The Sierpinski-Mazurkiewicz Paradox (is really weird) 

Szczęście mają że Grzegorz Brzęczyszczykiewicz nie był matematykiem 😂

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

Doesn’t look that hard to pronounce

@@michaelblankenau3129 but it is.

@@chessematics maybe for you

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

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.

@@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...

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.

@@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.

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

lmao i feel like you're right

😂

I absolutely would

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

Dude....

Same lol

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

Agreed

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

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

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.

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

@@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.

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

@@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.

Ikr

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

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

@@hurktang It is true for IEEE754

@@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.

@@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.

@@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.

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.

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

@@fullfungo4476 Oh, right, your points p0, p1, p2,... go counter-clockwise around the unit circle

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.

@@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.

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

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

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.

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

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

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.

@@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.

@@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.

@@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.

@@valasfar1557 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.

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. :)

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.

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).

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.

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

Did you expect a geometric problem to go to algebra?

@@theunknown4834 Yes... Why not?

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

@@Boringpenguin The set S is countably infinite though

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

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

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.

@@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!

@@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.

@@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.

@@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!

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.

Great solution

Worth mentioning that this relies on pi being irrational

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

wtf

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.

That's really cool

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

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.

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.

@@imoldandyoureinmyway 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

No, because you have a hole there.