It is amazing how you gradually introduce all this abstractions. I have an engineering background, so some of the topology related concepts were blurry to me. Thank you for this nice presentation.
@@XylyXylyX Can you recommend any particular homework from a school or a text book for this manifold course? I am studying this partially out of self interest but would like some way to self test. In terms of the lessons in this series on manifolds which are the most important for manifold learning (I'm a bioinformatician and I see them used in unsupervised clustering quite a bit).
@@anguscampbell3020 There is a book by RWR Dowling called “Differential Forms and Connections” which is a great tool for these lectures. However, there are many other good texts as well, but start there.
I found this to be useful since I didn't understand what he was rambling about regarding density. From Wikipedia: The real numbers with the usual topology have the rational numbers as a countable dense subset which shows that the cardinality of a dense subset of a topological space may be strictly smaller than the cardinality of the space itself. The irrational numbers are another dense subset which shows that a topological space may have several disjoint dense subsets (in particular, two dense subsets may be each other's complements), and they need not even be of the same cardinality. Perhaps even more surprisingly, both the rationals and the irrationals have empty interiors, showing that dense sets need not contain any nonempty open set. ... Every topological space is a dense subset of itself. For a set X equipped with the discrete topology the whole space is the only dense set. Every non-empty subset of a set X equipped with the trivial topology is dense, and every topology for which every non-empty subset is dense must be trivial.
On (0,1) a subset is dense if for any point x and for any eps>0 (x-eps, x+eps) contains a point of a subset. If a subset is dense it doesn't matter how much you "zoom in", there always be points (infinitly many of them).
If the open sets of the topology are fine enough, we can find an open set containing a point in the sequence that dose NOT contain any other point of the sequence. If that is possible, then that point can not be a limit point.
A limit point of a sequence does NOT have to be a member of the sequence. For example, there is a sequence of rational numbers that approximates \pi and the approximation is better and better as the sequence becomes longer. However \pi is not part of the sequence.
"...which I'm gonna call n-hoods 'cause I can't spell it." I have to respect your honesty, I don't think I would have done the same. These videos are great by the way, they're way more clear than a lot of the lectures on this topic.
It's fine, I think everyone has a few words that they use all the time but have trouble spelling. It's one of those things where you don't realize it's a problem until it's too late.
Sir, at 31:11 : since the closure S_bar = S + limit points of S, then in your statement (S is dense in X if for all p in X, either p in S or p in S_bar) I think the "p in S" part is redundant ?
At 24:20 : topology could also be described by a closed set where the finite intersection rule becomes infinite.. does that mean that points can be subsets of a set? Earlier it was mentioned that a point cannot be a subset of an open set since points will require intersection of infinite sets!
A point is certainly a subset of any set that contains the point. However in the standard topology of the plane single points are not open because it would take an infinite intersection to create an open set with a single point. But....if we were to simply declare every point as open then we have simply asserted that the plane will have the discrete topology.
Hello, Around 22:30 you said the interior of any set is open. But I can imagine a CLOSED set inside a set. So this must be a problem with my language. Could you elaborate?
Anon x The interior of a set can never be closed because the interior is the complement of the closure of the complement of a set :) If you work through that chain carefully you will see that the last step is the complement of a closure. A closure always results in a closed set, so its complement is open! Alternatively if we lean on the neighborhood style of thinking, the interior of any set contains a neighborhood of every one of its members.This point is worth looking into. If you ever have a puzzlement over something I present, I encourage you to look it up somewhere else. These lessons can't possibly be detailed enough for the thinking student. I also encourage you to track down EVERY puzzlement you have. That is what I do and that attitude has served me well! Good Luck!
@4:15 by "neighborhood" did you mean "open neighborhood"? Or by "contains" did you mean "properly contains"? I am thinking of a neighborhood S of the point p where S has some boundary points, or for simplicity let's say S is closed. Let T be the same closed neighborhood as S. Clearly S contains T, but S is not a neighborhood for every point in t. The counterexample is any boundary point of T which doesn't contain an open ball that is fully within S.
No! At that timestamp we are defining a concept called "neighborhood" as an ALTERNATIVE to open sets, so we can't use the idea of "open" in the definition of a neighborhood. We are now discussing the defining properties of a "neighborhood". The example you cite uses "closed" which is also a topology-derived term, so that can't be used either. The situation you describe fails the neighborhood test that S MUST contain a neighborhood of each of the points in T. There are different ways to do "topology" and this one does not involve defining a topology, but it does involve defining neighborhood. The fact that these are ultimately the same is the point.
Around minute 10, when you state that, " *An open set is a neighborhood to all the points that comprise said open set* ", I become slightly confused. I read on wikipedia that another way of defining 'neighborhood' is the following " *A set V in the plane is a neighborhood of a point p if a small disc around p is contained in V* ". So my question is the following: while I understand that the points comprising the 'border' of the open set are NOT part of the open set, imagine you have a point p that is as close to the border as possible. How then is it possible to draw a disc around this point p without, at minimum, including points on the border (as the disc tries to 'wrap around' the point that is as close as possible to the open set's border). If the disc in question DOES touch the open set's border as it wraps around the 'super-duper-close-to-the-border point', doesn't this automatically *conflict* with the definition that *an open set is a neighborhood to all the points that comprise said open set* ? -Thanks
Samuel Cramer Your error is the idea “close to the border as possible.” There is no such thing. No matter how close a point is to the boundary, it can always get closer. For example, no matter how close a real number is to, say, the value 1...we can always find a point that is closer: a + (1-a)/2. Therefore we can always find a disc around any point in an open set of the plane that is inside the open set. THis is a very important point to understand and is worth thinking about until you get it!
You're probably not discussing these lectures anymore since you're doing the lie algebra ones now. I have a question however, if by chance you feel like answering: at 5:00 you said that a condition to define neighbourhoods is that if S and T are neighbourhoods of p and S contains T, then for every point r in T, S is a neighbourhood of r. But, using the construction of a metric space - defining open sets as open balls - if some closed set contains a ball around p (let s call this set T) then T is a neighbourhood of p. But T contains itself and - once again pertaining to the construction under an "usual" metric space, say (R, | . | ) - if T isn't R or the empty set, since it is closed then it isn't open. Thus there exist points r in T of which T isn't a neighbourhood. I might be doing something wrong, but in any case any clarification would be greatly appreciated. Love your videos btw
You construction will reduce to the discrete topology. You are mixing the n-hood definition with the standard topology definition buy using the notion of “open” and “closed”. If you have a topology then you can discuss “open” neighborhoods. Otherwise our definition *defines* neighborhoods and if you allow S and T as in your plane, then T (as you say) is a neighborhood. But if T is a neighborhood then EVERY POINT is a neighborhood of itself. This is the neighborhood version of the discrete topology!
No, that is not what the material at the timestamp you mentioned. This is only a property of a *sequence* of points. Any limit point is arbitrarily close to another point of the sequence. That fact is NOT true of most points of the sequence. In a line, EVERY point is a limit point.
SepehrBLK No. When we discuss the limit points of sequences we must have a point where the sequence "clusters" around the point. In fact some references call such a point a "cluster point". The usual topology of the plane is what drives the identification of a sequences limit point. We need a point where ANY open set containing the point also contains a member of the sequence. Such points are extremely special and not every point in S will satisfy that condition. ONLY the point the set converges to is a limit point.
Jan Stout Yes, This is covered at minute 19, I think. Maybe I didn't make it clear that the actual members of a set are always limit points because every neighborhood of members of the set always contain at least one member of the set! And yes, about density also.
XylyXylyX I don’t think you’re right, maybe it’s just convention, but a set with an isolated point wich is in the set is an example of a set whose elements are not always limiting point for the set. Take S={0}union(1,2) as a simple example: 0 is in the set and it’s not a limiting point, in fact there is no sequence of S with 0 as a limit.
I'm going through your previous lectures on tensors and manifolds, and noticed what appears to be an error in your definition of neighborhood at 3:08. To define N as a neighborhood of P (not S), don't you need set S to be contained in N, also containing P, where S is possibly "open"? That is, this defines N, not S, as a neighborhood of P.
I just reviewed it and I don't see a problem with how I presented it. I asserted that S was a n-hood of p by assumption and then said N is also an n-hood by definition because S is a subset of N.
That may work. In effect you're assuming that every set containing P is a n-hood of P, but the definition I presented above from Wiki seems to better express the concept of closeness or nearness for P in set N, where S is asserted as containing P and inside N. This defines N as a n-hood of P.
Frank Bennett THe only sets that are n-hoods of p are those that meet all of the properties I listed. That is far from "any set". I checked the Wiki article and it defines n-hoods in terms of open sets which isn't what I am doing here. Go to the section of the Wiki titled "Topology from N-hoods" to capture what I am trying to demonstrate.
As I see it, you haven't defined what a N-hood IS, but rather if you have one, some additional properties exist by definition. I will now go to your Wiki reference.
Frank Bennett Indeed! That is the same with "open sets" in the more canonical definition. You define the relationship between them and any collection of sets satisfying the rules can make up a topology. You never specify exactly what a characteristic open set is. When you do, then.you have *specified* a particular topology. For example, if you pick a single n-hood carelessly, then you can easily be driven to the conclusion that every singly element of the set is a n-hood of itself. That would be akin to choosing a region with its boundary to be "open". That decision reduces to the discrete topology! Same thing!
I've always thought of topological balls as a specific type of neighborhood, To define a ball, one needs already implicitly to use a concept of distance in order to define a radius. So what about non-metrizable spaces? Still with that caveat these are a great review and I love your work!
It is true that the basis of “balls” can only be used for metric spaces, but topologies can be defined for non-metric spaces too. For example ANY set has the discrete topology.
I thought of the boundary of a set S as being the "region" which does not intersect with at least one n-hood of the points that are in the exterior of set S and which also does not intersect with at least one n-hood of the points that are in the interior of the set S. Is that correct?
Mohit Varshney Topology is filled with alternative definitions for the same concept. If you were taking a real course in topology you might find yourself assigned a problem to prove that the definition you offered is implied by some other definition. Many ppl, including mathematics undergraduates, seem to think that certain definitions are universally used. That is not true. There are many many equivalent definitions of things and it is fun to connect them through proofs, but I don’t do that. In these lectures. The alternative definition you offer is fine. I suspect that is it not good for constructing a boundary in most cases but it certainly is true as long as the n-hoods are forced to remain in the interior and exterior of the set in question.
A set S is dense in X if the Closure(S) = X. In the discrete topology, there are no dense sets because there are no limit points. The only way for the Closure(S) = X is for S = X.
This is absolutely great. I've been trying to learn (gain an understanding) about math and physics on my free time but I'm too lazy to read. Khan academy was great, but it only took me so far. This is one of those few channels that explains in a wonderfull way.
Please keep trying to work with a good textbook. These lectures are NOT intended to be the sole material for a course in GR/Differential geometry. These lectures are to supplement good coursework or to supplement any decent text! Keep trying my friend! Good luck.
@@XylyXylyX Yeah I'm aware of that. Im just trying to gain a basic understanding lol. I'm still in high school so I will eventually take these courses in university. So I maybe exaggerated with "too lazy" as it is more accurately "to broke to buy right now".
Can somebody give me an example, how do we define a neighborhood mathematically in an example? Do we give a specific interval? Because if we do not define a certain neighborhood, a point has infinite number of neighborhoods, hasn't it?
A n-hood of a point is any set that contains a point and also contains an open set that contains the point. In most topological spaces each point has many n-hoods, yes. However topology is a wierd and deep subject and there are almost certainly topological spaces that have points with only one n-hood!
I think an alternate way to look at it, from a limit point perspective in the usual topology, is that a closed set is a set which contains all of its limit points, and that an open set does not. The {interior, boundary, exterior} of a set S is the set of all points which are limits points to {only S, S and S complement, only S complement}.
Topology is soooo rich. There are many ways of understanding every concept and it is good that you compare each way and understand the equivalence of each way.
At around 25:16, you say that S(bar) is often used to denote the complement of S, and then later, around 29:00-30:00, you use the same notation, S(bar), to refer to the closure of S. I believe the latter notation is the most commonly used, but I know it's always one or the other, never both. As I'm sure you're aware, these are very different sets, the closure and the complement. I'm also sure you get consistent with this in later videos, but perhaps an annotation here is needed. So, S(bar) for closure, and maybe S^c for the complement?
You are lucky to have something that strikes sparks with you! Many people never find such a thing or the thing they find is small and quickly exhausted. Good Luck!
I wrote a few functions that (should) encapsulate the concepts described in this video. I hope they make sense: Complement(S, X) = X - S Closure(S) = Union(S, LimitPoints(S)) Exterior(S, X) = Complement(Closure(S), X) Interior(S, X) = Complement(Closure(Complement(S, X))X) Interior'(S, X) = Exterior(Complement(S, X), X) Boundary(S, X) = Intersect(Closure(S), Closure(Complement(S, X)) Boundary'(S, X) = Closure(S) - Interior(S, X) IsDense(S, X) = Closure(S) is X
In your definition of a point q being a limit point of a set S, you should say that every neighborhood of q contains a point of S that is different from q, as otherwise isolated points would be limit points.
IN YOUR CHANNEL YOU SHOULD WRITE ABOUT YOURSELF ALSO. YOUR LECTURE IS VERY GOOD. I WAS IN SEARCH OF SEARCH LECTURE IN MANIFOLD. AFTER LISTENING TO YOUR LECTURE IT BECOME VERY OBVIOUS TO KNOW ABOUT YOURSELF.
Your alternative definition of topological space seems incorrect. One of the axioms states that every superset of a neighborhood of some point is also a neighborhood of that point. That means that in your system of axioms neighborhoods of a point are sets (not necessarily open), for which the point is an inner point. Like [0; 1) is a neighborhood of 0.5. But your last axiom states that if there are two neighborhoods S and T of some point, such that T is a subset of S, than S contains neighborhoods for all points of T. But that is true only for open neighborhoods! Consider p = 0.5, T = [0; 0.75), and S = [0; 1). Obviously, S does not contain a neighborhood of 0.
Btw, it seems that all the axioms you need are that intersection of two neighborhoods is again a neighborhood, every superset of a neighborhood is again a neighborhood, and every point has at least one neighbourhood. Then, if you define an open set as a set that is a neighborhood of all its points, then all standard properties of open sets can be recovered. UPD. It turns out, that while all the properties of open sets could be proven in such a system, it couldn't be proven that neighborhoods of a point are precisely supersets of some open set containing the point. To prove that one more axiom is needed: every neighborhood contains an open neighborhood. UPD 2. Well, I just realized that this is almost the last axiom you stated. The only difference is that in my version I say that there exists such a T_p \subset S_p that this property holds, while your version states that this is true for all T_p.
Can you cite a time stamp for me to review? I think the resolution is in the fact that not all “neighborhoods” can be considered open. I think I say that some point. I didn’t focus much on the n-hood definition of a topology, but it is interesting as you have noted.
Thanks for the video. For a slightly clearer definition of boundary points (this video calls them limit points) please take a look at this video: th-cam.com/video/FHL4udeLf9Q/w-d-xo.html
FWIW: I have gone through videos 1-9 and am now going through again. This is just great. OK.... so... One thing concerned me. You set down rules for open sets to define a topology. But you never defined an open set. I understand an open set is one that does not contain its boundary. OK... But here, in THIS second video, you finally address boundaries. And this makes me wonder if Video one and its discussion on open sets was intended to be such that you did NOT have to define what an open set is. Could you elaborate?------------------------------I suppose the confusion set in because you said the complement of a single point is the entire plane minus the point and that set is open. And that made me lost, too. Because such a set -- the plane minus the point -- seems closed to me since there sort of is a boundary (yes, if you remove the boundary, you do not get this "infinite approach to the boundary" You get everything except ONE point.
JT Coriolis We can pick our open sets any way we wish. THere is no "definition" per se except that an open set be a member of the topology. It is those rules for a topology that matter and whatever you define your open sets to be, they had better follow those rules! THe level of abstraction here is high, no doubt :). As you pointed out before, we can define open sets to be balls that include their boundaries, but that is just another way of choosing the discrete topology! When I say "an open set does not contain its boundary" I am specifically referring to the "usual" topology of the plane with the base of open balls. The plane minus a point is an open set in the usual topology because you can construct an infinite union of open balls that included the whole plane without the single point and open balls are, by definition open sets. THerefore the single point is a closed set because it is the complement of an open set.
Pros: easy to follow, lots of examples
Cons: Story developes slowly
Good lecture.
It is amazing how you gradually introduce all this abstractions. I have an engineering background, so some of the topology related concepts were blurry to me. Thank you for this nice presentation.
cara I am so glad you are finding this all useful. I have been a bit distracted, but will continue the series early next year.
@@XylyXylyX Can you recommend any particular homework from a school or a text book for this manifold course? I am studying this partially out of self interest but would like some way to self test. In terms of the lessons in this series on manifolds which are the most important for manifold learning (I'm a bioinformatician and I see them used in unsupervised clustering quite a bit).
@@anguscampbell3020 There is a book by RWR Dowling called “Differential Forms and Connections” which is a great tool for these lectures. However, there are many other good texts as well, but start there.
1:47 "...neighbourhoods, which i will call n-hoods, 'caus i cant spell it" i did not think i would laugh this hard during a video on manifolds.
precious lectures
This is by far the best topology lecture "out of nowhere" LOL
I found this to be useful since I didn't understand what he was rambling about regarding density.
From Wikipedia:
The real numbers with the usual topology have the rational numbers as a countable dense subset which shows that the cardinality of a dense subset of a topological space may be strictly smaller than the cardinality of the space itself. The irrational numbers are another dense subset which shows that a topological space may have several disjoint dense subsets (in particular, two dense subsets may be each other's complements), and they need not even be of the same cardinality. Perhaps even more surprisingly, both the rationals and the irrationals have empty interiors, showing that dense sets need not contain any nonempty open set.
...
Every topological space is a dense subset of itself. For a set X equipped with the discrete topology the whole space is the only dense set. Every non-empty subset of a set X equipped with the trivial topology is dense, and every topology for which every non-empty subset is dense must be trivial.
On (0,1) a subset is dense if for any point x and for any eps>0 (x-eps, x+eps) contains a point of a subset. If a subset is dense it doesn't matter how much you "zoom in", there always be points (infinitly many of them).
37:40 Exteriors are all open, also. I think you mean to say that the set of boundary points (and the set of limit points) is closed.
At 16:39, every point of S could be a limit point of S. Why not? Or the considering point is omitted from n-hood of that point?
If the open sets of the topology are fine enough, we can find an open set containing a point in the sequence that dose NOT contain any other point of the sequence. If that is possible, then that point can not be a limit point.
At 19:49, how the empty points are limit points? Because they might not contain points of the subset.
A limit point of a sequence does NOT have to be a member of the sequence. For example, there is a sequence of rational numbers that approximates \pi and the approximation is better and better as the sequence becomes longer. However \pi is not part of the sequence.
@@XylyXylyX , thanks a lot. I thought it was an empty area, not just one single point. Then closure of the point is itself( that point)
"...which I'm gonna call n-hoods 'cause I can't spell it."
I have to respect your honesty, I don't think I would have done the same.
These videos are great by the way, they're way more clear than a lot of the lectures on this topic.
Racoon in the closet I’m not proud of the fact that I basically can’t spell.
It's fine, I think everyone has a few words that they use all the time but have trouble spelling.
It's one of those things where you don't realize it's a problem until it's too late.
Sir, at 31:11 : since the closure S_bar = S + limit points of S, then in your statement (S is dense in X if for all p in X, either p in S or p in S_bar) I think the "p in S" part is redundant ?
alan c Yes, I think you are correct about that. We could do with just “p in S^bar”.
At 24:20 : topology could also be described by a closed set where the finite intersection rule becomes infinite.. does that mean that points can be subsets of a set? Earlier it was mentioned that a point cannot be a subset of an open set since points will require intersection of infinite sets!
A point is certainly a subset of any set that contains the point. However in the standard topology of the plane single points are not open because it would take an infinite intersection to create an open set with a single point.
But....if we were to simply declare every point as open then we have simply asserted that the plane will have the discrete topology.
Hello,
Around 22:30 you said the interior of any set is open. But I can imagine a CLOSED set inside a set. So this must be a problem with my language. Could you elaborate?
Anon x The interior of a set can never be closed because the interior is the complement of the closure of the complement of a set :) If you work through that chain carefully you will see that the last step is the complement of a closure. A closure always results in a closed set, so its complement is open! Alternatively if we lean on the neighborhood style of thinking, the interior of any set contains a neighborhood of every one of its members.This point is worth looking into. If you ever have a puzzlement over something I present, I encourage you to look it up somewhere else. These lessons can't possibly be detailed enough for the thinking student. I also encourage you to track down EVERY puzzlement you have. That is what I do and that attitude has served me well! Good Luck!
@4:15 by "neighborhood" did you mean "open neighborhood"? Or by "contains" did you mean "properly contains"?
I am thinking of a neighborhood S of the point p where S has some boundary points, or for simplicity let's say S is closed. Let T be the same closed neighborhood as S. Clearly S contains T, but S is not a neighborhood for every point in t. The counterexample is any boundary point of T which doesn't contain an open ball that is fully within S.
No! At that timestamp we are defining a concept called "neighborhood" as an ALTERNATIVE to open sets, so we can't use the idea of "open" in the definition of a neighborhood. We are now discussing the defining properties of a "neighborhood". The example you cite uses "closed" which is also a topology-derived term, so that can't be used either. The situation you describe fails the neighborhood test that S MUST contain a neighborhood of each of the points in T. There are different ways to do "topology" and this one does not involve defining a topology, but it does involve defining neighborhood. The fact that these are ultimately the same is the point.
Around minute 10, when you state that, " *An open set is a neighborhood to all the points that comprise said open set* ", I become slightly confused. I read on wikipedia that another way of defining 'neighborhood' is the following " *A set V in the plane is a neighborhood of a point p if a small disc around p is contained in V* ".
So my question is the following: while I understand that the points comprising the 'border' of the open set are NOT part of the open set, imagine you have a point p that is as close to the border as possible. How then is it possible to draw a disc around this point p without, at minimum, including points on the border (as the disc tries to 'wrap around' the point that is as close as possible to the open set's border).
If the disc in question DOES touch the open set's border as it wraps around the 'super-duper-close-to-the-border point', doesn't this automatically *conflict* with the definition that *an open set is a neighborhood to all the points that comprise said open set* ?
-Thanks
Samuel Cramer Your error is the idea “close to the border as possible.” There is no such thing. No matter how close a point is to the boundary, it can always get closer. For example, no matter how close a real number is to, say, the value 1...we can always find a point that is closer: a + (1-a)/2. Therefore we can always find a disc around any point in an open set of the plane that is inside the open set. THis is a very important point to understand and is worth thinking about until you get it!
You're probably not discussing these lectures anymore since you're doing the lie algebra ones now. I have a question however, if by chance you feel like answering: at 5:00 you said that a condition to define neighbourhoods is that if S and T are neighbourhoods of p and S contains T, then for every point r in T, S is a neighbourhood of r. But, using the construction of a metric space - defining open sets as open balls - if some closed set contains a ball around p (let s call this set T) then T is a neighbourhood of p. But T contains itself and - once again pertaining to the construction under an "usual" metric space, say (R, | . | ) - if T isn't R or the empty set, since it is closed then it isn't open. Thus there exist points r in T of which T isn't a neighbourhood. I might be doing something wrong, but in any case any clarification would be greatly appreciated. Love your videos btw
You construction will reduce to the discrete topology. You are mixing the n-hood definition with the standard topology definition buy using the notion of “open” and “closed”. If you have a topology then you can discuss “open” neighborhoods. Otherwise our definition *defines* neighborhoods and if you allow S and T as in your plane, then T (as you say) is a neighborhood. But if T is a neighborhood then EVERY POINT is a neighborhood of itself. This is the neighborhood version of the discrete topology!
At 17:00, is the sequence of points becomes literally line(very close points)?
No, that is not what the material at the timestamp you mentioned. This is only a property of a *sequence* of points. Any limit point is arbitrarily close to another point of the sequence. That fact is NOT true of most points of the sequence. In a line, EVERY point is a limit point.
At 16:30, don't all the points within the set S become a limit point (other than the point the set converges to)?
SepehrBLK No. When we discuss the limit points of sequences we must have a point where the sequence "clusters" around the point. In fact some references call such a point a "cluster point". The usual topology of the plane is what drives the identification of a sequences limit point. We need a point where ANY open set containing the point also contains a member of the sequence. Such points are extremely special and not every point in S will satisfy that condition. ONLY the point the set converges to is a limit point.
Isn't the closure of S just the set of limit points of S? And likewise, isn't S dense in X if each point of X is a limit point of S?
Jan Stout Yes, This is covered at minute 19, I think. Maybe I didn't make it clear that the actual members of a set are always limit points because every neighborhood of members of the set always contain at least one member of the set! And yes, about density also.
XylyXylyX I don’t think you’re right, maybe it’s just convention, but a set with an isolated point wich is in the set is an example of a set whose elements are not always limiting point for the set.
Take S={0}union(1,2) as a simple example: 0 is in the set and it’s not a limiting point, in fact there is no sequence of S with 0 as a limit.
Just to expand on the other answers given, limit points of S aren't necessarily a superset of the S, as shown at 34:50.
Very nice playlist
I'm going through your previous lectures on tensors and manifolds, and noticed what appears to be an error in your definition of neighborhood at 3:08. To define N as a neighborhood of P (not S), don't you need set S to be contained in N, also containing P, where S is possibly "open"? That is, this defines N, not S, as a neighborhood of P.
I just reviewed it and I don't see a problem with how I presented it. I asserted that S was a n-hood of p by assumption and then said N is also an n-hood by definition because S is a subset of N.
That may work. In effect you're assuming that every set containing P is a n-hood of P, but the definition I presented above from Wiki seems to better express the concept of closeness or nearness for P in set N, where S is asserted as containing P and inside N. This defines N as a n-hood of P.
Frank Bennett THe only sets that are n-hoods of p are those that meet all of the properties I listed. That is far from "any set". I checked the Wiki article and it defines n-hoods in terms of open sets which isn't what I am doing here. Go to the section of the Wiki titled "Topology from N-hoods" to capture what I am trying to demonstrate.
As I see it, you haven't defined what a N-hood IS, but rather if you have one, some additional properties exist by definition. I will now go to your Wiki reference.
Frank Bennett Indeed! That is the same with "open sets" in the more canonical definition. You define the relationship between them and any collection of sets satisfying the rules can make up a topology. You never specify exactly what a characteristic open set is. When you do, then.you have *specified* a particular topology. For example, if you pick a single n-hood carelessly, then you can easily be driven to the conclusion that every singly element of the set is a n-hood of itself. That would be akin to choosing a region with its boundary to be "open". That decision reduces to the discrete topology! Same thing!
I've always thought of topological balls as a specific type of neighborhood, To define a ball, one needs already implicitly to use a concept of distance in order to define a radius. So what about non-metrizable spaces? Still with that caveat these are a great review and I love your work!
It is true that the basis of “balls” can only be used for metric spaces, but topologies can be defined for non-metric spaces too. For example ANY set has the discrete topology.
I thought of the boundary of a set S as being the "region" which does not intersect with at least one n-hood of the points that are in the exterior of set S and which also does not intersect with at least one n-hood of the points that are in the interior of the set S. Is that correct?
Mohit Varshney Topology is filled with alternative definitions for the same concept. If you were taking a real course in topology you might find yourself assigned a problem to prove that the definition you offered is implied by some other definition. Many ppl, including mathematics undergraduates, seem to think that certain definitions are universally used. That is not true. There are many many equivalent definitions of things and it is fun to connect them through proofs, but I don’t do that. In these lectures.
The alternative definition you offer is fine. I suspect that is it not good for constructing a boundary in most cases but it certainly is true as long as the n-hoods are forced to remain in the interior and exterior of the set in question.
It might work well with Euclidean topology on R^n. I don't think it can be extrapolated to other topologies.
A set S is dense in X if the Closure(S) = X. In the discrete topology, there are no dense sets because there are no limit points. The only way for the Closure(S) = X is for S = X.
you explain everything very well! Keep it up. Good work. Ans thanks a lot :)
This is absolutely great. I've been trying to learn (gain an understanding) about math and physics on my free time but I'm too lazy to read.
Khan academy was great, but it only took me so far. This is one of those few channels that explains in a wonderfull way.
Please keep trying to work with a good textbook. These lectures are NOT intended to be the sole material for a course in GR/Differential geometry. These lectures are to supplement good coursework or to supplement any decent text! Keep trying my friend! Good luck.
@@XylyXylyX Yeah I'm aware of that. Im just trying to gain a basic understanding lol.
I'm still in high school so I will eventually take these courses in university. So I maybe exaggerated with "too lazy" as it is more accurately "to broke to buy right now".
Can somebody give me an example, how do we define a neighborhood mathematically in an example? Do we give a specific interval? Because if we do not define a certain neighborhood, a point has infinite number of neighborhoods, hasn't it?
A n-hood of a point is any set that contains a point and also contains an open set that contains the point. In most topological spaces each point has many n-hoods, yes. However topology is a wierd and deep subject and there are almost certainly topological spaces that have points with only one n-hood!
if we consider the intersection of (0,1] and (1-e,1+e) for some e>0, it is (1-e,1].
Is this intersection a neighborhood of 1?
No, because this intersection is not an open set (at least in the standard topology on R).
I think an alternate way to look at it, from a limit point perspective in the usual topology, is that a closed set is a set which contains all of its limit points, and that an open set does not. The {interior, boundary, exterior} of a set S is the set of all points which are limits points to {only S, S and S complement, only S complement}.
Topology is soooo rich. There are many ways of understanding every concept and it is good that you compare each way and understand the equivalence of each way.
At around 25:16, you say that S(bar) is often used to denote the complement of S, and then later, around 29:00-30:00, you use the same notation, S(bar), to refer to the closure of S. I believe the latter notation is the most commonly used, but I know it's always one or the other, never both. As I'm sure you're aware, these are very different sets, the closure and the complement. I'm also sure you get consistent with this in later videos, but perhaps an annotation here is needed. So, S(bar) for closure, and maybe S^c for the complement?
Nevermind, this is cleared up in the next video.
I just realized I'm doing my Calc 4 homework while listening to more maths, damn I love this field.
You are lucky to have something that strikes sparks with you! Many people never find such a thing or the thing they find is small and quickly exhausted. Good Luck!
I wrote a few functions that (should) encapsulate the concepts described in this video. I hope they make sense:
Complement(S, X) = X - S
Closure(S) = Union(S, LimitPoints(S))
Exterior(S, X) = Complement(Closure(S), X)
Interior(S, X) = Complement(Closure(Complement(S, X))X)
Interior'(S, X) = Exterior(Complement(S, X), X)
Boundary(S, X) = Intersect(Closure(S), Closure(Complement(S, X))
Boundary'(S, X) = Closure(S) - Interior(S, X)
IsDense(S, X) = Closure(S) is X
I just love the scenery.
can you say S is dense in X iff every open set in X intersects S?
zassSRK yes.
In your definition of a point q being a limit point of a set S, you should say that every neighborhood of q contains a point of S that is different from q, as otherwise isolated points would be limit points.
Exactly. In this way we can get a sequence of points of S distinct from q that converge to q.
great videos!
Benjamin V Thanks! I hope you still feel that way by the end of the series, if you stick with it...:)
Sir may i have your know about?
Zeeshan Sayed Mohammed I don't understand your question.
IN YOUR CHANNEL YOU SHOULD WRITE ABOUT YOURSELF ALSO. YOUR LECTURE IS VERY GOOD. I WAS IN SEARCH OF SEARCH LECTURE IN MANIFOLD. AFTER LISTENING TO YOUR LECTURE IT BECOME VERY OBVIOUS TO KNOW ABOUT YOURSELF.
Your alternative definition of topological space seems incorrect.
One of the axioms states that every superset of a neighborhood of some point is also a neighborhood of that point. That means that in your system of axioms neighborhoods of a point are sets (not necessarily open), for which the point is an inner point. Like [0; 1) is a neighborhood of 0.5.
But your last axiom states that if there are two neighborhoods S and T of some point, such that T is a subset of S, than S contains neighborhoods for all points of T. But that is true only for open neighborhoods! Consider p = 0.5, T = [0; 0.75), and S = [0; 1). Obviously, S does not contain a neighborhood of 0.
Btw, it seems that all the axioms you need are that intersection of two neighborhoods is again a neighborhood, every superset of a neighborhood is again a neighborhood, and every point has at least one neighbourhood. Then, if you define an open set as a set that is a neighborhood of all its points, then all standard properties of open sets can be recovered.
UPD. It turns out, that while all the properties of open sets could be proven in such a system, it couldn't be proven that neighborhoods of a point are precisely supersets of some open set containing the point. To prove that one more axiom is needed: every neighborhood contains an open neighborhood.
UPD 2. Well, I just realized that this is almost the last axiom you stated. The only difference is that in my version I say that there exists such a T_p \subset S_p that this property holds, while your version states that this is true for all T_p.
Can you cite a time stamp for me to review? I think the resolution is in the fact that not all “neighborhoods” can be considered open. I think I say that some point. I didn’t focus much on the n-hood definition of a topology, but it is interesting as you have noted.
Really cool video, thanks a lot! At the very last sth understandable, not like wikipedia xd
Great lectures!
Nguyen Loc Thank you for watching. Consider joining the new forums to discuss.
Great lecture! Just noticed that your voice reminds me of Steven Wilson's ahahahah
Lorenzo Nioi I’m not sure who Steven Wilson is, but thank you for the compliment:)
thank you!
Are you Physicist Sir?
31:10 So then couldn't we just simply say S is dense in X if
Closure of S = X?
@@jojowasamanwho Yes! I never did quite say that very simple point. Thanks.
I want to make a t-shirt that shows a diagram of a closed set and have it say "Respect Boundaries."
@@kingassassin7953 I’d buy one
11:35 *immature snicker*
Thanks for the video. For a slightly clearer definition of boundary points (this video calls them limit points) please take a look at this video: th-cam.com/video/FHL4udeLf9Q/w-d-xo.html
FWIW: I have gone through videos 1-9 and am now going through again. This is just great. OK.... so... One thing concerned me. You set down rules for open sets to define a topology. But you never defined an open set. I understand an open set is one that does not contain its boundary. OK... But here, in THIS second video, you finally address boundaries. And this makes me wonder if Video one and its discussion on open sets was intended to be such that you did NOT have to define what an open set is. Could you elaborate?------------------------------I suppose the confusion set in because you said the complement of a single point is the entire plane minus the point and that set is open. And that made me lost, too. Because such a set -- the plane minus the point -- seems closed to me since there sort of is a boundary (yes, if you remove the boundary, you do not get this "infinite approach to the boundary" You get everything except ONE point.
JT Coriolis We can pick our open sets any way we wish. THere is no "definition" per se except that an open set be a member of the topology. It is those rules for a topology that matter and whatever you define your open sets to be, they had better follow those rules! THe level of abstraction here is high, no doubt :). As you pointed out before, we can define open sets to be balls that include their boundaries, but that is just another way of choosing the discrete topology! When I say "an open set does not contain its boundary" I am specifically referring to the "usual" topology of the plane with the base of open balls. The plane minus a point is an open set in the usual topology because you can construct an infinite union of open balls that included the whole plane without the single point and open balls are, by definition open sets. THerefore the single point is a closed set because it is the complement of an open set.