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Antoine's necklace is a pathological embedding of the Cantor set into $\Bbb R^3$. The second iteration looks like this:

Antoine's necklace

Interestingly, the complement $\Bbb R^3\setminus\rm A$ is not simply connected. This property is preserved by ambient isotopies. (Thanks, @MikeMiller!) Anything with an ambient isotopy to what's defined in the article, then, should be considered to be an Antoine's necklace.

What happens when you project the necklace onto a plane? That is, if $\pi:\Bbb R^3\to\Bbb R^2$ is a projection, what is $\pi({\rm A})$? I'm sure you don't get another Cantor set. In fact, it seems like it would always be connected. My guess is that it must be homeomorphic to the Sierpiński carpet. Is it?

Akiva Weinberger
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    An ambient isotopy induces a homeomorphism of the complement of $A$, so yes, that's correct. –  Mar 02 '16 at 15:46
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    I think that a projection of an Antoine necklace onto the plane might be homeomorphic to the Sierpinski carpet, but I don't believe that it must be. Keep in mind that Antoine's necklace is a topological entity - unless you specify a specific configuration using, for example, an iterated function system. Thus, there's quite a bit of flexibility as to how it is laid out in space. Furthermore, it is certainly possible for the projection of a Cantor set in $\mathbb R^3$ to contain a solid disk. I imagine that a projection of an Antoine necklace might contain a solid disk as well. – Mark McClure Mar 04 '16 at 23:30
  • @MarkMcClure I'll define "an Antione's necklace" to be anything with an ambient isotopy to the one defined in the article (which I'll make more explicit in the question) I'm not sure how a projection of such a thing would contain a disk, though I suppose it might be possible. – Akiva Weinberger Mar 05 '16 at 23:46
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    On a computer screen, isn't any drawing of a 3D object just a projection of that object onto a plane? So if you want intuition about your projection, drawings of Antoine's necklace from different angles with higher iterations may be helpful. This is a video with enough iterations to reach 160,000 tori. – Zubin Mukerjee Mar 06 '16 at 00:03
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    @AkivaWeinberger For every $n$, there is a Cantor set in $\mathbb R^n$ whose projection onto every $n-1$ dimensional subspace contains an $n-1$ dimensional ball. See, for example, This sets with fat shadows in the Monthly. Now, I'm not assuring you that an Antoine's necklace can be so configured but, if you're going to allow it to be twisted any which way, I don't see any reason to think that it can't. – Mark McClure Mar 06 '16 at 00:38

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The answer is no - it is quite possible for an Antoine necklace to be configured in such a way that some projection of it is not homeomorphic to the Sierpinski carpet. In particular, it's possible that some such projection has a cut point. This is impossible if it's homeomorphic to the Sierpinski carpet, which has no cut points.

To see this, let's take a look at a specific Antoine necklace:

enter image description here

And here it is from the side in orthographic projection:

enter image description here Note that in the middle, there is always a torus on its side. As a result, the height of that torus is decreasing down to zero and the projection in that direction will intersect a vertical line in a single point. That point is a cut point.

Mark McClure
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    Thank you! It's much simpler than I thought. May I ask where you got your images? Did you make them? – Akiva Weinberger Mar 06 '16 at 02:01
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    @AkivaWeinberger I did make the images with Mathematica, yes. And yeah, it did turn out to be not too hard, once I looked at it from the right angle. I can see how you thought that might have a structure like the Sierpinski carpet, though, as most projections will likely have a lot of holes punched out. And thanks for the very interesting question!! – Mark McClure Mar 06 '16 at 02:06
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    "…once I looked at it from the right angle." Literally! – Akiva Weinberger Mar 06 '16 at 02:08