At 11:00, it certainly is possible to remove a point from that interval without disconnecting it. Just take an endpoint! I think the more usual topological invariants used here include "can remove two points without disconnecting it" or "the number of points you can remove that do not disconnect it" or "the number of points you can remove that do disconnect it".
@5:16 Remembers pate a modeler but not play-doh that was excellent! Ive heard rubber sheet geometry as well, would say "modelling clay" if wanting to avoid the brand name haha
Dr. Peyam, you are one of the greats on math youtube. I am studying topology right now and some concept can be hard to grasp. Thank you for making videos like this, it really helps! Also you seem like such a fun guy to be around, the energy you give off is amazing. Keep up the good work!
You wouldn't believe it but I had to learn and apply this notion in literature for a project haha... So thanks for making it easy enough for me to understand!
I really like all the motivating examples you give (e.g. the continuous bijection whose inverse isn't continuous)! I am a little curious about the proof of the property at 9:14.
I wasn't interested in maths but watching 3b1r bprp and some other CZcams channel including yours has completely changed my view.... Now I want to do MSc in mathematics... It's an interesting subject
if you were to curve out the real line into a circle does that mean circles are homeomorphic to the real number line and subsequently any interval on the real number line could you also map the xy plane/the complex plane to a sphere mapping x to a circle generated by theta and y to the semi circle generated by angle psi?
So far as I understood it... - The real line has two open ends, but a plain circle hasn't, so they aren't homeomorphic. - If from the plain circle you'd take away a single point though, what'd be left of the circle would be homeomorphic with a line. - Those semi circles need to be open ended, too. So the complex plane won't be homeomorphic with the full surface of a sphere. The way of projection you propose leaves open the poles, as shared end points of the semi circles, and also doesn't include a continuous curve on the sphere surface connecting these poles and that's nowhere parallel to the equator. If you 'wish' to project the complex plane on the surface of a sphere, I think a sort of Riemann sphere would do better: - where the equator equals the unit-circle, - one pole equals the origin, - the other pole equals infinitely big, which is the point that is not part of the complex plane. - Longitude is just the argument or phase of the complex number. - Latitude is just dependent on the modulus.
20:23 Compact subspaces are not always closed subsets, so this proof does not work. Also, the proof cannot work because it is not true that continuous maps from a compact space are homomorphism, one needs the target to be Hausdorff
Sir at 0:43 you said that in homeomarphism the function can be from one matric space to another space and at 3:20 you said topology does not see distances. My question is, metric spaces cares about distances so how can we take Metric space as a function in homeomarphism definition?
I recommend Munkres Topology for this. Metric spaces are how topological spaces are constructed, and if the inverse of a bijective mapping from one topological space to another is continuous, you have yourself a homeomorphism.
Is it enough to find one homeomorphism f, so that M and N are homeomorphics ? or do we have to say they are homeomorphics for the specific homeomorphism f ?
Imagine being able to transform any object into any other object as long as they are toplologically homeomorphic in real life(like for example being able to transform a torus into a coffee mug)? How would that be as a superpower ?
13:28 ...but both (0,1) and [0,1] are open in themselves, so this doesn’t prove they aren’t homeomorphic. You just showed there’s no homeomorphism of R that sends (0,1) to [0,1], which isn’t the same thing. You need the compactness again, or the fact that there are points of [0,1] you can remove and have the remainder be connected, while this is false for (0,1).
Well you may not know the the difference between a donut and a cup of coffee but I do...I can eat a donut . Did I pass the test...it was a test, wasn't it ? Sorry for the levity...I gave up on maths after calculus 3. Have a good day sir.
In the category of topological spaces it is. The notion of isomorphism is that you can exactly match two objects and their structure, whatever the structure in question might be.
Casuals: *homomorphism*
Dr P: *homeomorphism*
I've waited for this for a long time - it's quite the treat.
he said that they are different
At 11:00, it certainly is possible to remove a point from that interval without disconnecting it. Just take an endpoint! I think the more usual topological invariants used here include "can remove two points without disconnecting it" or "the number of points you can remove that do not disconnect it" or "the number of points you can remove that do disconnect it".
WoW !, Thanks, Dr. Peyam. " NEVER ENDING LEARNING"
topology is one of my favourite subjects.
What the hell happened here
@@francaisdeuxbaguetteiii7316 Also my favorite subject...
Please share your Whatsapp no.
Okay Adam
@@gmjammin4367 who is adam
@5:16 Remembers pate a modeler but not play-doh that was excellent!
Ive heard rubber sheet geometry as well, would say "modelling clay" if wanting to avoid the brand name haha
Yes I’ve been waiting for this!!! Thank you :)
'Coffee cup is like a donut' well so much for my donut cravings
A good video. Good selection of properties and examples. Congratulations.
I enjoyed this video really much. You explained it clearly, while you have such an good welcoming attitude. Keep going!
Dr. Peyam, you are one of the greats on math youtube. I am studying topology right now and some concept can be hard to grasp. Thank you for making videos like this, it really helps! Also you seem like such a fun guy to be around, the energy you give off is amazing. Keep up the good work!
Thank you so much :3
You wouldn't believe it but I had to learn and apply this notion in literature for a project haha... So thanks for making it easy enough for me to understand!
im learning this to make math jokes in ceramics class
I did not expect an Animorph’s reference. Excellent video.
Wow, awesome and concise presentation.
crystal clear! thanks!
In short: Homeomorphisms are just relabelling the points and getting the same topology.
Precisely
I love those "In short" comments, they give further insights.
a very good video and explanation , thank you very much
Great explanation!
Thank you Dr Peyam!
Thanks sir for such explanation
This guy is a legend!!!
Thanks for such a great content with love from India
I really like all the motivating examples you give (e.g. the continuous bijection whose inverse isn't continuous)!
I am a little curious about the proof of the property at 9:14.
Continuity and Compactness czcams.com/video/6Ql6TpnpwDE/video.html
@@drpeyam thanks!
You made topology interesting
I wasn't interested in maths but watching 3b1r bprp and some other CZcams channel including yours has completely changed my view....
Now I want to do MSc in mathematics... It's an interesting subject
Congratulations :)
Very nice sir
maths is just playing with some pâte à modeler after all ;)
Thanks!
Thank you so much for the super thanks, I really appreciate it!!!
@@drpeyam absolutely! I love the way you teach. Less boring and more by example 👍
Hi Dr., If you find time, can you make a video about the first homology group? Thanks.
No way haha
@@drpeyam I’m fairly sure you must have taken algebraic topology and you took it more recently than I because you’re still suffering from ptsd.
Animorph fans represent!
I don't know much about topology, is there a way to define the limit of a sequence in a topological space without a metric?
Yes, sn goes to s if for all neighborhoods of s there is N large enough such that for n > N, sn is in that neighborhood
if you were to curve out the real line into a circle does that mean circles are homeomorphic to the real number line and subsequently any interval on the real number line could you also map the xy plane/the complex plane to a sphere mapping x to a circle generated by theta and y to the semi circle generated by angle psi?
So far as I understood it...
- The real line has two open ends, but a plain circle hasn't, so they aren't homeomorphic.
- If from the plain circle you'd take away a single point though, what'd be left of the circle would be homeomorphic with a line.
- Those semi circles need to be open ended, too. So the complex plane won't be homeomorphic with the full surface of a sphere. The way of projection you propose leaves open the poles, as shared end points of the semi circles, and also doesn't include a continuous curve on the sphere surface connecting these poles and that's nowhere parallel to the equator.
If you 'wish' to project the complex plane on the surface of a sphere, I think a sort of Riemann sphere would do better:
- where the equator equals the unit-circle,
- one pole equals the origin,
- the other pole equals infinitely big, which is the point that is not part of the complex plane.
- Longitude is just the argument or phase of the complex number.
- Latitude is just dependent on the modulus.
20:23 Compact subspaces are not always closed subsets, so this proof does not work. Also, the proof cannot work because it is not true that continuous maps from a compact space are homomorphism, one needs the target to be Hausdorff
Fun fact, JRPG maps are the same as a donut; not a sphere. This is another interesting example of a homeomorphism.
Interesting!!
Sir at 0:43 you said that in homeomarphism the function can be from one matric space to another space and at 3:20 you said topology does not see distances. My question is, metric spaces cares about distances so how can we take
Metric space as a function in homeomarphism definition?
I recommend Munkres Topology for this. Metric spaces are how topological spaces are constructed, and if the inverse of a bijective mapping from one topological space to another is continuous, you have yourself a homeomorphism.
Is it enough to find one homeomorphism f, so that M and N are homeomorphics ? or do we have to say they are homeomorphics for the specific homeomorphism f ?
One is enough
Will Lord Peyam have differential geometry videos on 2021? Would be amazing
I’m planning on doing a miniseries on differential forms, sometimes later this year
does (0,1) homeomorphic to R imply that any interval in R is homeomorphism to R
I think so, at least any open one
Imagine being able to transform any object into any other object as long as they are toplologically homeomorphic in real life(like for example being able to transform a torus into a coffee mug)? How would that be as a superpower ?
Not very effective…
11:50 how is it that it is both not homeomorphic and homeomorphic at the same time?
No they are not homeomorphic. If I said they are, I misspoke
13:28 ...but both (0,1) and [0,1] are open in themselves, so this doesn’t prove they aren’t homeomorphic. You just showed there’s no homeomorphism of R that sends (0,1) to [0,1], which isn’t the same thing. You need the compactness again, or the fact that there are points of [0,1] you can remove and have the remainder be connected, while this is false for (0,1).
Sir, 1/2x is not continuous at 0 but apne [0, 2] liya h?
?
Well you may not know the the difference between a donut and a cup of coffee but I do...I can eat a donut . Did I pass the test...it was a test, wasn't it ? Sorry for the levity...I gave up on maths after calculus 3. Have a good day sir.
By the definition, i wonder:
Is R^N Homeomorphic to any interval?
No if n >= 2 because if you remove a point from R^n it’s still connected but if you remove a point from an interval it becomes disconnected
@@drpeyam i wonder now. What its still true in R^1 and 2 ?
is there a special name for homeomorphisms which are uniformly continuous?
unimorphisms
Isnt homeomorphism the same as isomorphism ?
In the category of topological spaces it is. The notion of isomorphism is that you can exactly match two objects and their structure, whatever the structure in question might be.
@@mikhailmikhailov8781 aight thank you for your answer, But is there a case in which they are actually different ?
@@Caleepo isomorphism is just a generic term for any sort of equivalence between mathematical objects.
But is there an explicit formula to go from a coffee cup to a donut?
I bet
donuts at home
The morphism Is something new. Like Stokes theorem. I think we'll find a profound use.
What is your IQ sir?
Ok. So (f)-1 is continuous on the circle of radius 1 to the (0,2pi] because she's one to one and not onto sorry.
Thank you very much.