What amazing times we live in that such complex topics can be made so accessible. Thank you ☺
@lucaug102 жыл бұрын
I love how mind-bending it can be to work with "different" topologies, so fun! I would like to thank you once again for delivering this amazing content, it is such a joy to watch your videos! It feels great to refresh on some subjects or learn new ones with every new video, and in such an enjoyable way. There is a lot of effort put into these videos and it shows!
@brightsideofmaths2 жыл бұрын
Thanks again!
@StratosFair2 жыл бұрын
Man, I wait videos from this series like I wait for new episodes of my favourite shows. Thanks for your amazing work !
@brightsideofmaths2 жыл бұрын
That is what learning should be :) Thank you very much!
@lucaug102 жыл бұрын
Glad I'm not the only one who feels this way! :D
@zathrasyes12872 жыл бұрын
Same here!
@Duskull6662 жыл бұрын
Yes! As an undergraduate physics student who tries to self study extra math courses on the side, this and the functional analysis series are perfect! Thank you :)
@obaidurrehman24642 жыл бұрын
Same is the case with me
@arisioz2 жыл бұрын
They’re both awesome courses fr dude
@Syrian.Coffee Жыл бұрын
Finally a great video on the topic ! Shurkan 🙏🏻🙏🏻
@brightsideofmaths Жыл бұрын
Glad you liked it!
@scollyer.tuition2 жыл бұрын
I like your method of defining the various types of points one after the other, with the associated Venn diagrams; it makes the definitions very easy to compare (and easier to remember than from the typical topology text, IMHO)
@shirleymoon9934 Жыл бұрын
It took me some time to understand the boundary of S, but it really deepens my understanding after it!
@shirleymoon9934 Жыл бұрын
Also one little question: can I say that the boundary of S is every thing in its left hand side "in this topological space (X,T)"?
@brightsideofmaths Жыл бұрын
You mean for the example here?
@shirleymoon9934 Жыл бұрын
Yes @@brightsideofmaths
@ahmetfurkanemrehan2 жыл бұрын
That's amazing lesson. I want to cry in front of such pure form for teaching. Congrats !!!
@brightsideofmaths2 жыл бұрын
Thanks :)
@NewDeal19172 жыл бұрын
0:00 Intro 0:25 Quick recap: topology definition 1:27 Important points (interior, boundary etc.) 5:36 Important sets made of points (including closure) 8:05 Example. Non standard topology on R
@arpit_prajapati112 жыл бұрын
Thanks for this amazing series
@brightsideofmaths2 жыл бұрын
Glad you enjoy it!
@PunmasterSTP2 жыл бұрын
Dang, your examples make my brain melt, but I find them fascinating and I learn so much! Thanks again for taking the time to make and share everything.
@miguelsantiagoradasanchez96239 ай бұрын
Thanks
@brightsideofmaths9 ай бұрын
Thank you :)
@Dede-qq8qv2 жыл бұрын
Thank you so much!! A great effort ♥️
@natepolidoro45652 жыл бұрын
Good stuff, good channel
@brightsideofmaths2 жыл бұрын
Much appreciated
@carl32603 ай бұрын
The following seems fit the given axioms of topology (@ 0:48): X = [0,3] and T = {ø, X, [1,2]}, but I thought closed interval [1,2] couldn't be an open set (except in edge cases). I'm probably missing something but can't see how this violates the axioms. Or is it that that *is* a topology, and an arbitrary closed interval can be an open set, and it's more that it's not of interest and, in particular, the collection of *all* closed sets gives the discrete topology so isn't of interest?
@brightsideofmaths3 ай бұрын
If you say "closed" interval, you always mean "closed with respect to the standard topology of R". This is interval might not be closed with respect to another topology.
@tensorfeld2952 жыл бұрын
I like this manifold series.
@christianharriviktorreibol55218 ай бұрын
Very good example at the end
@arrowmancy2 жыл бұрын
Hi! Sorry it's a silly point on notation/aesthetics, but I notice you sometimes use the ":" (or :\iff in latex) when, in contrast to the assignment/definition symbol for variables (e.g. a := 69) you'd use ":" to define properties rather than variables. Is that a correct reading of it? If so then it's a really nice shorthand for writing down "we call x something if and only if P(x) is true".
@brightsideofmaths2 жыл бұрын
Yes, the colon in ":" is just a reminder that this is actually the definition of the thing on the left.
@zazinjozaza61932 жыл бұрын
Why does the definition of a topology require finite intersections but does not put any restrictions on unions? Would it work if it was the other way around or are there problems with such a definition?
@brightsideofmaths2 жыл бұрын
We want to conserve the essence of open sets in metric spaces. Intersections could shrink the set to a non-open set. Unions don't make a problem.
@mgostIH2 жыл бұрын
Consider for example in the euclidean topology of R the intersection of all open sets of the form (-1/n, 1/n) for all natural numbers n, this is {0}, which isn't an open set
@TheSandkastenverbot2 жыл бұрын
One example: imagine the open intervalls ]-a, 1+a[, a>0 being a real number. The intersection of all these sets is [0,1], which is a closed set
@zazinjozaza61932 жыл бұрын
@@brightsideofmaths Thank you, and everyone who replied. I understand it better now.
@mgostIH2 жыл бұрын
@@TheSandkastenverbot And what does it being closed tell me about it not being open? 😄 (Nevermind I did the same mistake before, shhh)
@sidharthd440010 ай бұрын
All your lectures are just awesome!!! On the important names: What about points of S that are not in any open set. Can you check the definition of boundary points. Are all p belonging to U the way it is defined boundary points?
@brightsideofmaths10 ай бұрын
Thanks a lot! I don't get your questions exactly. Do you have a problem with the names?
@sidharthd440010 ай бұрын
After studying the example thoroughly at the end of the video, my doubts are cleared. Thank you for the amazing explanations.
@dqrksun2 жыл бұрын
Helped me alot, Thanks!
@medounendiaye31232 жыл бұрын
Hello folks, I don't get why (0,1) is told to be not open. Any single point grabbed in this interval has an interval around it included in (0,1). I missed something ?
@brightsideofmaths2 жыл бұрын
You miss the important part of topology: the sets we call open are fixed in T.
@medounendiaye31232 жыл бұрын
Thanks for replying and the good work you do day in day out. Does it mean that (0,1) needs to enumerated in the collection of the open sets ? I'm confused because if we set a=0, (0,1) belongs to (a,+inf). So the statement : " there is no such interval in (a, + inf) " doesn't hold anymore. Edit : i get myself wrong. By definition, we decided that only (a; +inf) intervals are open in our topology. So, (0,1) is not open in our topology. For the record : the answer to my question is YES. The open sets has to be explicitly said to be open.
@TBXu2 жыл бұрын
@@medounendiaye3123 “More generally, one defines open sets as the members of a given collection of subsets of a given set” on wikipedia it says :-)
@mastershooter642 жыл бұрын
yayyy more manifolds :D
@StratosFair2 жыл бұрын
Hmmm it seems to me that being an accumulation point of S is equivalent to being either an interior or boundary point of S, but I can't manage to prove it only with these definitions... I guess that depends on the choice of topology ?
@StratosFair2 жыл бұрын
@Luca Zz What the heck, you just destroyed my brain. Can't wait for the rest of this series :)
@heimrath0072 жыл бұрын
@@StratosFair Think about the interior and boundary points of the set {1}u{2} as a subset of the real line with the usual topology.
@zaccandels66953 ай бұрын
Hi, what texts would you recommend for someone who wants to study this in a bit greater depth?
@brightsideofmaths3 ай бұрын
Do you mean topology or manifolds?
@angtrinh649527 күн бұрын
For the boundary point p of S, let's say it's not included in S (a little shifted to the left), but it belongs to U (an element of the topology) whose intersections with both S and complement of S are not empty. Can we still call point p the boundary point of S?
@brightsideofmaths27 күн бұрын
Why not?
@angtrinh649527 күн бұрын
@@brightsideofmaths In my intuition, any points that belong to the line of S is the boundary points. But in this case, although U contains p and intersection of U and S (and complement of S) are not empty, but p doesn't lie in the line of S (but still in U). Can you please explain more about it?
@brightsideofmaths27 күн бұрын
@@angtrinh6495 For such discussions, the community forum is better :)
@angtrinh649527 күн бұрын
@@brightsideofmaths Thank you! I'll take a look there!
@xxoloveitt Жыл бұрын
Hi, thank you so much for your videos! I have a question. On the example the exterior of S does not include the interval (-inf,0] because it is a closed set right? Then why is the boundary point of S (-inf,1] ? Isn't it supposed to be an open set? Why is (-inf,0] closed while (-inf,1] is open?
@brightsideofmaths Жыл бұрын
Thanks for the questions! Who says that (-inf, 1] is open? The exterior is always an open set.
@xxoloveitt Жыл бұрын
@@brightsideofmaths oooh okay, then if I understood correctly that means that the boundary of S does not need to be an open set? Btw thanks again for answering! I'm learning this in uni but I don't understand the professor 😭
@brightsideofmaths Жыл бұрын
No, the boundary is always closed.@@xxoloveitt
@xxoloveitt Жыл бұрын
@@brightsideofmaths got it, thanks!
@kingshukdutta20642 жыл бұрын
shouldn't a circle be S(superscript)1?
@brightsideofmaths2 жыл бұрын
That would be the notation we use later for the circle as a manifold. Here we are on an abstract level :)
@ihavezerofriends8 ай бұрын
Thanks manyfolds!
@c.m.1392 жыл бұрын
Hi, I am 14 and I’d really like to learn advanced physics, but I can’t find high quality videos like yours. Can you please start a Classical or Quantum physics series? (or Relativity). Thx
@santiagoarce56722 жыл бұрын
You may enjoy teaching yourself from textbooks. For Classical Mechanics I like David Morin's Intro to Classical Mechanics which you can prob find online. That should be a good start.
@sossupummi Жыл бұрын
may I suggest using the search function to look up the International Winter School on Gravity and Light 2015
@deansmith7163 Жыл бұрын
kzbin.info/www/bejne/pqq7aZ6HYq-tfqs
@alijoueizadeh2896Ай бұрын
Thank you.
@brightsideofmaths29 күн бұрын
You're welcome!
@smolfish34709 ай бұрын
Is it true that if T = P(X) then there is no boundary point and accumulation point, since {p} is contained in T?
@brightsideofmaths9 ай бұрын
This is the discrete topology, where every singleton is an open set. So you can show that a set cannot have boundary points.
@guigomartins9 ай бұрын
so, always it can be said that ∂S = X \ (S° ∪ Ext(S)) or is this a result just for the case of this example?
@brightsideofmaths9 ай бұрын
Yes, this follows directly from the definition.
@guigomartins9 ай бұрын
@@brightsideofmathsthanks! this is my first contact with this subject formally and your videos have been of great help
@wenanyaugustine3311 Жыл бұрын
Hi just thinking, isn't (0,1) an open set? you just said it isn't and I was wondering how prof. Thanks
@brightsideofmaths Жыл бұрын
"open" is always with respect to a given topology.
@HeirOfCuba2 жыл бұрын
Does this mean it would be impossible for a subset of a space with a discrete topology to have an accumulation point?
@brightsideofmaths2 жыл бұрын
Indeed!
@shobhitkhajuria74642 жыл бұрын
Can u please tell me why you used (a,inf) as topology and said it’s important, why not [a,inf)? It will still satisfy the 3 conditions and will be called open sets. Where am i wrong?
@brightsideofmaths2 жыл бұрын
If you have [a,inf), you could write the union of [a+eps, inf) for eps > 0 and get (a,inf).
@dangthanhtuanit Жыл бұрын
In example, S(0,1) is an openset.
@bobbybannerjee51562 жыл бұрын
Brilliant lecture. May I know what software (app) you are using to write on the screen with your (stylus) pen?
@brightsideofmaths2 жыл бұрын
Xournal :)
@roozbehr932 жыл бұрын
I hope the next one is about Charts, Atlases, Compatibility, Differentiable Manifolds etc. ;)
@brightsideofmaths2 жыл бұрын
First, we finish topologies and then we start with differentiable manifolds :)
@养兔大户 Жыл бұрын
nice video!
@brightsideofmaths Жыл бұрын
Thank you! Cheers!
@trondsaue78602 жыл бұрын
You state that a boundary point of S should be neither in the interior nor in the exterior of S, but formally define such points by an open set whose intersection with both S and its complement is not the empty set. However, X is in the topology of X, hence an open set, so what if I choose U = X ? In the definition you do not require U to be a subset of X....
@brightsideofmaths2 жыл бұрын
The point is that the property of the intersection holds for *all* open sets U.
@trondsaue78602 жыл бұрын
@@brightsideofmaths Got it ! Thanks ! I very much enjoy these videos.
@SamSarwat902 жыл бұрын
I got the accumulation points of S to be { x | x
@brightsideofmaths2 жыл бұрын
Nice work!
@shakesbeer002 жыл бұрын
So accumulation points are the union of interior points and boundary points.
@shakesbeer002 жыл бұрын
In the example, the derived set S' should be the same as the closure of S.
@brightsideofmaths2 жыл бұрын
No, not all boundary points are accumulation points.
@brightsideofmaths2 жыл бұрын
@@shakesbeer00 In this example, it's correct.
@shakesbeer002 жыл бұрын
@@brightsideofmaths You are right. It seems that S' contains such points of S that each has a sequence in S converging to it. Thanks for the prompt clarification!
@fabiangn80222 жыл бұрын
gracias.
@loden567711 ай бұрын
I don’t get how S can be a subset of X but not necessarily in T, given X itself is in T???
@brightsideofmaths11 ай бұрын
You have to be careful when talking about elements and subsets relation. We have S ⊆ X and S ∈ P(X) but also T ⊆ P(X). However since T is not equal to P(X), it can happen that we have S ∉ T. In short: a set can be an element or a subset of another set. These two notions are different.
@loden567711 ай бұрын
@@brightsideofmaths Thanks for clarifying, I see what you mean!
@samlai57152 жыл бұрын
interesting example
@jacobadamczyk33532 жыл бұрын
Does anyone know what note-taking/recording software this is?
@brightsideofmaths2 жыл бұрын
I know it!
@jacobadamczyk33532 жыл бұрын
@@brightsideofmaths haha, I'd sure hope so! Do you mind spilling your secrets? P.S. Thank you for making all these great videos!
@brightsideofmaths2 жыл бұрын
@@jacobadamczyk3353 Of course. I read your question like you just asked about existence.
@brightsideofmaths2 жыл бұрын
I use Xournal :)
@jacobadamczyk33532 жыл бұрын
@@brightsideofmaths I should've expected that much 🤣 thank you!
@nazmulhasanfahim731 Жыл бұрын
Where is Quiz for video part-2?
@brightsideofmaths Жыл бұрын
Oh, you can find it here: tbsom.de/s/mf
@michuosas2 жыл бұрын
so (0,1) is NOT in the topology, right?
@brightsideofmaths2 жыл бұрын
It is not an element of T. :)
@marcelo4303 Жыл бұрын
What if a=0 -> (0, infinity) in TAu. (0, 1, 2, 3... infinity). The elements 0,1( and the infinity elements between, when it comes to the elements of the set of reals) in this interval is not already contained in tau ?
@keeperofthelight96812 жыл бұрын
I can choose arbitrary u then any p can be shown as boundary interior exterior rip :’(
@ativjoshi10492 жыл бұрын
Definitions were simple; the example, not so much...
@PunmasterSTP2 жыл бұрын
I really like that he makes his videos that way. They start by laying a foundation, and then they take it in a whole new direction!
@ativjoshi10492 жыл бұрын
@@PunmasterSTP definitely true.
@cristhiangalindo48002 жыл бұрын
Hola como esta? hablaré esta vez en español. Surge una importante pregunta de clases peculiares de variedad de curvas Mg en algún 3-pliegue de CY. La pregunta en cuestión es probar que para n=3 de muchos pliegues todo CY\times{1}= CY\{1\}, que prueba para tal parte unitaria como es de proyectiva una variedad Mg, en Mg\times{} CY\{1\}= Mg\times{} \{-\prime}, esto pues la parte primitiva de una variedad Calabi-Yau con muchos pliegues, es capas de generar "puntos" de curvas muy parecido al espacio Q-racional. Las preguntas de investigación que yo con otros investigadores (basado en la obra de Dirchilf y Joyce) es entender cómo para este espacio Q-rational sustituimos el unitario CY\{1\}, por un space-modulo de Hodge que son todas las bases de un diagonal D^{\prime{} - 1}= D\times{} C. Aquí se construye un módulo D(1) para incrustar los 3-pliegues de un CY en su única y constante parte primitiva. De hay por ejemplo se podría entender muy bien con las curvas conjeturas por Dirchilf-Joyce son muy altas en una superficie-Enrique, que sea capas de producir curvas semi-estables no necesariamente degenerada (calculo "estable" para pliegues de el invariante DT), que puede ser escrito también como un grado de la curva, general al módulo D(1)-Hodge incrustado. Esto pues todo reflejo de la superficie-Enrique es cuadrática y a semeja cualquier curva a un esparce alto G-global muy próximo.
@antonellomascarello4698 Жыл бұрын
💛🙏
@zSliz111 ай бұрын
You do a miserable job with examples. Your examples suck, and you don't help with intuition. You just explain rules