For the defintion, I struggle with continuity being enough. I think it would be most precise to adopt f: x--->y such that the function is bijective implies a homeomorphism iff there is a continuous inverse. Excellent video, good job fleshing it out.
So far - thank you very much for your lectures. I have been enjoying them. I have a question tho, I hope you are able to answer it. In order to prepare for my analysis 2 class in germany I took measure theory course and they said that in case of the borel-sigma-algebra that for the infinite-norm of a unit-ball we can think of it as a unit-square. Are these two topics related? I have difficulties to understand why that is. I mean - we are talking about measurable open sets. And in metric spaces as far as I understand there is no such thing as a homeomorphism between a unit circle and a square, or is there one?`
The unit ball B and unit square S are in fact homeomorphic to one another by the continuous map h: S -> B that sends s |-> s / ||s||. This is just the last example in the video. They are, however, not isometric to one another because the map h does not preserve distances. When one talks about spaces with additional structure, one always needs to be specific about what structures one cares about preserving for an isomorphism. In general, an isomorphism of measurable spaces is weaker than a homeomorphism (isomorphism of topological spaces) which is weaker than an isometry (isomorphism of metric spaces). In the case of the Borel sigma-algebra, measurability means that the preimage of any Borel set is Borel. This is weaker than continuity, since a preimage of an open set like (0,1) could be some closed set in the Borel sigma-algebra like [0,1]. The point you mentioned with the euclidean and infinity-norms is also interesting. R^n with euclidean norm is different as a metric space from R^n with the infinity norm. However both norms generate the same topology (since unit balls and unit squares are homeomorphic). You can also see this more showing that the sets in which you can find open balls around any point are the same ones you can find open squares around any point. In summary, while the two versions of R^n are different as metric spaces, they are the same as topological spaces. In a similar fashion, different topological spaces could generate the same Borel sigma-algebra.
I think you are getting the sphere confused with the balls from the previous example. In this case both sets are closed. The example is talking about the surfaces of the shapes not their interiors.
Excellent!! just wanted to clearify one thing..In the second example, can u please explain the function which you have defined? the x used at three places .. will carry same values??
Yes, the notation x |-> x / (1 - |x|) means F(x) = x / (1 - |x|), that is any x is mapped to x / (1 - |x|). The symbol |-> is pronounced "maps to". Now in the two dimensional case, x is a vector x = (x_1,x_2), so F(x) = x / (1 - |x|) = x / (1 - \sqrt( x_1^2 +x_2^2 ) ) just gives a scaling of x by the factor 1 / (1 - \sqrt( x_1^2 +x_2^2 ) ).
The examples are just showing bijections between sets, and presuming bijection to be enough to assert homeomorphism or continuity. But the the definitions of continuity and homemorphism presume topological spaces, not just sets. A homeomorphism is a bijection between neighborhoods of elements.
In the examples I was assuming that we already know how to identify continuous functions between euclidean spaces with the standard topology. I believe I mentioned this, but I might not have been clear enough. This is usually part of an analysis or calculus course. For instance, we would use that the identity and constant functions are continuous, and that sums, products and quotients of continuous functions are again continuous. You are right that the things would have to be checked in the examples to show that they are homeomorphisms. I skipped it here because it would take a fair amount of work to establish these things and many people taking topology will have already seen it.
Yes! Homoemorphic spaces share all topological properties. This includes the set of closed subsets and the set of subsets that are both open and closed.
I have a doubt regarding to the following: Can we take as the argument of F all B^2 or open subsets of B^2? I ask you because you used points to show bijection. I think I'm confused on that.
@@ehrenfest9458 To show that F is a homeomorphism, one has to show that it is continuous and has a continuous inverse. In the video I argue that F is bijective by showing injectivity and surjectivity. In fact, G is the inverse to F, as one can check by showing that F(G(x))=x and G(F(x))=x by plugging in the definitions. As for continuity, both F and G are continuous using the following standard results from real analysis which I have assumed here: The norm |x| is continuous, sums of continuous functions are continuous, and quotients of continuous functions are continuous (as long as the denominator is never zero). These results are usually proved using the epsilon-delta definition of continuity, which is equivalent to requiring the preimages of open sets to be open. If you haven't seen proofs of these facts, it might be a good exercise to think about why the are true. In summary, I did not give a complete proof because I am assuming these results on continuity and could have stated this clearer in the video.
No (try to find a bijection from a point to the unit ball). Another way to think about this intuitively is that homeomorphisms preserve dimension. The point has dimension 0, while the unit ball is 3-dimensional. However, your intuition is not far off since one can continuously retract (shrink) the unit ball to a point. There is a weaker notion of equivalence of topological spaces called homotopy equivalence that focuses on these types of relationships. It is the case that the unit ball is homotopy equivalent to a point.
@@engelsteinberg593 You are right that you can always map any set to a point using a continuous function. The definition of homotopy equivalence requires more than this though. For example, a hollow sphere is not homotopy equivalent to a point. Neither is the disjoint union of two balls. Properly defining homotopy equivalence takes some space, so I won't do it here. It requires first defining homotopy between continuous functions, which intuitively is a way to continuously transform one continuous function into another. Then one uses this to define homotopy equivalence. You can find the details e.g. on wikipedia en.wikipedia.org/wiki/Homotopy, but this may be a bit overwhelming for a first read. I will eventually cover this in a video, but the topic a still a while away.
This requires some knowledge of real analysis which is why I didn't go through it in the video. You would first observe that taking the absolute value of a number is continuous. This can be proved directly from the epsilon/delta definition of continuity for real functions. Next, taking the maximum of two numbers is also continuous: For example, we can express max(x,y) = 1/2*(x+y+|x-y|) and use the fact that sums and products of continuous functions are again continuous (along with the fact that compositions of continuous functions are continuous for |x-y|). Alternatively, you could again prove this directly from the definition. Next observe that max(|x|,|y|,|z|) = max(|x|,max(|y|,|z|)), so by the fact that compositions of continuous functions are continuous, the former is continuous. Finally, F = (x,y,z)/max(|x|,|y|,|z|) = (x / max(|x|,|y|,|z|)), y / max(|x|,|y|,|z|), z / max(|x|,|y|,|z|)) is continuous because every component consists of a continuous function (dividing a continuous function by a non-zero continuous function is again continuous). So you see that the proof of this is a bit involved and uses almost all of the standard results for the preservation of continuity from real analysis. I hope this helps.
@@mariusfurter So max function is continuous in everywhere, but no differentiable everywhere, so I am confusing continuity with differentiability. And thanks.
@@engelsteinberg593 for you never get confused look at function Stietjes built: a function that is continuous in every point and non differentiable in no point. Also you need to revise your analisys course.
Finally an exciting stuff
awesome series thanks
For the defintion, I struggle with continuity being enough. I think it would be most precise to adopt f: x--->y such that the function is bijective implies a homeomorphism iff there is a continuous inverse. Excellent video, good job fleshing it out.
16:20, if we defined U=f^{-1}(V) by (*) U is open, then f^{-1}(V) is open; why writing f^{-1}(V)=f^{-1}(f(f^{-1}(V)))=...?
Agreed. Even I felt that the proof was more convoluted than it should've been
So far - thank you very much for your lectures. I have been enjoying them.
I have a question tho, I hope you are able to answer it. In order to prepare for my analysis 2 class in germany I took measure theory course and they said that in case of the borel-sigma-algebra that for the infinite-norm of a unit-ball we can think of it as a unit-square. Are these two topics related? I have difficulties to understand why that is. I mean - we are talking about measurable open sets. And in metric spaces as far as I understand there is no such thing as a homeomorphism between a unit circle and a square, or is there one?`
The unit ball B and unit square S are in fact homeomorphic to one another by the continuous map h: S -> B that sends s |-> s / ||s||. This is just the last example in the video. They are, however, not isometric to one another because the map h does not preserve distances.
When one talks about spaces with additional structure, one always needs to be specific about what structures one cares about preserving for an isomorphism. In general, an isomorphism of measurable spaces is weaker than a homeomorphism (isomorphism of topological spaces) which is weaker than an isometry (isomorphism of metric spaces). In the case of the Borel sigma-algebra, measurability means that the preimage of any Borel set is Borel. This is weaker than continuity, since a preimage of an open set like (0,1) could be some closed set in the Borel sigma-algebra like [0,1].
The point you mentioned with the euclidean and infinity-norms is also interesting. R^n with euclidean norm is different as a metric space from R^n with the infinity norm. However both norms generate the same topology (since unit balls and unit squares are homeomorphic). You can also see this more showing that the sets in which you can find open balls around any point are the same ones you can find open squares around any point. In summary, while the two versions of R^n are different as metric spaces, they are the same as topological spaces. In a similar fashion, different topological spaces could generate the same Borel sigma-algebra.
At 32:40 you should not assume that the points are not colinear.
I think G in the last example is not continuous because the sphere is open and the cube is not. You should change the max to be < 1
I think you are getting the sphere confused with the balls from the previous example. In this case both sets are closed. The example is talking about the surfaces of the shapes not their interiors.
Excellent!! just wanted to clearify one thing..In the second example, can u please explain the function which you have defined? the x used at three places .. will carry same values??
Yes, the notation x |-> x / (1 - |x|) means F(x) = x / (1 - |x|), that is any x is mapped to x / (1 - |x|). The symbol |-> is pronounced "maps to". Now in the two dimensional case, x is a vector x = (x_1,x_2), so F(x) = x / (1 - |x|) = x / (1 - \sqrt( x_1^2 +x_2^2 ) ) just gives a scaling of x by the factor 1 / (1 - \sqrt( x_1^2 +x_2^2 ) ).
@@mariusfurter Thank u so much.. your videos are just AMAZING.. it makes the concepts much much clear!!
The examples are just showing bijections between sets, and presuming bijection to be enough to assert homeomorphism or continuity. But the the definitions of continuity and homemorphism presume topological spaces, not just sets. A homeomorphism is a bijection between neighborhoods of elements.
In the examples I was assuming that we already know how to identify continuous functions between euclidean spaces with the standard topology. I believe I mentioned this, but I might not have been clear enough. This is usually part of an analysis or calculus course. For instance, we would use that the identity and constant functions are continuous, and that sums, products and quotients of continuous functions are again continuous. You are right that the things would have to be checked in the examples to show that they are homeomorphisms. I skipped it here because it would take a fair amount of work to establish these things and many people taking topology will have already seen it.
Does this mean there is also a one to one correspondence amongst closed sets and sets that are both closed and open?
Yes! Homoemorphic spaces share all topological properties. This includes the set of closed subsets and the set of subsets that are both open and closed.
I have a doubt regarding to the following: Can we take as the argument of F all B^2 or open subsets of B^2? I ask you because you used points to show bijection. I think I'm confused on that.
What part of the video are you referring to?
@@mariusfurter When you show that the unit ball is homeomorphic to R^n. My doubt is that you prove it with points and not with open sets.
@@ehrenfest9458 To show that F is a homeomorphism, one has to show that it is continuous and has a continuous inverse. In the video I argue that F is bijective by showing injectivity and surjectivity. In fact, G is the inverse to F, as one can check by showing that F(G(x))=x and G(F(x))=x by plugging in the definitions. As for continuity, both F and G are continuous using the following standard results from real analysis which I have assumed here: The norm |x| is continuous, sums of continuous functions are continuous, and quotients of continuous functions are continuous (as long as the denominator is never zero). These results are usually proved using the epsilon-delta definition of continuity, which is equivalent to requiring the preimages of open sets to be open. If you haven't seen proofs of these facts, it might be a good exercise to think about why the are true. In summary, I did not give a complete proof because I am assuming these results on continuity and could have stated this clearer in the video.
how will we show the bijection between the 2
To which 2 are you referring?
12:49 does this implcitly use canter-bernardstein theorem?
nvm im being dumb sorry
Is the Unit ball Homeomorphic to the point?
No (try to find a bijection from a point to the unit ball). Another way to think about this intuitively is that homeomorphisms preserve dimension. The point has dimension 0, while the unit ball is 3-dimensional.
However, your intuition is not far off since one can continuously retract (shrink) the unit ball to a point. There is a weaker notion of equivalence of topological spaces called homotopy equivalence that focuses on these types of relationships. It is the case that the unit ball is homotopy equivalent to a point.
@@mariusfurter Is no the function f(x) = 0 a continous transformation of any shape into a point, so Evey shpe is homotopy equivalent to the point?
@@engelsteinberg593 You are right that you can always map any set to a point using a continuous function. The definition of homotopy equivalence requires more than this though. For example, a hollow sphere is not homotopy equivalent to a point. Neither is the disjoint union of two balls.
Properly defining homotopy equivalence takes some space, so I won't do it here. It requires first defining homotopy between continuous functions, which intuitively is a way to continuously transform one continuous function into another. Then one uses this to define homotopy equivalence. You can find the details e.g. on wikipedia en.wikipedia.org/wiki/Homotopy, but this may be a bit overwhelming for a first read. I will eventually cover this in a video, but the topic a still a while away.
@@mariusfurter Thanks, and I don't think it to be overwhelming, but I think that I may have problems remembering or understanding the implications.
40:25 How dammed in the world the F will be continuous?
This requires some knowledge of real analysis which is why I didn't go through it in the video. You would first observe that taking the absolute value of a number is continuous. This can be proved directly from the epsilon/delta definition of continuity for real functions. Next, taking the maximum of two numbers is also continuous: For example, we can express max(x,y) = 1/2*(x+y+|x-y|) and use the fact that sums and products of continuous functions are again continuous (along with the fact that compositions of continuous functions are continuous for |x-y|). Alternatively, you could again prove this directly from the definition. Next observe that max(|x|,|y|,|z|) = max(|x|,max(|y|,|z|)), so by the fact that compositions of continuous functions are continuous, the former is continuous. Finally, F = (x,y,z)/max(|x|,|y|,|z|) = (x / max(|x|,|y|,|z|)), y / max(|x|,|y|,|z|), z / max(|x|,|y|,|z|)) is continuous because every component consists of a continuous function (dividing a continuous function by a non-zero continuous function is again continuous).
So you see that the proof of this is a bit involved and uses almost all of the standard results for the preservation of continuity from real analysis. I hope this helps.
@@mariusfurter So max function is continuous in everywhere, but no differentiable everywhere, so I am confusing continuity with differentiability. And thanks.
@@engelsteinberg593 for you never get confused look at function Stietjes built: a function that is continuous in every point and non differentiable in no point.
Also you need to revise your analisys course.