Explain this xkcd comic to me [Oct. 11, 2010: "Pumpkin Carving"]

Cafe Society
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It may be helpful, in understanding the Banach-Tarski paradox, to first consider some much simpler examples of the same phenomenon:

Suppose you had a stick of some particular length. If we allow ourselves to cut it up into infinitely many pieces of a certain sort, and then move those pieces around, we could re-arrange this into a stick of twice the original length; specifically, we could carve the original stick into one piece for every point upon it, and then move each of those points twice as far from the end of the stick as they started. We’d end up with all the points on a stick twice as long. That is, we could turn [0, 1] into [0, 2] by carving [0, 1] into infinitely many single points, then translating each single point appropriately. In this way, we could carve a body up, then move each piece, then end up with something larger. (Note that this depends on conceptualizing bodies as just collections of discrete points…)

“Well, c’mon”, you may say. “Of course if you could split things into one piece for each point, you could pull that off. That’s an ungodly number of pieces; uncountably infinite, even. What if I restricted you to using larger pieces and less of them?”.

Well, let’s try another example. By a rational angle, I mean an angle which is a rational multiple of 360 degrees [that is, an angle which, when rotated by repeatedly, eventually brings something back to where it started]. Suppose I have a circle. I’ll consider two points on the circle to be “in the same club” if the angle between them is a rational angle. This organizes the points of the circle up into a number of disjoint clubs. Next, select one (just one) point from each of these clubs, and drop some red ink on it. (Go ahead, just randomly pick one point from each club and drop red ink on it). The resulting collection of red points we’ll call V. Note that, since every point on the circle is in the same club as one and only one point in V, every point on the circle is a unique rational angle away from a point in V.

Thus, we can carve the circle up into one piece for every rational angle, each piece being a correspondingly rotated copy of V. So, since there are countably infinitely many rational angles, a circle can be carved into a collection of countably infinitely many rotated copies of V.

But any countably infinite collection can be split into two countably infinite collections (think of splitting the integers into the evens and the odds). So we can split a circle into a countably infinite collection of rotated copies of V, which is as good as two countably infinite collections of rotated copies of V, which is as good as two circles. In this way, we can carve up one circle into countably infinitely many pieces, then re-arrange them to get two circles. [Again, this depends on our viewing the body of a circle as nothing more than a collection of discrete points]. In that way, we can again increase the length of string with suitable cuts and re-arrangements.

“Well, alright, you managed to use only countably many pieces there, but that’s still an infinite number of pieces. Can you do the same sort of thing with only finitely many pieces?”

Yes, you still can, and along similar lines to the last example, but with some more technical cleverness. That’s the sort of cleverness that Banach and Tarski used to re-arrange one sphere into two using only finitely many cuts [but, again, as above with the circle, depending on A) being able to make infinitely many random choices and, even more importantly, B) conceptualizing the sphere as nothing more than a collection of discrete points].

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Am I correct in saying the bolded part is the part requiring the Axiom of Choice mentioned in the strip?

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Yes, that is correct.

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Basically, Banach and Tarski figured out a neat way of extending to three dimensions the fact that the integers can be divided into two subsets, each of which is the same size as the integers. That is the basic reason why the proof works at all: a set has a “paradoxical decomposition” into two subsets that are the same size as the original set. Most of that lego webcomic page was explaining the construction of how you are able to divide the sphere into a set of points that could be paradoxically decomposed. One can very easily do this by rotating each of the uncountably infinitely many radii of the sphere by twice the angle they form with some chosen meridian, but that’s not particularly interesting and requires uncountably many pieces. The real brilliance of the result is that it can be done with a finite number of subsets. But, and this cannot be emphasized enough, the theorem is a purely mathematical construct that works only on ideal objects that can be divided arbitrarily.

In another thread involving the conquering of the Earth by various fictional characters, someone brought up this theorem as a solution to the problem of Voyager’s never-ending supply of shuttlecraft, and that was the last thread I read before reading this one. Eeeeeerieeee…

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The group SO(3) of orientation-preserving rotations of S^2 happens to contain the free group on two generators F = F_2. It turns out that F can be written as the union of several copies of itself. (Since F is free, the Cayley graph of F is a tree, and the required decomposition is easy to write down in this case.) Each of these copies then acts on S^2, and the quotient gives the required decomposition. There are some geometric details and a lot of technical set-theoretic matters to flesh out, but the point is that the Banach-Tarski decomposition comes from the fact that S^2 has a group acting on it nicely that has this sort of paradoxical decomposition. As the “recent progress” section of that Wikipedia page suggests, a similar paradox is likely to occur for any space acted on nicely by F. The Tits alternative suggests that there should be many such subspaces of R^n, but they’re not necessarily interesting or easy to work with. I don’t know much about analogues of the Banach-Tarski paradox in higher dimensions, unfortunately.

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The first panel (“I carved a pumpkin!”) reminds me of a Drabble (I think) comic strip I read years ago. It showed the preschool-aged girl character in the strip sitting at the kitchen table with some paper and crayons. Her teenaged brother entered the room, saw her, and the conversation went like this:

Brother: Whatcha doing?
Sister: Drawing a pitcher!
Brother: No no no, you’re drawing a “picture”. A “picture” and a “pitcher” are two different things.
Sister: Oh…
Brother: So, what are you drawing a picture of?
Sister: Hideo Nomo.

(Hideo Nomo was a pitcher for the Los Angeles Dodgers at the time.)

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They reused that one?

I originally read it back in the early 80s, way before Nomo was even playing pro in Japan, let alone for the Dodgers. I can’t remember who they used for the punchline then…Koufax, maybe.

And, yes, it was a Drabble.

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Seriously? That’s hilarious. Yeah, I read it in the Seattle P-I in the early 90s. I imagine the original version did it with different characters, though, since the preschool-aged daughter was “born” in the early 90s (I remember the whole pregnancy storyline - there were already the teenaged son and his pre-teen brother, and then their mom turned up pregnant again. Surprise!)

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I would think the fact that a line is continuous would mean that you could not ever subdivide it sufficiently to reach a single point. Or, to put it another way, pick any two points on a line, and there will be another point between it. So the idea that you take only the odd numbered points or whatever means that you are creating new points between the even numbered ones, as well as in the new line’s odd numbered ones.

The reason this can’t be done in real life is the same as the reason that one can move between points at all. It can be illustrated in that problem where you cover half the remaining points in half the amount of time. We cannot cross the infinite barrier–every move we make is going through some infinite number of points.

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Maybe I am missing something. The link that Qadgop provides divides our sphere into sets of points, but those points were generated by many rotations of an irrational angle. Am I wrong to assume that this leaves out all the points that could be generated using rational angles? And if you try to argue that those points are included, then I would say your irrational angle isn’t irrational. So, he is defining a set of points that does NOT include every point in the sphere. Neither “copy” produced at the end of the exercise is complete.

Similarly, Idistinguishable’s example follows the same principle, but uses only rational angles. None of the points of his circle that represent an irrational angle are included in any “club”, so again, his duplicate circles are not complete circles.

Also, while I understand the logic behind “stretching” a stick defined as [0,1] to a stick twice as long, [0,2] by defining a one to one relationship with the infinite number of points in both sets, can’t the same logic be applied to the number sets [0,a] and [0,b] for any values of a and b >0? Can’t I make the stick as long or as short as I want? by that reasoning, could I make as many copies of the sphere as I want?

And back to the sphere question… It all seems to come down to the idea that dividing infinity into a finite number of sets of infinity is totally kosher, but adding those sets back together somehow gets you more then you started with. No, you still have infinity!

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Every point is in a club. Two points are in the same club if the angle between them is rational.

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Yeah, I think it was the two boys…your description mentioning the younger one as a girl struck me as odd. Hell, checking out the character descriptions (I haven’t read Drabble since before the girl was introduced) it could very well have been Ralph and Norman. :stuck_out_tongue:

… How did we end up discussing Drabble in an xkcd thread? <_<

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Ok, I think that makes more sense. If I pick point q on the circle, which is some irrational angle away from 0˚, It gets put into the 360˚ club, because it is a rational angle away from itself.
ETA: Oh, and I suppose that even if some point is an irrational angle away from some other point, it can still be a rational angle away from a different point. My brain is slowly starting to get all this.

That still leaves the idea that we are splitting one infinite set in half, and just calling it two infinite sets. Adding them back together doesn;t give us 2xinfinity. But I do get the pumpkin joke, at least. :slight_smile:

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Don’t get me started on Drabble. The esoteric references in that strip always go over my head.

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Or is it a Pumpkin of Theseus, and an identical pumpkin was assembled out of pieces that had been replaced in the pumpkin?