The Sierpinski-Mazurkiewicz Paradox (is really weird)

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[@Sinful_Citrus]

As a polish person, I respect Zach's refusal to say the name of this paradox haha

[@Bemajster]

Szczęście mają że Grzegorz Brzęczyszczykiewicz nie był matematykiem 😂

[@zachstar]

😂 the attempt would've turned this into a comedy.

[@michaelblankenau3129]

Doesn’t look that hard to pronounce

[@chessematics]

@@michaelblankenau3129 but it is.

[@michaelblankenau3129]

@@chessematics maybe for you

[@murmurmerman]

Sometimes I feel like the term "paradox" really means "this situation works because of a concept I don't fully understand yet"

[@KoenZyxYssel]

I was going to ask in what sense this is a paradox but I think that's what they mean here.

[@cHAOs9]

I Downvoted this crap because this is NOT a paradox. It's a mathematical question with an unintuitive (i guess) infinity manipulation based solution. Infinity minus 100 is still infinity. So i guess 100=0. Infinity manipulating is bs.

[@KoenZyxYssel]

@@cHAOs9 Agree with the sentiment but I figure Infinities imply dimensionality so 100=0 is fine but the 100 and the 0 are just coordinate components...

[@VioletCatastrophe]

A veridical paradox is a situation that seems unlikely or absurd that is nonetheless true, highlighting a disconnect between intuition and reality. This is that situation. So yeah, veridical paradoxes are indeed 'this situation works because of a concept I don't fully understand yet'. It's a legitimate class. More famous paradoxes might be unresolvable situations (often as a result of time travel in fiction) that feel more satisfying due to their unknowable situation, but they are not the only type.

[@VioletCatastrophe]

@@cHAOs9 This is a clear lack of understanding. You are trying to view infinity as a number, which is almost certainly is not. To highlight this, lets take the idea of shifting to the left by 1 to get back to the original set. We aren't going for anything crazy here, or counter-intuitive mind you. We are going to use the set of whole numbers, positive and negative. We create a new set from every element of that original set, minus 1. So 0 in the original set has corresponding element -1 in the new set, 5 has corresponding element 4 in the new set, and so on. So A={..., -3, -2, -1, 0, 1, 2, 3,...} becomes B={..., -4, -3, -2, -1, 0, 1, 2,...}. From our narrow perspective, it looks like we 'gained' -4, and 'lost' 3. But we haven't. We can always get the next element by adding one to it. That's just how numbers work. That's how infinities work. But lets suppose there exists some extremely large number N. N is in our original set A, however was so big that it was the last number, so N-1 becomes the final element in our set B. But this means that there exists no N+1, because N+1-1=N, which would be in B. However there is no such integer that exists such that the next integer does not exist. There is always another integer. Whatever perspective we choose is inherently limited. Because it is finite, and these are infinite sets, because there is no limit to the integers as we just said. We could write A as {..., -1000000000000000000000, ..., 1000000000000000000000,...}, and B as {..., -1000000000000000000001, ..., 999999999999999999999,...} and there would STILL be a 1000000000000000000001 in A that became the 1000000000000000000000 in B that got rolled into those ...'s. Equally, we did not 'gain' -1000000000000000000001. A always had a -1000000000000000000001, and -1000000000000000000002, and -1000000000000000000003, and so on, *literally* ad infinitum. Saying something is an 'infinity manipulation' simply comes from trying to understand infinites from the framework of the finite. Which is understandable, because our pea brains cannot grasp infinity. The purposes of veridical paradoxes like these is to show us the flaws in our intuition to better respect and comprehend how bizarre infinity is, so that we don't fuck up when our monkey brains try to understand the incomprehensible. Also as an addendum, your example of '100=0' is completely flawed, just take the example I laid out above but with a -100 instead of -1. What it means is that for every element in the set of integers N, we can define some N' in a derived set such that N' = N-100. We know that N-100 must also be an integer, so N-100 belongs in the set of integers, and we also can map every N in the set of integers to some N'=N-100 in this new set which is also in the set of integers AND we can map every N' back to the original N. So we can one-to-one match every integer N in the integers with every integer N' in this derived set, which means they must be the same. What you messed up is not that infinity-100=infinity, because you DO NOT do arithmetic with infinites. It is that (using said incorrect process since as I said, you don't do arithmetic with infinites) infinity - 100 -> infinity, and 100->0. 100 doesn't equal 0, 100 is the element that gets mapped to 0. This is a VERY important distinction.

[@NotHPotter]

It really is fascinating the difference in tone between your videos explaining concepts here and the just raw madness of your humor on your personal channel. I also feel like Zach Star Himself would give Zach Star a wedgie.

[@absobel]

AAAAAH I DIDN'T RECOGNIZE HIM I was wondering since when was I subscribed to a math channel

[@muenstercheese]

lmao i feel like you're right

[@zachstar]

😂

[@ZachStarHimself]

I absolutely would

[@wrathfulstorm7174]

@@ZachStarHimself He sounds like someone who believes that the pyramids were not built by the aliens

[@johnchessant3012]

Putting the sponsorship before the solution so that we'd have more time to try the problem was real sneaky. Zach is so good at getting us to not fast forward through those :D

[@DocSeville]

Dude....

[@rev1601]

This paradox is basicly just Hilbert's Hotel all over again, right? You take an infinite number of objects, apply a specific transformation to all of them, and you end up with all the objects you started with, plus infinetely more new ones.

[@gcewing]

It's Hilbert's Hotel with a revolving restaurant on top!

[@dumnor]

@@gcewing infinite hotel has infinite revolving tops...

[@eduardoguthrie7443]

That's it. If all Hilbert's guests in rooms divisible by 2 check out, the remaining guests can add 1 to their room number and divide by 2, and switch to that room number. The departing guests can also fill another infinite hotel, with no vacancies in either hotel. It's different math, but the same infinity phenomenon.

[@whatno5090]

it is an example of that. fun fact though, without the axiom of choice, it is impossible to prove that you can do hilbert's hotel with all infinite sets. There may be some infinite sets (called dedekind-finite infinite sets) which cannot be cut up into two smaller pieces where one has the same size as the larger whole. In fact, there can even be some infinite sets called *amorphous* sets, where if you cut it up into two pieces, one of them has to be finite! However, the axiom of choice (or even a small fragment called "countable choice") bans any such collections from existing

[@mishie618]

​@@eduardoguthrie7443that was the first thing I thought of as well.. it's so similar, but with a rotating rooftop lol

[@trymbruset3868]

I am so happy this was suggested to me, thank you so much for making this. It may seem a bit silly, but I found a real passion for mathematics in university and examples like this were my driving motivation for learning. I have since moved into development work, and been promoted into a managerial role where I work on interesting things, but have lost touch with this, lovely, interesting and beautiful side of mind play. This felt like discovering your favourite album growing up had an extra track you just hadn't heard, and it being exactly as rewarding as you recall it used to be. Thank you

[@LegendBegins]

I appreciate the paradox, although whenever something unintuitive/apparently contradictory happens because we used an infinite set, it sorts of feels like cheating =P

[@RobertBlair]

yeah, like take the positive integers, split it into two, do some math, and now you have the positive integers twice. Infinity does weird things with our intuitions.

[@guest_informant]

Agreed. Before the whole thing started I was thinking, this will just be an infinite set.

[@Vgamer311]

@@guest_informant that’s not really saying much, since any shape or line segment you could possibly draw on a Cartesian plane is an infinite set. Given the prompt it would almost be weird if you *didn’t* choose an infinite set.

[@seanshameless0]

I feel like the real part about this that’s cheating is the use of complex numbers.

[@mittunsudhahar634]

@@seanshameless0 Why? He just said the statement is on a 2d plane - you can easily take the real coefficient to be x and the imaginary to be y.

[@Jakub1989YTb]

It's really hard to focus on the science stuff after binge watching "zach star himself" :D

[@redjr242]

Same lol

[@Jakub1989YTb]

@@redjr242 It's like God himself is speaking to me while trying to get through TSA to see his ex.

[@ZachStarHimself]

Agreed

[@marcosgutman6349]

I could see his face when he said "the answer better be no for everyone."

[@tomasstana5423]

Quite the opposite here. I like zach star himself, but i love zach star :D

[@Mutual_Information]

That's wild. Any ostensibly geometric problem whose solution comes from algebraic geometry.. count me out.. It's mind bending stuff.

[@chessthejameswei]

Few are capable of reading such a title and not clicking instantly.

[@CommandoTM]

not a maths major, but just Sierpinski in the title got me

[@nHans]

*The essence of paradoxes,* which I've distilled after studying a large number of them, is that there are 2 types: 1. Arising from a *negated self-reference.* In this case, the proposition you started with is false. It's just a subset of proof-by-contradiction. 2. *Involving infinity,* in which case, it's not really a paradox; it's just counterintuitive to those not versed in the magic of infinities.

[@hurktang]

The paradox is the amount of people who still believe that ∞+1=∞ and 2∞=∞ after ALL these counter example. They are not the same...

[@pierrecurie]

@@hurktang It is true for IEEE754

[@hurktang]

@@pierrecurie I'm fully aware that even renowned mathematician believed that. IF you never EVER refer to another infinity, you can get away with this. In this example, the problem arise when he attempt to move ALL the infinity. It was the second time he referred to THAT infinity. I'm well aware that the field of mathematic still cling to that concept of "all infinities are infinite and therefore equal". But I'm fairly confident that while I'm in the minority group who object that concept, history will give me reason on the long run.

[@pierrecurie]

@@hurktang Last I checked, there are at least 2 sequences of infinity, which most math completely ignores. It's rare to go beyond the cardinality of the reals.

[@flambambam]

@@hurktang A friend and I defined 1/0 to be some "infinite" constant. Doing this lets us distinguish between infinities by simply writing that infinity as 3k+7, where k is 1/0. Not very practical for most things, but we found it useful in determining how quickly some functions approach infinity.

[@apowergso]

“Don’t think about this visually” Me, an aphantasic: I was born for this very moment

[@fullfungo]

(before watching the answer) I found a solution for B’ = S. Since we deal with rotations, let’s start with a unit circle U centred at the origin. Let’s start with A = {(1,0)} and B = U\A. Now B is a circle with a hole on the right. If we rotate B by 1 radian, the hole will land on a point 1 radian along the circle (p1), which is not in B’, but is in S. So let’s remove p1 from S. Now B and B’ have two holes and p1 doesn’t appear in either. However the new hole now lines up with a point 2 radians along (p2), which is in S, but not in B’. So let’s remove p2 from S. Continuing this logic we will remove point that are n radians along the circle (p1, p2, …) from U. So A = {p0} = {(1,0)} and B = U\{p0, p1, p2, …}. Now every hole in B’ is p1, p2, … and as such B’ = U\{p1, p2, …} which is exactly S. p0 is not removed, i.e. not in the list p1, p2, …, since if it corresponded to a point pn, this would imply n=2πm, and thus π=n/(2m) making π rational, which it is not. I’m sure this can be extended by adding all point +1 along the x-axis, rotating and removing point, and so on.

[@jonipaliares5475]

How interesting! The solution I came up with for B' = S was, in a way, the opposite of yours. I considered S to be all points (x, y) such that x = cos(n) and y = sin(n), where n is a natural number (starting at 0). I then took A = {(cos(0), sin(0))} = {(1, 0)} and B = S\A. You can see that A and B are disjoint because otherwise it would imply that there exists a non-zero integer k such that cos(k) = 1, but that can only be true if k = 2πm, that is, if π is rational, which is known to be false. You can also see that, by definition, S is the union of A and B, therefore {A, B} is a valid partition of S. Okay, so we got: A = {(cos 0, sin 0)} B = {(cos 1, sin 1), (cos 2, sin 2), (cos 3, sin 3), (cos 4, sin 4), (cos 5, sin 5), ...} And rotating B by 1 radian clockwise we obtain: B' = {(cos 0, sin 0), (cos 1, sin 1), (cos 2, sin 2), (cos 3, sin 3), (cos 4, sin 4), (cos 5, sin 5), ...} = A U B = S My set S consists precisely of the points you labelled p0, p1, p2, etc in your solution.

[@fullfungo]

@@jonipaliares5475 cool I think they would correspond to p0, p-1, p-2,… in my notation actually

[@jonipaliares5475]

@@fullfungo Oh, right, your points p0, p1, p2,... go counter-clockwise around the unit circle

[@nathanderhake839]

Wouldn’t A have to be more than just a single point since moving (1,0) one over to the left gives you a set with just one element: (0,0), which is not the same as S.

[@fullfungo]

@@nathanderhake839 as I said, it’s a solution with B’=S, since a solution with A’=S was already presented (infinite half-plane). If you want to satisfy both conditions simultaneously, you need a complete solution like the one in the video.

[@eustachiomartinez7298]

I was waiting for this kind of videos of yours for months. I love your work as a mathematical content creator. U re one of the bests.

[@skyscraperfan]

Is that really a paradox? It would be a paradox if a set was equal to one of its proper subsets, but with shifting or rotating that should not be impossible. It is like the hotel with an infinite number of rooms, where all rooms are occupied, moving an infinite number of guests to new rooms creates room for new guests. Moving the guests is the key.

[@mironhunia300]

Paradox has 2 meanings. The first is an impossible, contradictory situation. The second is something that defies common sense/is counterintuitive, but isn't actually contradictory. A lot of uses of word "paradox" in maths refers to the second definition.

[@angelmendez-rivera351]

@@mironhunia300 Yes, but it is worth noting that the second usage originated entirely by mistake, because people could not tell the difference between genuine contradictions and counterintuitive theorems. If a result takes on the latter result, we should call it a theorem, not a paradox.

[@mironhunia300]

@@angelmendez-rivera351 Source? Because it's the first time I'm hearing of this and, in fact, when browsing dictonaries it was more common for them to skip the first of the meanings I listed than it was to skip the second meaning.

[@vlix123]

@@mironhunia300 At least in math, I don’t think paradox should be used to describe something “counterintuitive”. Different people (especially mathematicians) have different standards for counterintuitiveness. Mathematics is a place of rigour and absolutes. I don’t believe that something as subjective as “paradox” should apply in this world.

[@mironhunia300]

@@vlix123 On the other hand, labeling actual contradictions as "paradoxes" is even less useful, because they are simply contradictions, no fancy word needed. Sure, being counterintuitive is subjective, but historically stuff like Banach-Tarski paradox, Hilbert's hotel and Zenon paradoxes made its impact exactly by being counterintuitive and thus showing the need for rigour in mathematics. Clearly, the word paradox has been used like it for a long time. In fact, "unexpected" *is* what the word means in greek.

[@amaarquadri]

Here's another set that you can take a subset of, rotate that by 1 radian, and get the original set back: e^(ni) for integers n ≥ 0. The set is a bunch of points on the unit circle. If you remove the first point which is at 1, then rotating everything clockwise by 1 radian makes all the other points take each other's places except for e^i which goes to 1. And we know that the chain goes on forever since there are 2*pi radians in a circle which is irrational.

[@linoarenz548]

A part of the set is actually nice to visualize. Just imagine the positiv integers wrapped around the complex unit circle A supset of the Projektion points has this magnificent subset rotational property

[@Andrew90046zero]

I would still be cool to see an animation of the points on the complex plane moving around. I know theres infinite of them, but just the ones that fit on screen :P Because I'm still trying to wrap my mind around what the rotation looks like on these points.

[@TheEternalVortex42]

The set S is dense in the complex plane so there's no way to draw the points, it would just look like a solid square. (Of course S is countably infinite and C is uncountable).

[@guardingdark2860]

If it helps at all, the fact that it's a rotation isn't really all that important and, I would argue, something of a deceptive tactic. The "shift" and "rotation" are just the visual equivalent of mathematical operations; what we're really doing is the subtraction by 1 and the multiplication by p^(-1). It seems to me that the use of those terms is meant to throw you off by priming your brain with the idea that, a transformation is a "movement", and thus they should intuitively end up in different spots, making it seem more strange when they both give the original set. That's why I said it's deceptive, though I would note that I don't mean to imply malicious intent, of course it's just for the show of it.

[@pmcate2]

@@TheEternalVortex42 Must sets that exhibit this behavior always be dense? Seems like the case

[@DavidPysnik]

Very clever in that it contains a traditional translation encoded by a shift of one to the left that drops all of the constant terms by one and a sort of “power translation” encoded in the rotation that drops all of the powers of each term of the polynomial by one. I could imagine this perhaps being discovered in reverse, though. That is, someone studying sets that do the aforementioned types of transformations along with their associated operations and then realizing they could form the weird “paradox” presented.

[@DanGRV]

I'm pretty sure the proposed set S is dense in the plane (haven't proved it yet). The building blocks required for that proof would be something like this: 1. The set of powers of p is dense in the unit circle. Therefore, you can approximate any direction with arbitrary precision. 2. You can make points arbitrarily close to the origin (for some n,m such that p^n and p^m have almost opposite direction, in which case p^n+p^m is "almost" 0, has very small size). 3. You can multiply p^k by integers so you can get a rough approximation of any point in the plane (for example, choose the floor or ceiling of the size of the target point). 4. Add to that rough approximation appropriate small numbers of the form p^n+p^m in order to get an arbitrarily good approximation.

[@user-tn2dk2pg2p]

Yeah, in the end, it's enough to show that you can get points arbitrarily close to the origin so that scaling and rotating gives that the set is dense in the plane. That's true by again noting that you can get points arbitrarily close to -1 (by the density in the unit circle), and so adding 1 to such points gets you arbitrarily close to 0. :)

[@eriksteffahn6172]

All you need to do to formally prove this: 1. The powers of p are dense in the unit circle, so the unit circle is contained in the closure of S. 2. The closure of S is closed under multiplication with a positive integer and addition, because S itself is closed under those operations. 3. Any point in the unit disc can be written as the sum of two points of the unit circle (z = (z/2 + iz/|z| * (1-|z|^2/4)) + (z/2 - iz/|z| * (1-|z|^2/4))), so the unit disc is contained in the closure of S. 4. Any point in the complex plane is an integer multiple of a point in the unit disc, so the entire complex plane is contained in the closure of S, which means that S is dense.

[@modolief]

Wow, great video! And I love the proof-sketch style, extremely clear. I understood all of this perfectly (minus the fact that I'm forgetting how to do rotations in the complex plane, but I remember having done those). Thanks !!!!

[@pyglik2296]

That's a very nice result, but in mathematics involving infinite sets it's not *that* surprising. For example you can divide the set of all natural numbers into two sets, odds and evens, that are both the size of the original set and can be mapped into it. I also wonder about the nature of the set S. Is it like rational numbers, that you can always find a new point between any two others, or does it have empty places where there is no point?

[@joshuazelinsky5213]

To answer your second question: the property you are referring to is the rationals being dense in themselves (or dense in the real line which has a slightly different meaning). But there are a bunch of different generalizations of this to the plane. In this case, the definition you want is whether for any given real number x and any r >0, is there necessarily an s in S with |x-s|< r. That is, for any little circle you draw in the complex plane, no matter how small does it necessarily have an element of S. The answer to this question is yes. Here's a proof sketch: Our starting point p is irrational and has absolute value 1. So the set of points p, p^2, p^3, p^4 ... are dense on the unit circle (in the sense that for any point on the unit circle there are elements of this list arbitrarily close to them). So we can then by taking sums from that set get as close as you want to anywhere you want. (This second step takes a little to see.) Since those are precisely the elements of S, we are done.

[@pyglik2296]

@@joshuazelinsky5213 Thanks! I forgot it was called a dense set and wasn't sure how to generalise it. The second step makes sense, for any point there is at least one way to get to it by line segments of length one and since we can get arbitrarily close to any angle we want we can get to this point as closely as we want. Nice reasoning!

[@joshuazelinsky5213]

@@pyglik2296 Your question brings up another question though which I don't know the answer to: is any set S with this property (that is it can be partitioned into A and B where a 1 unit translation of A gives all of S and a rotation of one radian of B gives all of S) necessarily be dense? It is the case for this construction, but it isn't obvious to me that any such set must have this construction.

[@bob53135]

@@joshuazelinsky5213 I'd say yes. If you have a starting point Z, then you can similarly construct a bunch of polynomials in p that have to be in S (these are non-negative integer polynomials + Z·p^n, where n≥degree of the polynomial). With n big enough, you can then find a polynomial that is close to any point minus Z·p^n.

[@joshuazelinsky5213]

@@bob53135 Yeah, good insight there. The key observation that you are making is that morally any set that has this property has to end up algebraically looking a lot like the originally constructed S set. Thanks!

[@kirimaru73]

This is a brilliant way to explain the original mind-boggling paradox related to two furry spheres and nasty NESW notations.

[@horrorspirit]

> furry spheres I hate the way my mind visualized that phrase

[@miklov]

I got sidetracked half way through, researching the benefits of various productivity tools and then I wasn't in the right frame of mind to focus on the remaining video. But I hope to get back to it since it seems fascinating. I enjoyed the part of the presentation I did focus on and you have earned a new subscriber. Thank you!

[@MichaelRothwell1]

Great video and fascinating paradox. After the example for A, I realised that this example could be "simplified" by letting S=N (natural numbers on the real axis of the complex plane), B={0}, A=N\{0}, so translating A left by 1 gives S. I then realised that if you wrapped this set around the unit circle, it would give an example for B. So we take S to be the non-negative integer powers of e^i,A={1}, B=the positive integer powers of e^i, so rotating B clockwise by 1 rad gives S. It turns out I was on the right track!

[@scruffyy]

**Me enjoying math class **Kid saying "We are never gonna see this in life" **Me binge watching Zach Star videos 20 years later 😁

[@freddynovember5842]

It feels like this is some kind of explanation of 3d without someone thinking about perspective being the new dividend

[@BabaBabelOm]

Struggling to imagine how a logical fallacy, starting with a contradiction, can create a paradox… partitioning a paradox would create logical fallacies though

[@TechnocratiK]

I feel as though this is, with only slight modification, entirely equivalent to Banach-Tarski if instead of using polynomials over the complex numbers we use polynomials over the unit quaternions.

[@rackyphyr]

that's why he brought it up at the end of the video lol

[@3moirai]

Great to see a new math video! I always learn something new

[@Arkalius80]

This makes me think of the Banach Tarski paradox which is even weirder (if easier to say)

[@TheLuckySpades]

This reminds me of the infinite dictionary and how removing the first letter in a section gives you the whole dictionary back

[@ratzou2]

10:40 I think I disagree there ? A-1 = S means that all the elements of A-1 are in S and vice-versa. Yet the polynomial -1 is definitely in S but isn't in A-1, there is no element of A to which you can substract 1 to obtain -1, the only element of S that would apply is 0, and 0 isn't in A. Am I wrong ?

[@KarlFredrik]

Zacks explanation is so clear that I don't even notice it to be paradoxical

[@martijnonderwater]

Nice video, it made me reminiscent about my time as math-student! I did have to think a moment about whether I'd name this a paradox, but yeah I can live with that :-) Also, when proving that A+1=S with both being infinite sets, it's probably necessary to show A+1 is in S, and that S-1 is in A. At least, it was not precisely clear to me that the examples you gave formed a 1-1 mapping between the two sets. Ahum, I hope my math memory has not failed me here :-)

[@vibinpenguin7021]

What's wrong with this paradox? I don't really understand, infinite sets are strange, we know that already :) For example, when you shifted all Re(number) >=1 to the left by 1 and ended up with all non-negative numbers as your new set (AKA A turned to A and B), that wasn't really too surprising.

[@theunknown4834]

Did you expect a geometric problem to go to algebra?

[@fearstrofic8837]

@@theunknown4834 Yes... Why not?

[@Boringpenguin]

Yeah I think these kinds of "paradoxes" are just showcasing some of the inherent properties of (uncountably) infinite sets. They are definitely strange but that's how they work based on the axioms we've chosen. Great name btw

[@angel-ig]

@@Boringpenguin The set S is countably infinite though

[@cobalt3142]

This is a veridical paradox, i.e., it seems counterintuitive when you first hear it, but once you know the solution or how it works the paradox is resolved

[@toniokettner4821]

naming e^i = p and then looking at polynomials over p is a war crime

[@josefstr]

New video by Zach Star! Day is saved! 😍

[@shipwreck9146]

"Do not think about this visually" me, a physics major: *panics*

[@neuvocastezero1838]

This is cool, sort of like the "Where can you travel 1 mile north, 1 mile west, and 1 mile north and end up in the same place?" riddle, but in a higher dimension.

[@TmOnlineMapper]

That's the fun part of playing with infinity. Quite a few basic conservation of properties principles just don't apply anymore.

[@ckq]

Is this similar to that Banach tarski paradox vsauce posted like 6 years ago and took me watching it 5 times in like 5 years to finally understand it. Oh nice you talked about it at the end.

[@danielrhouck]

One where you can say yes to the second question (but not the first), with a somewhat visual representation, would be to take the point (1, 0), then that point rotated about the origin by 1 radian, and 2 radians, and all other positive integer numbers of radians. This is a countably infinite set and no later point will ever correspond with the original. Let the set A be the original point, and B be all the others. This has a *somewhat* visual representation because if you try to do draw it it looks like youʼre just drawing the unit circle, but itʼs still easier to comprehend that than the mess where you go with all polynomials instead of just (when you see it as the complex plane) powers of p.

[@maxmuller445]

Great solution

[@AssemblyWizard]

Worth mentioning that this relies on pi being irrational

[@danielrhouck]

@@AssemblyWizard Yeah; so does the main set in the video.

[@DerekSLedbetter]

You can extend this solution to construct countably-many disjoint sets that solve the problem. Instead of just polynomials over p, consider all power series with nonnegative integer coefficients. You can't actually use these, because they don't converge at p if they go on forever. But some of them are expressible as rational functions (one polynomial with integral coefficients divided by another). Then the analytic continuation is defined at p, and each series gives a distinct point. You can partition these series into equivalence classes; two series are equivalent if they are equal after removing i elements from the start of one series and j elements from the other series. Then each of these classes is a solution to the problem. One of these is your solution, which is all series that eventually become all zeroes. I don't know if these are the only possible solutions.

[@pirsabel]

A useful way to visualise the sets would be to cap the degree and coefficient range - the full sets would be dense in the complex plane, but countable.

[@pirsabel]

(proof of denseness follows from wellknown uniform distribution of the powers of p on the unit circle imo)

[@MrRoyalChicken]

Ok, I needed to pause the video at 2:40, but after four or five minutes I found a suitable set S and proper disjoint subsets A and B with their union being S, such that both the left shift of A by 1 and the rotation of B by 1 radiant are equal to S. You simply start from S_0={0} and work your way back, iteratively taking into account points which you need to satisfy either the property of a subset A or B. Here's how it goes (Haven't watched the video any further yet, wouldn't be surprised if this is precisely the construction presented in it): Observe that for any set T, the set obtained as a union over all right shifts of T by whole numbers n >= 0 is covered by the left translation of the set where you require n >=1. The latter one is a good candidate for A if the first set is S. A similar thing can be done with counterclockwise rotations by n radiants, again looking at n >=0, and n >= 1, giving a larger S with suitable B. Since the set got bigger in this step, you have to redo the first step to find an appropriate A again. Iterate this process and take the union of all A's, B's and S's obtained in finitely many steps as the final answer. Haven't seen this paradox before. Feels a bit like Tarskis paradox. Been studying mathematics at uni for five years now though... :)

[@evanwright365]

I fuck with the fact that you posted this. Probably wont get as many clicks, but i really enjoyed it

[@kinyutaka]

3:32 the set that works with is the Set S containing unit squares in the top left, and top right quardrants Subset A us the top right, Subset B is the top left. Move A to the left and it completely overlaps B, turn B to the right and it moves to the spot of A. Because the two Subsets are the same size and dimension, they become the same Set S

[@marcoleung5078]

I think you misunderstand something. It's not that both A and B cover the S after both transformations are applied. It's that they individually must cover S. i.e TA = S and T'B = S where T is the translation left by 1 and T' is the rotation by 1 radian. NOT TA + T'B = S. So unless rotating the top left quadrant gives you the entire top half of the plane, then this solution is incorrect. Also, 1 radian is not 90 degrees. 1 radian the angle where the arc length equals the radius.

[@ГеоргиГеоргиев-с3г]

A set which after rotation gets back to the original set would need to get the new points from infinity so just like the set x=positive you can move it back by 1 it turns back, so the rotational set would be any set that loses points to infinity after rotation. I would use a circle with radius infinity around the origin point, cut in half down OY and shifted up by it's radius to be able to be visualised(you can construct such a circle) the other end will be at infinity at which point you can and and substract points as needed only with rotation because you get a line version of the translational problem, only due to infinity being infinity.(a shadow of a circle on the number line from the point opposite it) you can add and subtract as needed rotate it and get the original set(minus 1/0, the bit at infinity).

[@davidbeckham2715]

I didn't know about this paradox but the moment I heard of rotating a set of points I just thought of the complex plane and boom that's where it happens. Amazing video

[@cHAOs9]

In what stretch-of-the-imagination is this a paradox, in any way, what-so-ever?

[@ReifAndreas]

Axiom of choice was not involved (12:43)? Is that right? Please think over it. You chose polynomials with p=e^i (6:21). That is only guaranteed by the a. o. choice. You cannot claim such an set without it. (AoC is not only about choosing from "different", disjoint sets. Wikipedia illustrates that a bit misleading imho. You can also think of AoC as choosing elements of an set, which of course can be sets.)

[@Fazz321]

Simple explanation: 1. Moving infinite set to the left should raise a question about all those points at the inifinity. This operation is not "moving the set", it's more like "shifting left boundary of the set". 2. You can't just change your space from R^2 to space of complex polynomials, those are not equal. So it's not a "turn" in a R^2 sense of the word anymore, it's not "turn" at all. That's why you can't really show it on R^2 plane, if objects of the set were just points this wouldn't be a problem. So trick is in the replacement of operations. I don't think it's allowed in math but it's a fun paradox to think about.

[@ArionEquus]

Im not sure this is a paradox as much as it is something that just works because your selection of points is just vast and complex enough that you cover all possible places where the points could fall when you start doing the math.

[@jongyon7192p]

Not a problem to visualize at all! Easy enough that I'll describe the picture even. The set of all p^n forms a circle. It's more of a subset of a circle, but it's entirely contained in it. Next, we multiply this circle by int a, and you get a set of integer-radius rings going infinitely outwards (and 1 dot in the middle). This is our B subset. They're all rings so you can rotate them (by an integer radian amount) and nothing changes. For our A subset, take this ring pattern and copy paste it to the right 1 unit, then again to the right and again and infinity. So you can move this 1 to the left and get back S. And that's it. ... ...right...?

[@wyboo2019]

i finally finished the problem and the trick that i was missing seems so obvious in hindsight. to anyone else that can't figure it out: once you have one value that is definitely in A, you can figure out values in B, which are then in the union of A and B, which are then also members of A translated left by 1 and B rotated by 1 radian, which then yield new values in A and B, and etc. then its not hard to prove via induction that A and B contain the polynomials in p as described in the video. i'm not a huge fan of complex numbers and prefer a geometric algebra approach to these types of problems so i'm going to do it again tomorrow using vectors and rotors from Geometric Algebra

[@yurisich]

You can approximate the visual transformation of applying p^-1 as a spiral that shifts points from the current solution, to the "next" solution, which is identical to the starting position, minus the elements which contained only a degree-one term. Those would head to unity at each application of the rotation.

[@theultimatereductionist7592]

I love math problems that require non-measurable sets! However, you did it with countable (hence measurable) sets.

[@FranzBiscuit]

Great presentation! And TBH I don't think I would have been able to grasp the concept either had the explanation given not been so lucid. Well done, Sir. =)

[@polyrhythmia]

I think I blew a gasket trying to understand this, mainly, how they were able to think of this.

[@sharpnova2]

very nicely explained. you might have included the spoken definitions for A and B above their representations as you did for S. for some viewers, that would make it easier to follow. a paradox is usually something involving some subtle issue with semantics that results in something necessarily being simultaneously true and false until we define things more deeply. (like "the smallest number that can be unambiguously described in 14 words or less" or "is the set of all sets that are not subsets of themselves a subset of itself?") this, rather, is just "weird"/"unexpected" (and incidentally, quite beautiful)

[@tylerduncan5908]

My first thought was to take the set of every number (x,y) such that x=r(sin(mπ/n)+Z and y=cos(mπ/n) With Z,m, and n all being integers That way, every horizontal translation by 1 gives you a new point already in the set, and any rotation by any fractional value of 2π with numerator 1 again gives you another number in that set.

[@seneca983]

There are a few problems with that: 1) The goal wasn't to come up with a set that equals itself translated and rotated. Rather you should have two proper subsets of that set one of which equals the set when rotated and the other that equals the set when translated. 2) 1 radian is not any fraction times 2π. Their ratio is irrational (and transcendental as well). 3) Your expression for x has an unmatched parenthesis. 4) I don't think your choice works for rotations (by other than an integer multiple of π) work in any case. Your expression for x includes r (which presumably can take any real value) but y doesn't so y can only take values in the interval [-1,1] whereas x can have any value.

[@medexamtoolscom]

S is the set of (x,y) coordinates such that y

[@秋風英語足です]

This is the same as the example in the video, where half of the plane was translating on himself: it does not work with a rotation(your group has a triangular shape and the one in the example was rectangular, but basically it works in the same way); the group of polynoms used in the paradox, instead, remains the same both after a translation and a rotation.

[@pi4313]

I want to see the set!!!

[@juha-petrityrkko3771]

This presumes an agreement that rotation in complex plane is an operation with the same qualities as rotation in Euclidean xy plane.

[@unchaynd7266]

Zach: doesn't matter, just pick a subset of the x-y plane. Me: *picks a single point* Video: 0:27 Me: 😧

[@alecesne]

Thought I was following along until about 6:07, but suddenly hit the end of my mathematical education like a guy sliding down a rope only to discover there isn’t any more rope!

[@NikolayVityazev]

That is so cool! Thanks for bringing it.

[@arongil]

Thanks for a thought-provoking video here. No junk, just math well explained :)

[@aa01blue38]

My solution for a subset B that becomes the set S after a rotation by 1 radian: Rotation by 1 radian clockwise in the complex plane is just multiplying by e^(-i). We can reverse engineer the problem by finding some B such that multiplying every point by e^(-i) yields a new set with all the points of B, as well as some extra points not in B. We will call this new set S. We can then partition S into B, which we have already determined, and A, which is just the extra points not in B. A only needs to be non-empty, so as little as 1 point will do. We can choose a simple point for A, say (1,0) or if we use the complex number equivalent, 1. The number which becomes 1 when multiplied by e^(-1) is just e^i. This is our first point in B. The number which becomes e^i when multiplied by e^(-i) is e^(2i), and so on. Therefore, a simple solution for the subset B would be {e^i, e^2i, e^3i, e^4i, e^5i, e^6i, ...} and the set S would be {1, e^i, e^2i, e^3i, e^4i, e^5i, ...} note that this would not work if the rotation was by 180 degrees, because the subset B would have to be {e^πi, e^2πi, ...}, but e^(2πi) is just 1 again. In other words, rotating by 180° twice just puts you back where you were. On the other hand, rotating by any whole number of radians will never put you back to your original location, because that would require a multiple of 2π radians, an irrational number. I haven't watched the rest of the video, but I just wanted to try and figure out the problem myself, hopefully I did it right

[@aa01blue38]

I just realized it needs to satisfy the conditions of both A and B. Then I doubt this method works. The solution in the video is genius though.

[@aa01blue38]

I think you can actually think about the set in a somewhat geometrical way. A solution for only set A would be something like {1, 2, 3, 4, 5, ...} and S would be {0, 1, 2, 3, 4, ...} The structure of this is just a dotted line that extends infinitely far in one way. A solution that only works for set B would be what I mentioned above. The structure of this is like a dotted ring that winds up infinitely in one way, which is bizzare but possible in the mathematical world. Multiple concentric dotted rings will also satisfy the same property together as a set. Similarly, multiple parallel dotted lines will work with A. This might be an absolutely absurd idea, but we can then just construct a set that happens to simultaneously look like an infinite number of parallel dotted lines and an infinite number of concentric dotted rings. This is why it was needed to construct the polynomials of e^i with nonnegative integer coefficients. If we just look at the constants, it's just something+1, something+2, something+3, and so on, which basically creates a dotted line out of every something, where something is a point in B. For rotation, you can generate a dotted ring out of every point in A by multiplying them by e^i repeatedly. The most mind-boggling thing is how the points are all unique and stay in the set.

[@fk319fk]

A disk with a hole in it that is split in the middle, can be rotated.

[@NikkiTrudelle]

Please, help. My brain exploded on the floor and I’m trying to get its portions back into figure “S” … oh wait never mind. It’s all coming back together after watching your video, sorta

[@KillianDefaoite]

A very strange and satisfying paradox.

[@Strype13]

Pretty sure my brain just queefed.

[@AnotherPerson-ql6tt]

I was about to bring up Vsauce’s Banach-Tarski video, because that was definitely what inspired my approach to the problem: Let T translate one unit to the right, R rotate one radian counter-clockwise, and 0 represent (0, 0). Where x represents all finite sequences of T and R that either end with T or are empty (to avoid redundancy since R0 = 0), let S = {x0} A = {Tx0} B = {Rx0} It should be clear by definition that A shifted to the left by 1 unit (T’) and B rotated clockwise by 1 radian (R’) each recover S, and by observations analogous to Vsauce’s we find A and B do partition S. It, of course, remains to prove that distinct values of x produce distinct x0, as this video showed using e^i being transcendental.

[@HoSza1]

"This one weird trick will make mathematicians angry."

[@Pseudo___]

my set was all pairs of integers, and my subset was the pairs whos sums were even.

[@newwaveinfantry8362]

Should S really contain the zero degree polynomials though? Aka the integers - multiplying them by p^(-1) gives you something outside of S. I don't think you were supposed to include the integers.

[@edwardmacnab354]

just like the quadrant where you remove the partition A and move all of infinity left , you must choose the turf carefully where this will work . So in fact you've done a bait and switch , because the subject of infinities is still not well understood , and since there are different types of infinities , you must be very careful when working with them !

[@kjnoah]

In order for you to divide your shape into two pieces, you need four or more points, not just 2.

[@danielsegalman1347]

Am I correct that there is no finite set that can equal the translation or rotation of a proper subset of itself?

[@RobbieHatley]

Though I can't prove it, I'm pretty sure the answer to your question is "yes". For overlapping two sets to cause no membership differences, the ends would have to be infinity; finite ends will always have gaps and/or overlaps causing both membership and cardinality differences.

[@urban.p]

"Then this point is 2p" - now I got 2 pee as well.

[@htomerif]

I'm confused as to why this is called a paradox. I mean an easy way to visualize a valid set where S is regenerated by rotating it by one radian is to take the set of all non-negative integers and multiply them by e^i. There's a 1:1 correspondence between the non-negative integers and this new, countably infinite set that is contained in the unit circle. You just remove the first element and then wind everything back by 1 radian. That would be kind of be painful on the eyes but using a smaller transcendental number and winding everything back by that smaller amount would show that the rotational translation is mathematically equivalent to the linear translation. The only special thing here is making sure you don't end up with a set with duplicate points. Once you've proven that, you've proven that the rotation is equivalent to the subtracting 1.

[@j121212100]

A disc centered around the origin. Or a segmented disc in units 1/2 a radian.

[@amadexi]

Both the Banarch-Tarski paradox and the Sierpinski-Mazurkiewicz Paradox come from the fact that they manipulate infinite sets.

[@jorgeluiz1698]

Well, that was insanely cool, thx Zach

[@alan2here]

I'm kind of used to it. When a shape is infinitely complicated like a fractal and unlike a polygon, then start by throwing the rulebook out the window and forgetting all sense or assumptions. Ditto with Hilbert space, or fractional dimensional space.

[@marcelczerwinski2725]

I specifically watched this entire video just to hear you say Sierpiński-Mazurkiewicz.

[@rainerausdemspring894]

Wonderful. I have a master degree in mathematics but didn't know this. And the construction is so simple. No tricky application of the axiom of choice.

[@rainerausdemspring894]

@@imoldandurinmyway This is plain nonsense. Are you a mathematician? Do you have any book about set theory? Have you read it? Have you understood it? en.wikipedia.org/wiki/Axiom_of_choice

[@eielyreithstormwake1905]

I used to love math but this makes me remember why I stopped doing it on paper and self-taught myself to do everything mentally because it's just easier.

[@rogerkearns8094]

And after all that there was still room for new guests at the hotel.

[@ramlydiosa]

I think it has to do with how we understand numbers 1+2 = 3 1 is not 3 2 is not 3 . but move 1 to the right, it becomes 2 move 2 to a certain number line, it becomes whatever number lies in the number line . its like saying I can move 1 and 2 to whatever values I want on the cartesian plane without moving the original 3 then I get 3 because I moved it

[@cptn_n3m012]

First, I thought about (R\Q)[i], thinking that the conservation of irrationnality by a 1 rad rotation would help. That leas me to think about the actual effect of e^(-i) as a rotation, and then I figured out that we needed an e^i to destroy it with, which lead me right to polynomials ! Working out a few quirks (especially to get all that to agree with A), I stumbled upon the exact solution presentes in the video !

[@user_2793]

That's really cool

[@DrownedLamp]

Schrodinger's Cat, Roko's Basilisk and Pascal's Wager, Heisenberg's Uncertainty Principle. What do they all have in common? An alternate word for people who cant pronounce the names.

[@gheffz]

Brilliant! And yes, weird!!! _(Also weird like some infinite series summed in a different order approach a different limit!!!)_

[@martind2520]

So, I saw this cool maths video in my suggestions and clicked it. As I was watching I was like "Wow, the guy narrating this sounds exactly like that guy that does those skits I like, that's weird..." ... ... I worked it out eventually...

[@marloc2019]

When you start seeing a vid with a smug expression on your face expecting some kind of "trick" and slowly the expressione becomes a smile of admiration....

[@phillipotey9736]

It's cool that you can just keep pulling from infinity forever.

[@Abihef]

Nice how you pronounce p exactly the way we pronounce pi over here. Makes it real interesting following along without looking😂

[@db21071979]

The issue is that the first example was on the Cartesian plane, not in the complex numbers plane... If the question was limited to "real" non-complex numbers, then the answer is no.