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Antoine's necklace is an embedding of the Cantor set in $\mathbb{R}^3$ constructed by taking a torus, replacing it with a necklace of smaller interlinked tori lying inside it, replacing each smaller torus with a necklace of interlinked tori lying inside it, and continuing the process ad infinitum; Antoine's necklace is the intersection of all iterations.

Antoine's necklace

A number of sources claim that this necklace 'cannot fall apart' (e.g. here). Given that the necklace is totally disconnected this obviously has to be taken somewhat loosely, but I tried to figure out exactly what is meant by this. Most sources seem to point to this paper (which it must be noted contains some truly remarkable images, e.g. Figure 12). There the authors make the same point that Antoine's necklace 'cannot fall apart'. Nevertheless, all they seem to show in the paper is that it cannot be separated by a sphere (every sphere with a point of the necklace inside it and a point of the necklace outside it has a point of the necklace on it).

It seems to me to be a reasonably trivial exercise to construct a geometrical object in $\mathbb{R}^3$ which cannot be separated by a sphere, and yet can still 'fall apart'.

                                            Image1

In the spirit of the construction of Antoine's necklace, these two interlinked tori cannot be separated by a sphere (any sphere containing a point of one torus inside it will contain a point of that torus on its surface), but this seems to have no relation to the fact that they cannot fall apart - if we remove a segment of one of the tori the object still cannot be separated by a sphere, and yet can fall apart macroscopically.

                                            Image2

The fact mentioned here that the complement of the necklace is not simply connected, and the fact mentioned here that there are loops that cannot be unlinked from the necklace shouldn't impact whether it can be pulled apart either, as both are true of our broken rings

My question is this: Is it possible to let me know either:

  1. How I have misunderstood separation by a sphere (so that it may still be relevant to an object being able to fall apart),

  2. What property Antoine's necklace does satisfy so that it cannot fall apart (if I have missed this), or

  3. What is actually meant when it is said to be unable to fall apart (if I have misunderstood this)

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  • $\begingroup$ What does falling apart mean? $\endgroup$
    – Pedro A
    Apr 22 '20 at 13:07
  • $\begingroup$ Welp, if nothing else, I have a new "OC" gif $\endgroup$
    – corsiKa
    Apr 22 '20 at 15:06
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    $\begingroup$ @PedroA Even though the necklace is a totally disconnected set of points, no (deformed) sphere can be put around any proper subset of it, that doesn't touch some other point of the necklace, as pointed out by Alexander Gruber and proved in the linked paper. That means there's no way it can be 'disentangled', so it's like the interlocking rings, rather than the broken rings. (You could imagine a stretchy balloon around the C, but not the O, which is related to why you can pull the C out of the O, but not pull an O out of the O; in the necklace, all parts are like the O). $\endgroup$
    – Anon
    Apr 22 '20 at 15:46
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    $\begingroup$ @corsiKa Courtesy of my terrible paint skills $\endgroup$
    – Anon
    Apr 22 '20 at 15:47
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    $\begingroup$ @anon you seem to have misspelled "spectacular" - a common typo to be sure. $\endgroup$
    – corsiKa
    Apr 22 '20 at 20:41
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if we remove a segment of one of the tori the object still cannot be separated by a sphere, and yet can fall apart macroscopically.

Well, sure it can. The key observation is that "sphere" means "homeomorphic image of a sphere" in this context.

Turn the ring with the part removed (the "C") and pull the in tact ring (the "O") through the missing piece. Then you have two separate components, the C and the O, which it is easy to see could be placed inside and outside of a sphere. Let's put the C on the inside and the O on the outside.

Now, think about shrinking that sphere very tightly (but not touching) around the C, then putting the C back where it was. You've separated the two by the homeomorphic image of a sphere.

Think about if we tried to do this with two Os, like your first figure. If the Os weren't interlocked, sure, it's easy, just like when the C and the O were separated. But if two Os are interlocked, there isn't any way to fit a sphere around one of them in the same way as you could with the C. And if you can't do it with only two Os, you certainly can't do it with Antoine's Necklace. Thus, it will stay together.

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    $\begingroup$ +1 Thank you so much for this simple answer. I don't know how I missed that, but that simply and clearly shows why the broken rings differ, and I have a much clearer intuitive picture now of how lack of separation by a sphere implies it cannot fall apart. $\endgroup$
    – Anon
    Apr 21 '20 at 9:25
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The necklace is a topological space $X$, together with a natural embedding $i\colon X\to \Bbb R^3$. Since the verb "to fall apart" describes a process, it may best be described by adding time as a variable. So we ask whether there is a homotopy $H\colon X\times[0,1]\to\Bbb R^3$ such that $H(\cdot,0)=i$ and $H(\cdot1)$ is a embedding $X\to\Bbb R^3$ that is in an obvious fashion separated, say a non-empty part of (the image of) $X$ is in the $z>0$ region, a non-empty part in the $z<0$ region, but nothing in the $z=0$ hyperplane. Instead of a separating plane, a separating sphere would certainly also count (and in fact be slightly more general: two concentric spheres cannot fall apart in the first sense, but can in the second; both interpretations have a point as the inner sphere cannot really fall out, but it is not really linked to the outer sphere either ...)

I am unaware whether the referenced authors took this homotopy-and-sphere approach and you overread the homotopy part, or perhaps whether they had some argument that the sphere property of $i$ is sufficient in the given situation (but not in general, as your examples show).

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    $\begingroup$ +1 Thank you for this answer. I have accepted Alexander Gruber's answer because it hit closer to the source of my misunderstanding but this answer was useful as well. $\endgroup$
    – Anon
    Apr 21 '20 at 9:27

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