So I've been working on a post for a while that has to do with the metamathematical/philosophical concept of finitism, i.e., that the notion expressed by Russel's paradox, which he expressed as the barber's paradox (which can't be solved by saying the barber was a woman, or maybe it can, but more on that later), should be solved, rather than by the axiom of regularity, by an axiom of finitism, where infinite sets are treated as proper classes. I'm rather dragging my feet on that, since maybe it turns out that trying to do math drunk is almost as foolish an endeavor as doing it sober. Consider the Achaemenids, blah blah blah.
Now, you probably know the name Tarski from the Banach-Tarski paradox, that you can chop a sphere into a finite number of pieces (nine) and reassemble them into two spheres, which Iâm now going to explain in tiresome, prosaic detail for really no reason at all. Â First accept the axiom of choice, in the form that given a set of disjoint sets, you can infer the existence of a set containing exactly one member of each. Â Now draw two axes through a ball at a funny angle - pretty sure one radian will do - anything that leaves no finite, alternating string of half-turns around one axis and third-turns (in either direction) around the other that ends up get you back to where you started. Â (I think one radian will work because its sine and cosine are transcendental - this Iâm sure of - and theyâd have to be the solutions to a finite polynomial of algebraic coefficients to fail this criterion - this Iâm also fairly sure of but canât be arsed to work out.) Â This guarantees that every such string corresponds to a unique rotation, and itâs easy to see that you can invert them by just going backwards (thus failing to alternate, Iâm sure I donât need to say). Â Now, first, letâs split the surface of the sphere into four sets - the fourth will be all the points that donât move in one of these rotations, and youâll see why Iâm calling it the fourth very shortly. Â In fact, by the end of this sentence: every point belongs to whatâs known as an âorbitâ of all the points you can get to by a rotation in the set of rotations described above, which is a set since the relation is reflexive (the trivial rotation - zero degrees - is in the set), symmetric (the inverse of every rotation in the set is in the set), and transitive (the composition of two rotations in the set is in the set), which for every point not in the aforementioned fourth set (i.e., those not fixed in any rotation; note that the orbits of those in the fourth set will be entirely within the fourth set - in fact, I think but canât prove the fourth set itself - if it were finite, itâd follow from the bijection, so stay tuned - since the operation of bringing it to the fixed point, giving it an appropriate turn, and bringing it back will keep it fixed while being a nontrivial element the set of rotations) has a bijection to the set of rotations, and having CHOSEN (would that I could figure out how to add sparkles to that) one from each orbit you get a set that itself has an orbit thatâs the whole sphere bar that fourth set, and that orbit weâll divide into three unions of elements of the orbit by separating the rotations into three sets. Â Yes, I will call that a sentence, thank you very much. Â Anyway, the rotations, the trivial rotation is in the first set, if you turn a rotation around the third-turn axis in one direction it goes from the first to the second to the third and back, the other it goes from the third to the second to the first and back, regardless of whether the instructions alternate or not. Â Likewise, if you flip a rotation in the second or third set, you get one in the first set, always. Â However - if you flip one in the first set, it depends. Â Remembering that each rotation has a unique alternating representation of flips and turns (letâs call âem that because why not?), if the last instruction in that representation is a flip, you obviously go back to whichever brought you here. Â If itâs a turn, you go to the second. Â So now you take these four sets on the surface and use them to divide the sphere into four sets plus the center. Â First, take the second and third away. Â Now, you can flip the first set to get the second and third, break it up, turn both parts to get two copies of the first, and flip one to get the second and third again. Â Since thatâs keeping the fourth and the center, youâve now got the whole ball plus a nice chunk of it. Â (If youâre keeping score, only four pieces - the one corresponding to the fourth set and center, one for the preimage of each of the second and third sets under the mapping from the first - have actually been rearranged separately, and one for the points beneath the second and third sets themselves. Â Only the lattermost will be considered from here on out.) Â The next partâs tricky, inasmuch as it involves a hypothetical ball and its hypothetical rearrangement into a subset of the other two sets, but thatâs not going to be what weâre doing in the end: to wit, hypothetically rearrange the hypothetical ball in the reverse of the way we build the last ball to get just the points under the first and fourth sets (and the center), then turn the first set to become the second, turn the fourth to be entirely in the other three (since the set of rotations we built was countable, the points of the fourth set on the sphere - i.e., the surface - are countable, and since only two rotations can connect a given two and there are uncountably many rotations since there are uncountably many angles, there being are uncountably many reals between zero and tau, there must be such a rotation) and just like the entirety of the three sets before break it in two and rotate it totally into the third set, and shift the center to be somewhere else, anywhere else, in the third set (âchosen,â which didnât actually make my phrasing, but pretend it did, gets no sparkles because itâs only one arbitrary choice, not infinite - more on that nasty little-big word soon). Â The next bit to break off is the points under the second and third sets of the sphere that youâll never get to by doing this to the points outside the second and third, or to the points you get from doing that, or from doing that, ad inf. Â (Arguably a different sense, but really also sort of the same - Euclidâs sense, to the extent he had one.) Â These stay put, while the points that you will get you break into five according to where you get them, and send them there. Â Since each iteration is walked back one, you end up with another entire ball. Â So now, like the barber, youâve got two balls.
Anyway, slightly simpler: Vitali sets. So consider measure: if you have a finite number of disjoint intervals, the amalgamation has a "length" equal to their sum. If you have a sequence of lengths of disjoint intervals that converges (e.g., 1 to 1.5, 2 to 2.25, 3 to 3.125, &c.), that'll converge to the sum, and if you have such a sequence that diverges, we'll say that's infinite, like the real line itself. So let's consider just between zero and one, and divide up the numbers such that in each set, the difference between any two is rational. Now, every such set will be of measure zero, but there are "too many" to put into a sequence, to that's fine. Now let's CHOOSE one from every such set, and since they're disjoint, if you add a rational number, you'll get a different set that'll have the same measure, and if you take the part that's greater than one or less than zero, you'll get two disjoint sets whose sum has the same measure, and moving one will give you a set of the same measure between zero and one. These can be put into a sequence, since the rationals can be, and they take up the whole interval, and since they're disjoint, they add up to the whole interval, but since they have the same measure, that's impossible, so they have no measure, not even zero. The aliens who don't understand Euclidean geometry might have a different notion of "number," and in particular "infinitesimal" from us, but that would take more explanation, which I don't have on me.
