According to the Banach-Tarski Paradox, you can divide a sphere into five pieces and construct two spheres exactly like the original simply by moving the pieces around and rotating them. The article, however, notes that these “pieces” don’t have a definable volume. It seems that the “pieces” are just five sets of points that don’t have to have any connection with each other. So, by what definition of “piece” are there five of them?
It comes down to some weirdness from mathematical measures. The Lebesgue measure allows us to supposedly define the volume of a set of points in three-space (or space of any other dimension). The weirdness is that two sets in three-space with the same number of points, i.e. an uncountably infinite number of points, need not have the same volume.
When you originally divide the sphere into five “pieces”, all you’re saying is that you’re creating a partition of all the points in the sphere into five sets. Trying to visualize what the five pieces would look like geometrically would probably only drive you insane.
I’m not familiar with the proof, but I’m virtually certain that they mean each “piece” is topologically connected. A subset C of a ball is said to be connected if and only if there are open subsets U and V of the ball such that: 1. U and V both have nonempty intersection with C, and 2. C is contained in the union of U and V.
This leads me to suspect that something is fundamentally seriously wrong with set theory.
That was one of the motivations for (inventing or discovering, whichever you prefer) things like the Banach Tarski paradox–to demonstrate that something is seriously wrong.
The controversy was over the axiom of choice, which essentially says: Given any family of nonempty sets, you can construct a new set by choosing one element from each set in the family.
This axiom is independent of the other standard axioms of set theory (Zermelo-Fraenkel (ZF) set theory). It seems rather obvious, but it has some unexpected consequences, such as this BT-paradox.
Under ZF set theory, the axiom of choice has many equivalents, such as the well-ordering principle (any nonempty set can be well ordered) and Zorn’s lemma (Any nonempty partially ordered set in which each chain has an upper bound must have a maximal element), to name just a couple.
The fact that, while all three of these are equivalent, they are not equally intuitive, led Jerry Bona to state, “The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn’s Lemma?”
In mathematics, it’s often customary to state which theorems actually depend on the axiom of choice (AC) (or any of its equivalents), such as the BT-paradox.
However, since in addition to AC’s intuitiveness, it also makes things much nicer (and more natural, in my opinion), AC is a commonly accepted axiom. Without AC, things are much messier. A couple of examples: 1. The axiom of choice is necessary for an arbitrary vector space to have a basis, and 2. AC is necessary to insure that an arbitrary set even has a cardinality, which, in my opinion, is a fundamental property of a set.
Nah, that’s not weird at all. It’s perfectly straightforward to see that a big ball and a small ball both have the same number of points (that number being the first uncountable infinity). The weirdness is that some sets of points don’t have a volume at all, even though they’re subsets of a set which does have volume.
One interpretation of “piece” would be that all of the points in each “piece” stay in the same position relative to each other, without regards to connectedness. This seems much more intuitively plausible, but I’m not familiar with the details of the theorem, either.
And Cabbage, are you sure that’s the definition you meant to post? By that definition, every nonempty subset of the interior of the ball is connected. Take U and V to both be the entire interior of the ball. Then U and V are both open subsets of the ball, U and V both have nonempty intersection with C, and U union V is still the whole interior of the ball, so C is contained in U union V. If I’m not mistaken, I think you also need to add the requirement that U and V have empty intersection with each other.
Thanks for catching that, Chronos. I did mean to say that U and V are disjoint.
As I understand it, you can’t blame this one entirely on the axiom of choice. As long as you’re willing to accept that some subsets of R[sup]3[/sup] can’t be assigned a volume, there’s no problem.
I think you’re right; the BT paradox can be proven under weaker axioms than choice. Historically, though, the connection is there.
Maybe it can be proven under weak versions of choice (otherwise it must be equivalent), but that doesn’t detract from it as an argument for rejecting choice. In fact it makes it also an argument for rejecting even certain weak versions of choice.
You have (with the additional requirement that U and V be disjoint) given the definition of not connected.
Oops. Put a “not” up there in my original post, as well as the “disjoint”.
No, very much not. Two or three of the pieces are dusts.
IIRC (IANA Analyst), the construction of Lebesgue nonmeasurable sets depends on choice.
I am so glad for this thread. I’d heard of the Banach-Tarski paradox - described in brief, and was trying to wrap my head around how it was possible with traditional geometry. Now I know that’s not at all the perspective necessary. Whew. Load off of my mind.
You might be right about this, I’ve never read through the proof and was just going on my expectations. It would have been cooler if each piece was connected, though.
I don’t think choice is necessary. Here are a couple of cites:
As your last cite points out, Hahn-Banach is sufficient, which is weaker than AC itself. However things could go two ways from here:
(1) Hahn-Banach is itself at least as strong as the axiom of countable choice. Finite choice follows from ZF, countable and general choice are independent of ZF.
(2) Hahn-Banach is weaker than any non-ZF choice axiom, which means that B-T doesn’t really argue coherently against AC (since the weirdness is inherent in ZF analysis). I don’t think this is the case, though, since again I seem to remember that some form of choice is needed for non-Lebesgue-measurable sets and the common wisdom is that B-T has something to do with measure. Basically, if Hahn-Banach is weaker than countable choice, then B-T has very little to do with measure theoretic weirdnesses and more directly to do with set theoretic weirdnesses.