The Sierpinski-Mazurkiewicz Paradox (is really weird)

this video is sponsored by clickup okay for those who feel like participating think of a subset of the xy plane if you don't want to do any work then just follow what i pick but this subset can be whatever you want it can be a disc or a square or something more chaotic it can be something with infinite area like the first quadrant or everything outside the unit circle or it could be just some random points doesn't matter just pick a subset of the x y plane going to go with a disc for simplicity not centered at the origin though but whatever you picked call that set s now partition that set s into two smaller non-empty sets however you would like and call one of those sets a and the other b it doesn't matter how you do this partition like set a could be a single point and b is everything else it doesn't matter you just need to break your set s into two disjoint subsets a and b such that neither a nor b is empty and the union of a and b is s just like we can see here okay now in a sec i'm going to ask if two sets are equal and when i say that i mean are they exactly the same set of points in the exact same location like if i called these two circles u and v and said does the set u equal v the answer is no they're in different locations sure they're the same size but they're not the same sets they are a different set of points in the plane if i ask does set a equal b right here the answer better be no for everyone no matter what you picked because they have to be disjoint they can't be the same set if i ask does a equal s the answer is also no because a is a proper subset of s okay now move your set a to the left by one unit does your new set a after being moved i'm just going to keep calling it a does that now equal s and when i say s i mean the original set s which is the disk in our case that never changes well mine definitely does not i mean obviously s is the entire disk we see here a is not a disk they don't even overlap obviously not the same set i mean no translation could turn a into s in this case but just stay with me here now take your set b and rotate it by one radian clockwise well try your best now does your new set b after being rotated equal s i'm willing to bet that not a single person is saying yes right now unless you've seen this paradox before i mean mine definitely doesn't so i don't know what you guys all picked for s and how you partitioned it but i bet no one was able to say yes to each of the last two questions does a equal s after translation and does b equal s after that rotation but there does exist a set s where this is possible where you can say yes to both those last two questions there's a set that you can partition into subsets a and b where if you shift a to the left by one it becomes s you get the entire original set back and if you rotate b by one radian clockwise you also get the original set back that's insanely weird the shifting part is actually pretty simple though like look if i take the entire right side of the xy plane to be my set s and i partition it as such where subset a is everything to the right of the line x equals one then when i move set a to the left by one unit it becomes the entire original set s everything to the right of the y axis that we started with but taking the other set b and rotating it one radian clockwise definitely does not get us s back so that failed but again there is a set s where this is possible i might be missing something but i can't think of any set s visually where rotating a proper subset of it by one radian can get us back so the one hint i'll give if you want to make an attempt on your own is do not think about this visually this solution this subset of the xy plane is not something i can actually show so ponder that for a bit while i quickly thank the sponsor of this video click up click up is a productivity and work management platform designed to save you time on just about anything whether it be an upcoming school assignment to a large team project at work with click up you can do things like set goals and keep track of progress you can create dashboards that give you a visual breakdown of everything that needs to be completed for an upcoming project you can create calendars share documents tasks ideas and plenty more all in one place whether you're a team of one or a team of a thousand and what's great about this platform is how customizable everything is while still being user friendly so you can organize your workflow and even your teams however you'd like no coding or complicated setup required so whether you're in sales project management engineering marketing hr or you're just in school click up is a great easy to use solution that creates a more efficient work environment and to get started now use the code zackstar to get 15 off clickups massive unlimited plan for a year meaning you can start reclaiming your time for under five dollars per month okay now let's get to the solution instead of thinking of this plane as the x y plane we're going to change it to the complex plane real and imaginary axis then we're going to put a point at e to the i and call that point p e to the i can also be written as cosine of 1 plus i sine of 1 which is this here just a complex number so if i call that point p then this point here would be 2p just twice as far away from the origin on the same line at the same angle this would be p squared here's p squared plus two and we could keep going but just notice that these are all points any polynomial using p these are all just points now we're gonna let s our subset of the plane be the set of all polynomials in p with non-negative integer coefficients all polynomials in p just means all polynomials where you have p instead of a variable x so s would include p and two p plus five and three p squared plus six p plus one other crazy looking polynomials and even non-negative constants like four zero and so on just a bunch of polynomials using p instead of x which means again all of these elements of s are just points because p is still just e to the i there's an infinite number of these points scattered throughout the plane a few of which we saw earlier but this is why this is not really a visual problem because the subset of the plane we're choosing is an infinite set of very scattered points not a nice visual shape or anything like that now here's the question are these points all going to be unique because that's important for when we split s into a and b like is it possible that two or more points which are in s and seem different are actually the same point well e to the i is transcendental meaning it is not the root of any polynomial with finite degree that has integer coefficients that's actually all we need to know that all the points in s are in fact unique without getting too rigorous the idea is if let's say these points were in fact the same and this equality is true then we can move everything to one side and this must be true too if this is true then p which remember is just the number e to the i is the solution to this polynomial same thing as above but with x's but we just said e to the i is transcendental it cannot be the solution to this polynomial or any polynomial with integer coefficients so we get a contradiction and thus every element in s is unique that's the idea so now here's how we're going to partition s into subsets a and b a will be the set of all polynomials in p that have a non-zero constant term anything that ends with like plus 1 or plus 7 or whatever it's just got to have that constant term and this also includes just constants like 4 and 28 but not 0 itself because again it has to be a polynomial with a non-zero constant term the set b will be everything else those without a constant term so this is a valid partition since we know that a and b are disjoint everything in s is unique and of course the union of a and b is s so now we're gonna take set a which is again just a bunch of points and shift it to the left by one like before which means just subtracting one from every element and when we do this we get s the entire original set back that may take a second to see but just think how any element in s whether it has that constant term or not can come from something in a something with that constant term subtracted by 1. see since all the constant terms in a are positive integers then by subtracting 1 you'll still be left with something valid that is an s like look at this here this polynomial is in a because it has a constant term subtract 1 and we still have a constant term so the result is still in a subtracting didn't produce anything new but all the polynomials in a that specifically have a constant term of plus one once you do the subtraction you get a polynomial without a constant term something new that's in s but not in a so that's how everything in s can come from something in a minus one so we're halfway there we've shown that if you take all the points in a and move them left by one you get all the points in s now moving on to set b we need to rotate all of these points by one radian and hopefully get s back to rotate a set clockwise by one radian in the complex plane is the same as multiplying all of its elements by e to the minus i remember though that p is e to the i so e to the minus i is just p to the minus 1. thus to rotate b we are going to multiply all the elements in b by p to the minus one and this does give us back s again to see it just think of any elements in s any of those polynomials i'll use the same four as before and they can all be derived by something now specifically in b something without a constant term times p to the minus 1. in this case since b has no elements with a constant term then multiplying by p to the minus 1 just cancels things out and leaves us with still a valid polynomial in s but all those polynomials in b with a p to the first term once multiplied by p to the minus 1 give you a polynomial with a constant something outside of b but still an s so that's how you get everything else that's not in b from just a simple multiplication so once you multiply b by p to the minus 1 rotate all those points you get everything back in s so that's really it may take some time to process but you now know there exists a set where you can partition it and apply rigid motions to each partition and get the same set back twice and i want to mention for those who have seen the banach tarsky paradox which vsauce has a cool video on you'll notice some similarities to what we just saw so maybe this isn't as surprising however one big difference here is that in our case the axiom of choice was not involved but if you found this interesting definitely go look into that paradox as well but that's it for this video thanks as always to my supporters on patreon social media links to follow me are down below and i'll see you all in the next video

Channel: Zach Star

Views: 436,586

Keywords: zach star, zachstar, paradox, math paradox, banach tarski paradox, banach-tarski paradox, Sierpinski-Mazurkiewicz Paradox, sierpinski, serpinski

Id: hVaDTnrwCnk

Length: 13min 2sec (782 seconds)

Published: Thu Jul 28 2022