Error: thanks to @dani-rybe for pointing this out first. At 1:05 onwards I said that the circle with one point missing is equivalent to a segment of the real line that includes the point 0 but does not include 2pi so [0, 2pi). This was a mistake, it should not include the point 0. The circle with one point missing is equivalent to (0, 2pi). I apologize for this silly error but the good news is that the rest of the argument still works! (0, 2pi) is still uncountable and can map onto the entire circle.
1:00, shouldn't the circle without one point be equivalent to a line that doesn't include both 2pi and 0? Because 0 and 2pi are the same point on the original circle.
You are right, I made a mistake. Thanks for pointing this out. It should be (0, 2pi). The good news is that you can still cover the entire circle with this. I'll pin a comment to inform others
You say at ~ 1:00 that after removing a point, the circle ⭕️ has now become equivalent to the line segment [0;2pi). However, before removing the point, the points at 0 radians and 2pi radians are the same exact point (because of 2pi-periodicity of circle angles, by definition of a radian). This means that the circle ⭕️ is equivalent to the said line segment _even before_ the point is removed. Which also means that after removing the point, the circle ⭕️ cannot be equivalent to the said line segment anymore.
You had me up until "We've successfully duplicated a section of the sphere". How?? I'm not seeing how you made the leap from being able to rotate points on the sphere to cover the free groups to somehow duplicating them? It feels the video is missing an entire section
Consider all the points the make up the Free Group. Then, simply by rotating only those points in S(A) by A^{-1} and rotating only those points in S(B) by B^{-1} we get 2 copies of ALL the points of the Free Group. So we've duplicated all the original points after doing these 2 rotations.
@@thepro4805 Ok after some searching I think I understand better. We are not really duplicating anything, it is just poor/fancy word choice. The sphere is defined by the points, which we can also define by their free groups. We can define the free groups in such a way that if we accept the axiom of choice, then there are infinitely many reconstructions of the sphere. It is mostly because what constitutes a sphere is not a finite set of points, but an infinite group of points which we can relate to each other by rotation. If you imagine a sphere as a finite set of points, then this does not work. It speaks more about how we can reach all points in the infinite set of points if you have the right free groups. In this case there are 2 free groups that can reconstruct all infinite points for the sphere, so not duplicating in the literal sense but meaning we can construct two balls that have two different definitions even though we started with one ball.
@@Terszelit's kind of like eachbsphere has less.mass but the same.volume righht? Since you take some of the points from the first solo sphere to make the duplicate right? So in a sense each of the 2 new spheres is more porous..
choosing an irrational multiple of pi doesnt ensure the sewuences will be disjoint..why would.it? If pi is being multiplied by the same.irrational number 2 sequences will be the same?
Two polish Geniuses. For those who are interested in the biographies of great mathematicians, I recommend getting acquainted with the life of Stefan Banach, a story worth filming. Thx )))
S(A), S(B), S(A^{-1}), and S(B^{-1}) should be disjoint since any sequence in one begins with a different letter. However, an issue could arise if you had certain sequences that really are just epsilon (the empty sequence). For example, if A is PI/2 then the sequence AAAA would really just be epsilon. And if B is PI/2 then BBBB would be epsilon. This means that AAAABBBB or BBBBAAAA would also be epsilon and S(A) and S(B) would not be disjoint. By using the angle arccos(1/3) we get rid of this potential issue.
Why is the axiom of choice needed? I would get it for countable infinity, but I don't see a justification for the axiom of choice on an uncountably infinite set. You can specify any number in a countably infinite set, but we can only possibly specify an infinitesimal part of an uncountable infinite set. Literally 100% of numbers between 0 and 1 are impossible to specify. I just don't think the axiom of choice holds.
The axiom of choice is used to pick points on the sphere that were not in previous orbits. Rejecting the axiom of choice is certainly an acceptable position. I think the fact that one of its consequences is the Banach-Tarski Paradox is a good example of why one might reject it.
@AbideByReason Also, is not a necessary part of the construction ordering an uncountably infinite number of points? (Since you sequentially pick points until filling the whole sphere). But cantors diaganol argument proves it is not possible to order an uncountably infinite set.
@@samuelowens000 Where did that come from? It certainly is possible to well-order an uncountably infinite set (such that every two elements are comparable and every non-empty subset has a unique minimum). That every set can be well-ordered is a consequence of the axiom of choice (in fact, the two propositions are equivalent). Indeed, Banach-Tarski theorem (and the simpler theorem on the existence of the non-measurable Vitali set) is the consequence of the existence of well-ordering of real numbers. But that's a quite strong proposition; it's consistent in ZF that continuum can't be well-ordered, but Banach-Tarski theorem is true. What if I go from the opposite direction? Sure. There's a model of set theory where continuum can't be well-ordered, Banach-Tarski theorem is not true, and Vitali set doesn't exist. A necessary consequence of this is that a set with the cardinality of the continuum can be split into more than continuum-many disjoint and non-empty subsets. Let's visualize this: Imagine Cantor's Hotel (a competitor of Hilbert's Hotel), where rooms are indexed by real numbers. Of course, the hotel has infinitely many floors, and infinitely many rooms on each floor: on each floor there are numbers which differ from each other by a rational number. (So on one floor are all rational numbers, on another floor are all numbers which differ by a rational from π, and so on.) In absence of axiom of choice, it's consistent that there are more floors than rooms! ("Say what? Of course there are no more floors than rooms!" What does that mean? "It means that floors can be mapped one-to-one with a subset of the rooms." And can you do that? "Sure, just pick a single element from each floor..." Oops. That's precisely what axiom of choice says.) But on the other hand, it can be proven without axiom of choice that there are no less floors than rooms.
Error: thanks to @dani-rybe for pointing this out first. At 1:05 onwards I said that the circle with one point missing is equivalent to a segment of the real line that includes the point 0 but does not include 2pi so [0, 2pi). This was a mistake, it should not include the point 0. The circle with one point missing is equivalent to (0, 2pi). I apologize for this silly error but the good news is that the rest of the argument still works! (0, 2pi) is still uncountable and can map onto the entire circle.
1:00, shouldn't the circle without one point be equivalent to a line that doesn't include both 2pi and 0? Because 0 and 2pi are the same point on the original circle.
After watching that i went straight to the comments to see if anyone else commented about it. Arent the 2 points, 0 and 2pi, the same point?
Glad I could find someone with the same question
You are right, I made a mistake. Thanks for pointing this out. It should be (0, 2pi). The good news is that you can still cover the entire circle with this. I'll pin a comment to inform others
@@AbideByReason I hope that's not too discouraging 🤗
Wow, you did an incredible job!!! Well done! This video deserves millions of views!
You say at ~ 1:00 that after removing a point, the circle ⭕️ has now become equivalent to the line segment [0;2pi). However, before removing the point, the points at 0 radians and 2pi radians are the same exact point (because of 2pi-periodicity of circle angles, by definition of a radian). This means that the circle ⭕️ is equivalent to the said line segment _even before_ the point is removed. Which also means that after removing the point, the circle ⭕️ cannot be equivalent to the said line segment anymore.
You're right. I just pinned a comment so others can see.
Man sometimes math is so crazy.. super clean video, loved it 😄
this deserves more views, its rare to see high level rigorous math in nice tutorials on youtube
👏👏👏
So beautiful.
With luck and more power to you.
hoping for more videos.
You had me up until "We've successfully duplicated a section of the sphere". How?? I'm not seeing how you made the leap from being able to rotate points on the sphere to cover the free groups to somehow duplicating them? It feels the video is missing an entire section
Consider all the points the make up the Free Group. Then, simply by rotating only those points in S(A) by A^{-1} and rotating only those points in S(B) by B^{-1} we get 2 copies of ALL the points of the Free Group. So we've duplicated all the original points after doing these 2 rotations.
yes this! i dont get how is rotation = duplication
@@thepro4805 Ok after some searching I think I understand better. We are not really duplicating anything, it is just poor/fancy word choice. The sphere is defined by the points, which we can also define by their free groups. We can define the free groups in such a way that if we accept the axiom of choice, then there are infinitely many reconstructions of the sphere. It is mostly because what constitutes a sphere is not a finite set of points, but an infinite group of points which we can relate to each other by rotation. If you imagine a sphere as a finite set of points, then this does not work. It speaks more about how we can reach all points in the infinite set of points if you have the right free groups. In this case there are 2 free groups that can reconstruct all infinite points for the sphere, so not duplicating in the literal sense but meaning we can construct two balls that have two different definitions even though we started with one ball.
@@Terszelit's kind of like eachbsphere has less.mass but the same.volume righht? Since you take some of the points from the first solo sphere to make the
duplicate right? So in a sense each of the 2 new spheres is more porous..
choosing an irrational multiple of pi doesnt ensure the sewuences will be disjoint..why would.it? If pi is being multiplied by the same.irrational number 2 sequences will be the same?
Two polish Geniuses. For those who are interested in the biographies of great mathematicians, I recommend getting acquainted with the life of Stefan Banach, a story worth filming. Thx )))
Can this be done in countable? And can this done if there is a discontinuous.
I'm unsure what you are asking. Can you clarify what you mean by these questions?
You could duplicate the universe like this.
Sadly, this is only a mathematical theorem.
infinities gonna infinitize
Trolling is really kind of souless these days.
i dont really understand why exactly the 2 sets you mentioned are disjoint, can you help, thanks!
S(A), S(B), S(A^{-1}), and S(B^{-1}) should be disjoint since any sequence in one begins with a different letter. However, an issue could arise if you had certain sequences that really are just epsilon (the empty sequence). For example, if A is PI/2 then the sequence AAAA would really just be epsilon. And if B is PI/2 then BBBB would be epsilon. This means that AAAABBBB or BBBBAAAA would also be epsilon and S(A) and S(B) would not be disjoint. By using the angle arccos(1/3) we get rid of this potential issue.
Why is the axiom of choice needed?
I would get it for countable infinity, but I don't see a justification for the axiom of choice on an uncountably infinite set.
You can specify any number in a countably infinite set, but we can only possibly specify an infinitesimal part of an uncountable infinite set. Literally 100% of numbers between 0 and 1 are impossible to specify.
I just don't think the axiom of choice holds.
The axiom of choice is used to pick points on the sphere that were not in previous orbits. Rejecting the axiom of choice is certainly an acceptable position. I think the fact that one of its consequences is the Banach-Tarski Paradox is a good example of why one might reject it.
@AbideByReason
Also, is not a necessary part of the construction ordering an uncountably infinite number of points? (Since you sequentially pick points until filling the whole sphere).
But cantors diaganol argument proves it is not possible to order an uncountably infinite set.
@@samuelowens000 Where did that come from? It certainly is possible to well-order an uncountably infinite set (such that every two elements are comparable and every non-empty subset has a unique minimum). That every set can be well-ordered is a consequence of the axiom of choice (in fact, the two propositions are equivalent).
Indeed, Banach-Tarski theorem (and the simpler theorem on the existence of the non-measurable Vitali set) is the consequence of the existence of well-ordering of real numbers. But that's a quite strong proposition; it's consistent in ZF that continuum can't be well-ordered, but Banach-Tarski theorem is true. What if I go from the opposite direction? Sure. There's a model of set theory where continuum can't be well-ordered, Banach-Tarski theorem is not true, and Vitali set doesn't exist. A necessary consequence of this is that a set with the cardinality of the continuum can be split into more than continuum-many disjoint and non-empty subsets.
Let's visualize this: Imagine Cantor's Hotel (a competitor of Hilbert's Hotel), where rooms are indexed by real numbers. Of course, the hotel has infinitely many floors, and infinitely many rooms on each floor: on each floor there are numbers which differ from each other by a rational number. (So on one floor are all rational numbers, on another floor are all numbers which differ by a rational from π, and so on.) In absence of axiom of choice, it's consistent that there are more floors than rooms! ("Say what? Of course there are no more floors than rooms!" What does that mean? "It means that floors can be mapped one-to-one with a subset of the rooms." And can you do that? "Sure, just pick a single element from each floor..." Oops. That's precisely what axiom of choice says.) But on the other hand, it can be proven without axiom of choice that there are no less floors than rooms.