can be shrunk to a point within itself, like an n-disk or Euclidean space R^n.
W was discovered in 1935 by J.C. Whitehead, after he erroneously thought he'd proved that every contractible 3-manifold (without boundary) is topologically equivalent to 3-space R^3.
One way to define it is to create the infinite intersection J of tori as below inside the 3-sphere S^3; then the complement S^3 - J is the Whitehead contractible manifold.
The reason it's not topologically R^3 is that W is not "simply-connected at infinity" < http://en.wikipedia.org/wiki/Simply_connected_at_infinity >). -----
--Dan
RWG wrote: <<
I wrote: << (I wonder what happens in the case where there is only one torus in each stage, pulled longitudinally around the previous torus and made to link with itself.)
Ick! I think the nth one winds and unwinds 2^n times around inside the outermost one. But we lose the shrinking to a point effect.
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I'm not very familiar with the Whitehead manifold. But: With 4, or as few as 2, tori linked around (the interior of) a larger one T, clearly a meridian of T lying in R^3 - T (or S^3 - T) cannot be shrunk to a point in the complement of the linked tori. Yet with only 1 torus T_1 self-linked around a larger one T, then -- surprisingly -- a meridian M of T *can* be shrunk to a point in the complement of T_1. (Not through homeomorphisms, but through a continuous family of continuous maps f_t : S^1 -> R^3 - T_1, 0 <= t <= 1, where f_0 is a homeomorphism of S^1 onto M. There must be at least one t, 0 < t < 1, for which f_t is not one-to-one.) Note: The meridian M and the core circle of T_1 form what's known as the Whitehead link, with linking number = 0. (For a useful image, see fig. 16 at < http://www.math.cornell.edu/~mec/2008-2009/HoHonLeung/page2_knots.htm >. It's a symmetrical link: the two closed curves can be interchanged.) --Dan On 2012-12-08, at 1:32 PM, Andy Latto wrote:
On Sat, Dec 8, 2012 at 4:23 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I had a suspicion I knew that space [the intersection of nested tori, each pulled around the previous one so as to link itself], from a previous encounter with it.
Sure enough, it turns out that it's the *complement* of the amazing "Whitehead contractible manifold" W in the 3-sphere.
When you nest 4 linked tori inside each torus at each step, you've proved that the complement of the intersection is not simply connected, hence not contractible. Why doesn't the same proof work in the case where you place a single self-linked torus inside the torus each time?
participants (4)
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Andy Latto -
Bill Gosper -
Dan Asimov -
Fred lunnon