Wikipedia:Reference desk/Archives/Mathematics/2018 September 4

= September 4 =

Question Regarding Banach–Tarski Paradox - Perhaps a single point is defined more than once?
First, please be gentle! This is my first ever Wikipedia post and it's both: i) very late - means I may be prone to erroneous thinking; and ii) regarding a topic in which I am in no way an expert - means I may lack some fundamental understanding. Before I simply ask for clarification on the topic, I'd like to offer that I believe I have done my homework before jumping on here to ask a groundless question out of impatient and effortless ignorance. In addition to the main Wikipedia article on the topic (references hereafter referred to as FTA) and its Talk Page, I've reviewed the 18 (non-just-link) search results from within this section for "Banach Tarski" , read the paper "The Banach-Tarski Paradox by Avery Robinson" , and watched "The Banach–Tarski Paradox by VSauce" (the entrance to and beginning of my trip down this rabbit hole). I've come back to this topic several times, and I believe my thoughts and conclusions (though obvious and most likely incorrect) are at the limit of where I can get without asking for help.

As a nearly-century-old theorem, I've discounted any possibility for an error in the application of any of the rules of set operations, because someone else would (likely) have identified such an error. Clearly the part that makes this a Paradox, and what is unsettling to me is, FTA, "...in this case it is impossible to define the volumes of the considered subsets. Reassembling them reproduces a volume, which happens to be different from the volume at the start.". Even if it is impossible to define the volumes at any given point in time, it would seem (to me) reasonable to hold as fundamental that a new "amount" of volume wouldn't be created by non-generative operations that did not exist before the operations were applied as a non-stretched surface would (by definition) have the same surface area in all cases and the relationship between surface area and volume is known. This relies on the fact that identical components are used in the reconstruction and there is no double-layering or half-layering (or the sort) in the resultant sphere(s), which I understand to be the assumption of the Banach–Tarski Paradox. To this end, I have tried to identify where, within the concept FTA in the section "Obtaining infinitely many balls from one", an error in logic may be hidden that would allow such a result to follow. My only thought is that for there to be enough points in sub-sets that create more than the full sphere S, indeed enough to create at least another full sphere S2, then there must be some error in assignment of the points in the original set, such that there is - in fact - hidden duplication in the sets mapping the points of sphere S without it being noticed. I must admit, that I am having trouble following the definitions of the sets and their operations FTA, and so I am relying on the simpler description from the VSauce visualization. Hmm, as I am writing this (as I am being careful with my references so as to not infer too much without sources) I am beginning to worry that my first ever Wikipedia post may be obviously childish. Oh well, onwards and upwards!

From the Vsauce video, the breakdown is into, in addition to the (shared?/assumed?/hypothetical?) centre-point, all points along the surface of the sphere S mapped to six sets containing the sets of points of 1) "[O]rigin" (point) and those with rotation sequences ending with a final : 2) [U]p; 3) [D]own; 4) [L]eft; 5) [R]ight; and then the remaining 6) [P]oles (points). However, I would think that any rotation that ends with an U (or, inversely, a D) can really just have the number of Us in the set replaced by a number of Ds equal to the countable infinite set of Ds minus that number of Us (maximum of which Us is an infinite countable number). The same would be true for the relation between Ls and Rs; I would think that L and R sets, like U and D sets, are opposite-sign-pairs (pardon the incorrect mathematical term here) like an axis with positive and negative integers going in opposite directions of the origin. My thinking is that from any given point, if the sphere is rotated a needed number of times minus 1 in one of the L-R or U-D direction, then it would eventually reach back at 1 rotation less than the same given point, which would be exactly the same point as 1 rotation in the direction of the opposite-matched-pair set. This (oversight?) would result in a duplication of counting of the points, without being obviously noticed? For each needed count of rotations for a given point to return to itself being a circle - a semi-unique number of rotations since there are multiple identical-circumference circle slices in a sphere - about either the L-R or U-D axis, it would seem that L and R is really L (or R) + / - in a range, rather than only in the positive values, where the + infinity and - infinity meet at the mid-point on the opposite side of the sphere (circle) from the given point. Likewise for U and D. The remaining two of the six sets given by Vsauce are those of O and P, and it could be that these are identical sets since each O could be selected to be one of the four possible Ps. From just these four Os, we would still be able to traverse every single point of the sphere from either of the two Os on each axis since, for the O-axis, a rotation about the non-O axis, either +1 or -1, will get us to the first circle of >1 point where a pole has only 1 point. From this first >1 point circle, we can continue to rotate away from the O to the other side of the sphere. For each successive circle we reach by rotating the non-O axis, we will reach each point by rotating about the O-axis an increasing amount of times in size until the mid-point of the sphere and then will begin rotating about the O-axis a decreasing amount of times to reach each point in each such circle, until we rotate about the non-O axis one more time and hit the opposite side of the sphere, which will be a pole. In fact, we could follow this process from each of the Os, but (I believe) that would result in identical sets (just out of order).

There are (at least) three areas that I have considered for my logic to be in error: 1) my rationale for de-duplication of the six sets of Vsauce into three are not valid; 2) the six (of seven) sets presented by Vsauce are themselves not valid interpretations of the Banach–Tarski paradox and its application to the surface of a sphere; and 3) my general understanding of the Banach–Tarski paradox - either or both FTA references and Vsauce - are totally bonkers.

I can't self-identify where my logic is in error. If my reasoning is correct, and the the minimum number of sets needed to fully describe all points along the surface of sphere S are the three necessary sets listed above only and the members of which are all absolutely needed to re-create the sphere S, thus there would be no remaining elements left to create a sphere S2. That is, the sets of: 1) Origins, each of which happen to be a Pole; 2) Rotations ending with a final Up (or Down) from those Origin(s), and 3) Rotations ending with a final Left (or Right) from those Origin(s).

Incidentally, if it helps explain my thinking, I imagined that half of the rotations ending in L and half of the rotations ending in R (and so, too, for U and D) would be valid even without my reducing of the sets. It is only when the number of L rotations begins to exceed half of the infinite number of rotations that such a first sequence would "cover up" (that is, simultaneously occupy, which is not valid since each point must have only one mapping) the same point where a second sequence that is identical but with Ls replaced with Rs would begin to reach half of the infinite number of rotations of R rotations (about the L-R axis). It is in the moment where (the limit of) half of the infinite number of rotations is exceeded that the two L and R identical-except-L-R-mirroring sequences overlap. I saw the same image for the U-D axis, only more sparse for whatever points might not of (if any) been captured by this first pass. And, somehow, I thought of the international date line uses + / - hours but needs that "mid-point" divider, lest the timezones of the earth be duplicated as +0 / -24, +1 / -23, +2 / -22, etc... always referencing two different days each - two whole globes, overlapped, accidentally defined by timezones left of and right of Greenwich. It's what got me thinking about how we could be accidentally duplicating the number of points by not noticing that "left of" and "right of" are both about the same axis like how "hours ahead" and "hours behind" are both about the same Greenwich Meridian. In this way, the number of points would be (at least) doubled, which would explain the existence of enough points to generate (at least) S2.

Anyway, it may all be worthless babble. My apologies if it has been a waste of time for the reader to have got this far. I would appreciate any thoughts and guidance. Thank you for your consideration. And for your tolerance of any formatting, procedural, or other rules that I may be unintentionally violating!

chad.trytten (talk) 08:48, 4 September 2018 (UTC)
 * First, if you like educational math videos then you might enjoy 3Blue1Brown and Numberphile, just in case YouTube's algorithm hasn't pointed them out already. Second, instead of using "left of" and "right of" it's more correct to say clockwise and counterclockwise, or if you're talking about the globe, east and west. You can always go 5 degrees east, even if you have to cross the date line to do it. Third, don't get too hung up on the details of the construction, they have been checked many times already and they are valid except for one controversial point which I'll be getting to. Fourth, the Vsauce video takes a couple of liberties for the sake of accessibility (you don't get millions of views by being confusing), so don't take everything said there at face value. For example early on (about 4 minutes in) when they talk about why interval between 0 and 1 is uncountable they ask what comes after 0. You could ask the same question about the rational numbers between 0 and 1, but it turns out that the rationals are countable so there's a fallacy there. What they are actually demonstrating is that the rationals aren't well-ordered which isn't the same thing. Later though they do give Cantor's proof which is valid. Similarly, the bit near the beginning where they're tearing the dollar bill (about 1:45 min) is a bit misleading; the Banach–Tarski construction does a lot more than just tear up the sphere. The individual pieces you get are so finely divided they can't actually exist as objects in the physical universe. An actual sphere is made of atoms, and though it seems like there are infinitely many, the number is actually finite, and the construction assumes that the sphere can be infinitely divided. The video does eventually cover this near the end but the dollar bill thing is what they show first. Fifth, there are two kinds of paradox, the kind where you actually prove a statement is both true and false, and the kind where you prove something true which 'seems' like it should be false. The first kind, of which a famous example is Russell's paradox, is very bad and must be resolved. The second kind, of which the Banach–Tarski paradox is an example, would be more aptly called a counterintuitive result and does not need to be resolved. Such a result indicates that either something is wrong with your intuition, which does happen occasionally even to the best mathematicians, or that your assumptions for some reason don't capture the situation you're trying to model. The Banach–Tarski construction is a logically valid consequence of set theory, and there's no actual theorem that says it can't exist, so it falls under the category of a counterintuitive result rather than a true paradox. So if you find it paradoxical then either your intuition about set theory needs to be adjusted, or the assumptions of set theory need to adjusted to better fit people's intuition. Which brings me (finally) to sixth, what's actually going on with Banach–Tarski. Intuition is generally based on experience and most people's experience with geometrical sets is they have either length, area or volume. For Banach–Tarski we're talking about the surface of a sphere which has a certain area. Intuition says that if you cut up a sphere then the areas of the pieces should add up to the area of the original sphere, and if pieces can be assembled to form two spheres then the areas of the pieces should add up to twice the area of the original sphere. But intuition doesn't allow for pieces that don't have an area and where assigning an area to them is like dividing by zero. And the pieces Banach–Tarski constructs are exactly this type, so you can throw away any intuition you have what you get when they can be reassembled. The fundamental assumption in set theory which allows these sets with no area, or more generally non-measurable sets, is the Axiom of choice. Needless to say the axiom of choice is highly controversial, as least as axioms of set theory go. Those in the anti-choice camp can (and do) point to Banach–Tarski to support replacing the axiom of choice with something else. On the other hand those in the pro-choice camp can point to formulations of the axiom of choice which are just as intuitive as Banach–Tarski is counterintuitive. Not an issue that can be resolved here, but for bettor or worse the axiom of choice is accepted as one of the standard list of axioms for mathematics. That may change over time though. I'm not sure if anyone outside math cares one way or the other; the video points to some papers near the end which seems to say it may play a role in physics, but I get a bit skeptical when I hear a phrase like "There have been a number of papers published suggesting...". Anyway, a long answer but there was a lot to cover when you include the video and the links. I won't claim to have reviewed all the material but hopefully this helps with the main issues. --RDBury (talk) 22:22, 4 September 2018 (UTC)
 * Short answer: the "international date line" duplication does not happen because the rotations used in the construction are by irrational fractions of 360°. Since they are not commensurable with 360°, no integer number of rotations but zero brings you back to the starting point (although you can get arbitrarily close). So your "needed number of times" does not exist.
 * There is also no reason why volume should be conserved in this case. Let's consider: why do we believe (correctly) that volume is conserved in normal circumstances and their mathematical, idealised forms? Because each piece of the original then has a well-defined volume, so their sum is well-defined and unique. When you break a unit sphere in half, each hemisphere has volume 0.5; so the volume is 0.5 + 0.5 = 1 regardless of where you put those hemispheres (you could put them back together, or you could put one in your house and one on Mars), as you can't add up the same numbers and get different results. But in the Banach–Tarski decomposition the pieces are not measurable; simply put, they do not have well-defined volumes. So we're not summing up any numbers in the first place. It just so happens that these pieces, when united in two ways, give two sets that have well-defined volumes of 1 and 2 respectively. There is no contradiction because we are not adding any numbers to give these two different answers. Double sharp (talk) 02:32, 5 September 2018 (UTC)


 * You might (emphasis on might, it'd constitute QUITE the deep dive, and you already dove in quite deeply) wish to review Alex Simpson's paper "Measure, Randomness and Sublocales", available here: http://homepages.inf.ed.ac.uk/als/Research/Sources/mrs.pdf, which demonstrates how the Banach-Tarski 'paradox' cannot arise in absence of (specific instances of) the axiom of choice. (For why I say "specific instances of", I suggest you review "Computability Beyond Church-Turing via Choice Sequences" by Bickford et al., available here: https://www.cs.cornell.edu/~lironcohen/pubs/FCS.pdf, and more importantly, figure 3 of "Bar Induction: The Good, the Bad, and the Ugly" by Rahli et al., available here: https://vrahli.github.io/articles/bar-induction-lics-long.pdf. If that, or all of that, constitutes too deep a dive for you for now, you might wish to start out with Andrej Bauer's "Five Stages of Accepting Constructive Mathematics", available here: http://math.andrej.com/2016/10/10/five-stages-of-accepting-constructive-mathematics/). --No identd (talk) 23:16, 5 September 2018 (UTC)
 * There is a lecture version of "The Five Stages...":, just in case people would rather watch than read. --RDBury (talk) 00:13, 6 September 2018 (UTC)

Approximate density of safe primes?
I found a few interesting papers related to the subject, but unfortunately all were behind pay walls. Is the average density known? I did do a little experimentation and was able to ascertain that just as the number P of primes below N is approximately N/log(N) [and so a probablity of 1/log(N)] similarly the number of safe primes S seems to be roughly P/log(N), or in terms of just N, S=N/log(N)^2 [thus a 1/log(N)^2 probablility]. Does that sound about right? Earl of Arundel (talk) 17:39, 4 September 2018 (UTC)

Just realized that this article gives an estimate for the number of Sophie Germain primes as being roughly 1.32032*N/Log(N)^2. That would imply that the same would hold true for safe-primes, right? But in fact when I apply that constant to the experimental results the relative error actually grows which doesn't make much sense at all. I mean for every safe prime there must be a Sophie Germain obviously. So the error should shrink as N increases, not the other way around. Earl of Arundel (talk) 19:23, 4 September 2018 (UTC)
 * If p is a Sophie Germain prime then 2p+1 is a safe prime. 1.32032*N/Log(N)^2 is the expected number of Sophie Germain primes p < N. If you count safe primes 2p+1 below N then you need p < N/2. The expectation becomes 1.32032*(N/2)/Log(N/2)^2. Asymptotically it's 0.66016*N/Log(N)^2. PrimeHunter (talk) 21:04, 4 September 2018 (UTC)
 * Ah, nice! That helps a lot. Thanks again PrimeHunter. :) Earl of Arundel (talk) 22:52, 4 September 2018 (UTC)

(See Safe prime )