Just a note, we can use the IVT because S^1 is path-connected, but be careful because we are not working on intervals of R where the IVT is usually stated!
I remember the outline of the proof for the same pressure & the same temperature. You do the intermediate value theorem for temperature on all the "equator" of the sphere (between two arbitrary points). Everytime you'll get 2 antipodal point with the same temperature (or more). The crucial point is that if you color the antipodal points who have the same temperature let's say in yellow, it's gonna form a curvy yellow line around the sphere, and that's gonna be continuous. Then you apply the intermediate value theorem to, not an equator but that continuous curvy yellow line we discussed and there you go, you found two antipodal point that have the same temperature because they are yellow, but also have the same pressure because of the ivt.
This was cool, thank you! A question: When you take a point on the sphere, is it really on S^2 or S^3? I am not familiar with notation "S". And in the beginning, is the unit circle in 2D really S^1, or S^2? First I struggled understanding a continuous function from S^2 into (not necessarily onto, right?) R, but it was nice to try to figure it out. So, a map from S^2 to R is a closed track, the projection of which on xy-plane is the nit circle, when drawn in R^3, right? And this theorem says that there is a plane parallel to xy-plane, in which the two intersection points of the track are (x, y, z) and (-x, -y, z) for some x, y, z.
Revising this interesting video finally I've funny proved a beautiful corollary/application in single variable. Let F(x) continuos in [a,b] / F(a)=F(b) then exists at least one c included in [a, (a+b)/2] and its 'antipodal' value c+ (b-a)/2 with F(c) = F(c+ (b-a)/2). That it's one c and its antipodal at half distance of the length of the interval with the same function value. Calculus's great!! Thx DrPeyam !!
Stanislaw Ulam invented the hydrogen bomb with Edward Teller. The fact you illustrate at the end means that you can't map the globe to a flat map without omitting at least one point. I think?
Also, if temperature at points x and -x are the same, then symmetry requires that the two points at the ends of any line parallel to line joining points x and -x must also have same temperature (which could be different from temperature at point x). How? (By symmetry, I'm referring to sun's position.)
No Dr. Peyam, I have serious problem understanding this theorem. Take a'FIXED' point y, and consider g(x) = T(x) - T(y). Follow the same argument to arrive at the conclusion that the value is the same everywhere, not necessarily diametrically opposite. Also, why should it be S - put some arbitrary shape! I mean, there is no provision in the theorem that prevents removing S and inserting another shape.
How do you call a simple closed curve if no straight line intersects with it more than 2 times? I'd imagine that this theorem could be applied to such curves if you define (x, -x) as the intersection points between the curve and lines going through an arbitrary fixed point inside the curve. A mapping between the unit circle and our curve would serve to prove that. Just curious, I'm not a mathematician. What motivated my thought was the fact that Earth is not a perfect sphere.
Cool video. I guess the one dimensional version of this theorem is either an exercise or a theorem/corollary in some calculus/analysis books. Never knew about its multidimensional version. Secondly, I know this is not the right video to ask about this, but could you make a video on why the cross product of two vectors exists only in R^3 (and R^7 possibly) and nowhere else? I know there is something called exterior product of which I have absolutely no idea about (but will get to it in a month or so). But intuitively why is the cross product of R^3 special and exists nowhere else? Would really really appreciate it. :)
Proof of ham sandwich. Fix a point c and draw a line through it with a variable direction x (some vector on the unit circle). Under an arbitrary orientation, the line divides the plane into two halves. Let A(x) denote the area of our compact sets that lies in the “first” half. By Borsuk-Ulam there exists x such that A(x)=A(-x), but this means that there is a line with some direction x such that the area in one half (A(x)) equals the area in the other half (A(-x)). This concludes the proof
Cool! Does this result hold only for perfect spheres, or can "bumpier" shapes also be included? I ask only bc the spherical earth is ofc an approximation. I guess I mean more in the way that earth is squished at the poles a bit!
Yes, bumpier shapes can be included. Though I don't know the general proof, because I also don't know algebraic topology all that well. Regardless, since the proof is topological, then it would be the same for any topologically equivalent objects. Definitely for the proof given for the circle case, since notice that no mention was actually given to the circle being smooth.
In my case, for being secure and comfortable, for the unitary circle I've worked similary with polar coordinates: X: (1,alpha) X' (1,alpha + pi) and in cartesian coordinates: X: (x,y) X' (-x,-y) And prove it with IVT for single and multi-variable in each case
Wait... an sphere Sn is not homeomorphic to Rn But if I take away one point of the sphere (let's call it ... "a") maybe now it is homeomorphic to Rn (correct me if I'm wrong) BUT, if we want to put this point a back to the sphere, we must choose an point of Rn to be f(a), but it's already full (since Sn/a is homeomorphic to Rn)! Is this some kind of proof? idk, seriously...
No, it isn't a proof since the cardinality is not finite. To put an example, as you may know, N is homeomorphic to N^2. However, NUA, where A is any set of finite numbers, is also homeomorphic to N^2. More explicit, you may consider f:A-->B homeomorphic and A and B countable sets just to illustrate this idea; such that f(a_n)=b_n. Therefore, if you consider now AU{c}, there exists an homeomorphism g:AU{c}-->B such that g(a_n)=f(a_n+1)=b_n+1 and g(c)=f(a_0)=b_0 for all naturals. If this works for a countable set, even more for a non-countable set such as the sphere Sn for n>0. Just take a finite subset of Sn and repeat the same process, and you may end in a contradiction. In resume, the sphere Sn is not homeomorphic to Rn, but you can't conclude by this way.
Unfortunately, this proof doesn't really work; you are indeed correct that, for instance, the sphere S^2 without one of its points is homeomorphic to R^2 (to see this, just use stereographic projection to obtain a continuous and bijective association between R^2 and the sphere without the point in question, which you might as well assume to be the north pole), and your argument shows that this homeomorphism cannot be used to map R^2 to the entire sphere. However, it doesn't follow that such a homeomorphism cannot exist. If you've studied set theory before, what you've done is a bit like "proving" that the integers are of strictly greater cardinality than the natural numbers by showing that there are integers left over (namely, anything strictly less than 0) if you map the naturals to themselves via the identity (that is, 0 -> 0, 1 -> 1, etc.). This is a valid argument, but it doesn't prove that no bijection between the two sets exists at all, and in fact, there is one. Once you've studied some topology, however, it's pretty easy to show that the sphere S^n cannot be homeomorphic to R^n via the property of compactness. Recall that a set is called compact if every open cover of it has a finite subcover, and in R^n, this is equivalent to the condition of being closed and bounded (this is what is known as the Heine-Borel theorem). Now, it's pretty easy to show that the continuous image of any compact set is itself compact (in other words, compactness is a topological property), and since S^n is a compact subset of R^{n + 1}, any continuous image of it must also be a compact subset of the topological space of R^n. However, R^n is clearly not compact because it's not bounded, and so it cannot be homeomorphic to the sphere.
Actually it's not turkish name. Borsuk was Polish and Ulam was Polish-American btw participated in Manhattan project.
I'm happy to Know the truth .
Just a note, we can use the IVT because S^1 is path-connected, but be careful because we are not working on intervals of R where the IVT is usually stated!
Because of you I dream to be a mathematician🥰🥰, your student from Morocco 🇲🇦🇲🇦
Il est marocain?
Good luck!
و انا ايضا من المغرب
و احلم ان اصبح عالم رياضي وفيزيائي
I remember the outline of the proof for the same pressure & the same temperature. You do the intermediate value theorem for temperature on all the "equator" of the sphere (between two arbitrary points). Everytime you'll get 2 antipodal point with the same temperature (or more). The crucial point is that if you color the antipodal points who have the same temperature let's say in yellow, it's gonna form a curvy yellow line around the sphere, and that's gonna be continuous. Then you apply the intermediate value theorem to, not an equator but that continuous curvy yellow line we discussed and there you go, you found two antipodal point that have the same temperature because they are yellow, but also have the same pressure because of the ivt.
I'm mathematically delighted to see your videos .
I had one audacious individual pronounces the name as "Boar-suck and U-Lame"
Such colorful appellation
This was cool, thank you! A question: When you take a point on the sphere, is it really on S^2 or S^3? I am not familiar with notation "S". And in the beginning, is the unit circle in 2D really S^1, or S^2?
First I struggled understanding a continuous function from S^2 into (not necessarily onto, right?) R, but it was nice to try to figure it out. So, a map from S^2 to R is a closed track, the projection of which on xy-plane is the nit circle, when drawn in R^3, right? And this theorem says that there is a plane parallel to xy-plane, in which the two intersection points of the track are (x, y, z) and (-x, -y, z) for some x, y, z.
wikipedia has a good definition of S^n: en.wikipedia.org/wiki/N-sphere
@@FT029 Thank you!
That is actually cool !!
Agreed!
This video is from 2020 i guess another slow upload xD
Ok. Thanks.
But pressure at any other point x + ∆ is also the same as that at point x because everywhere it is 1 atm pressure. So, redundant!
Revising this interesting video finally I've funny proved a beautiful corollary/application in single variable. Let F(x) continuos in [a,b] / F(a)=F(b) then exists at least one c included in [a, (a+b)/2] and its 'antipodal' value c+ (b-a)/2 with F(c) = F(c+ (b-a)/2).
That it's one c and its antipodal at half distance of the length of the interval with the same function value. Calculus's great!! Thx DrPeyam !!
And you can go even stronger: for any distance d
@@breberky Great!! really I knew that for all d=(b-a)/2^ n / n=1,2,3 ...it's proved just using the theorem recursively..but for all d
Stanislaw Ulam invented the hydrogen bomb with Edward Teller.
The fact you illustrate at the end means that you can't map the globe to a flat map without omitting at least one point. I think?
The function from the globe to the flat map must either
* not be continuous
* not be 1 to 1, so 2 points on the globe map to the same point
Also, if temperature at points x and -x are the same, then symmetry requires that the two points at the ends of any line parallel to line joining points x and -x must also have same temperature (which could be different from temperature at point x). How? (By symmetry, I'm referring to sun's position.)
You’re overthinking this way too much lol
No Dr. Peyam, I have serious problem understanding this theorem. Take a'FIXED' point y, and consider g(x) = T(x) - T(y). Follow the same argument to arrive at the conclusion that the value is the same everywhere, not necessarily diametrically opposite. Also, why should it be S - put some arbitrary shape! I mean, there is no provision in the theorem that prevents removing S and inserting another shape.
K
How do you call a simple closed curve if no straight line intersects with it more than 2 times? I'd imagine that this theorem could be applied to such curves if you define (x, -x) as the intersection points between the curve and lines going through an arbitrary fixed point inside the curve. A mapping between the unit circle and our curve would serve to prove that. Just curious, I'm not a mathematician. What motivated my thought was the fact that Earth is not a perfect sphere.
S^n \congruent SO(n+1)/SO(n)
Do you think anything can be said about this regarding this theorem?
Cool video. I guess the one dimensional version of this theorem is either an exercise or a theorem/corollary in some calculus/analysis books. Never knew about its multidimensional version. Secondly, I know this is not the right video to ask about this, but could you make a video on why the cross product of two vectors exists only in R^3 (and R^7 possibly) and nowhere else? I know there is something called exterior product of which I have absolutely no idea about (but will get to it in a month or so). But intuitively why is the cross product of R^3 special and exists nowhere else? Would really really appreciate it. :)
Proof of ham sandwich. Fix a point c and draw a line through it with a variable direction x (some vector on the unit circle). Under an arbitrary orientation, the line divides the plane into two halves. Let A(x) denote the area of our compact sets that lies in the “first” half. By Borsuk-Ulam there exists x such that A(x)=A(-x), but this means that there is a line with some direction x such that the area in one half (A(x)) equals the area in the other half (A(-x)). This concludes the proof
Cool! Does this result hold only for perfect spheres, or can "bumpier" shapes also be included? I ask only bc the spherical earth is ofc an approximation. I guess I mean more in the way that earth is squished at the poles a bit!
Yes, bumpier shapes can be included. Though I don't know the general proof, because I also don't know algebraic topology all that well. Regardless, since the proof is topological, then it would be the same for any topologically equivalent objects. Definitely for the proof given for the circle case, since notice that no mention was actually given to the circle being smooth.
Amazing video. Something one thinks about.
Wait, is x a vector in this context?
Yes I think x is a vector of real numbers where norm is 1. For the S1 unit circle I think you could also view x as a complex number with norm 1.
In my case, for being secure and comfortable, for the unitary circle I've worked similary with polar coordinates:
X: (1,alpha) X' (1,alpha + pi)
and in cartesian coordinates:
X: (x,y) X' (-x,-y)
And prove it with IVT for single and multi-variable in each case
Wait...
an sphere Sn is not homeomorphic to Rn
But if I take away one point of the sphere (let's call it ... "a")
maybe now it is homeomorphic to Rn (correct me if I'm wrong)
BUT, if we want to put this point a back to the sphere, we must choose an point of Rn to be f(a),
but it's already full (since Sn/a is homeomorphic to Rn)!
Is this some kind of proof? idk, seriously...
No, it isn't a proof since the cardinality is not finite. To put an example, as you may know, N is homeomorphic to N^2. However, NUA, where A is any set of finite numbers, is also homeomorphic to N^2.
More explicit, you may consider f:A-->B homeomorphic and A and B countable sets just to illustrate this idea; such that f(a_n)=b_n. Therefore, if you consider now AU{c}, there exists an homeomorphism g:AU{c}-->B such that g(a_n)=f(a_n+1)=b_n+1 and g(c)=f(a_0)=b_0 for all naturals.
If this works for a countable set, even more for a non-countable set such as the sphere Sn for n>0. Just take a finite subset of Sn and repeat the same process, and you may end in a contradiction.
In resume, the sphere Sn is not homeomorphic to Rn, but you can't conclude by this way.
Unfortunately, this proof doesn't really work; you are indeed correct that, for instance, the sphere S^2 without one of its points is homeomorphic to R^2 (to see this, just use stereographic projection to obtain a continuous and bijective association between R^2 and the sphere without the point in question, which you might as well assume to be the north pole), and your argument shows that this homeomorphism cannot be used to map R^2 to the entire sphere. However, it doesn't follow that such a homeomorphism cannot exist. If you've studied set theory before, what you've done is a bit like "proving" that the integers are of strictly greater cardinality than the natural numbers by showing that there are integers left over (namely, anything strictly less than 0) if you map the naturals to themselves via the identity (that is, 0 -> 0, 1 -> 1, etc.). This is a valid argument, but it doesn't prove that no bijection between the two sets exists at all, and in fact, there is one.
Once you've studied some topology, however, it's pretty easy to show that the sphere S^n cannot be homeomorphic to R^n via the property of compactness. Recall that a set is called compact if every open cover of it has a finite subcover, and in R^n, this is equivalent to the condition of being closed and bounded (this is what is known as the Heine-Borel theorem). Now, it's pretty easy to show that the continuous image of any compact set is itself compact (in other words, compactness is a topological property), and since S^n is a compact subset of R^{n + 1}, any continuous image of it must also be a compact subset of the topological space of R^n. However, R^n is clearly not compact because it's not bounded, and so it cannot be homeomorphic to the sphere.
Vsauce music plays....
Haha There should be some support group for people who had to take algebraic topology.
Hahaha yes
Numberphile has a video on the ham sandwich theorem:
th-cam.com/video/YCXmUi56rao/w-d-xo.html