The example of openness, closeness and boundaries towards the end of the video is an absolute gem... shows that blind intuition (drawing analogies mainly from the continuous realm of reals) can deceive us more often than not...
For all the people who are confused why in the last example (1,3] is open.... thats because the epsilon-ball B(x) = { y element X | d(x,y) < eps) and when you consider the point at x=3, there are no points complementary to A between 3 and 4, is just not part of the metric space X by definition, so the eps-ball definition gives you a thumbs up and this is why its an open set
I'm having fun with these short lectures! I'm super curious to see how far this will go into the topic. Congratulations on your ability to explain these concepts so clearly!
0:30 if you fix a point x in a matrix space, you can look all the other points that have the same distance from the point x 0:44 the notion of the ball we want to generalize in a metric space 1:20 the epsilon ball is never empty 1:40 open set 3:15 boundary point 4:58 open set A is the set where all the boundary points of A do not belong to A 5:10 closed set define using the above logic 5:29 the definition of closed set 5:52 closure 6:33 example : X is a subset of real number 6:55 considering subset of X
I am afraid the statement at 4:58 is not true. The example explains exactly this. The boundary points can be part of the open set as long as the ball at the boundary dA does not include elements of the complement of A, A^c. A set can be both open and closed.
I'm a Korean student, and recently started studying college level mathematics for double-major. This is the part I've been having a hard time with these days, and this video was very helpful. Your other videos are very cool, too. Thanks alot! :D
If you want to work just in the real numbers, my real analysis series might be helpful. See here to learn about open sets: ruclips.net/video/Wqo4Svs4erw/видео.html
Dang that example with X := (1,3]U(4,∞) and A := (1,3] really threw me for a loop. If I understand correctly, the point x = 3 is *not* a boundary point because its epsilon-ball when d < 1 lies entirely in the region that is less than 3, since the region (3,4] doesn't exist in the set X. Hence the epsilon-ball is entirely contained in A, showing that x = 3 isn't on a boundary but rather is in the interior of A. Now that I (think I) understand that concept, it is *very* cool! Man, these videos are ballin!
Yes, totally correct! This example shows why "boundary", "open", "closed" and so on are always to be considered in the respective context. Here, the whole metric space has to be known.
@brightsideofmaths Would this be better stated as A is open with respect to X? Because with respect to the real numbers R, the set A is neither open nor closed.
The most important lessons of this great video are the followings: 1) closeness and openness are not mutually exclusive concepts 2) closeness and openness must be checked independently 3) the same subset A can have different properties depending on the structure of the set X it belongs to.
6:30 In the example, A is a subset of X and 3 is an element of A, so 3 is either a boundary point or not of A. If 3 is a boundary point of A then any open ball centered at 3 will contain elements of A and elements of X/A. In particular, a ball of radius 1 centered at 3 will only contain A and (3,4]. Since (3,4] is not in X/A = (4, infinity), then 3 is not a boundary point of A. Since none of the points of A are boundary points of A, we can say that A is open. Also, since the complement of A is open, then A is closed. I think that you could use a ball centered at 3 with any radius < 1 and come to the same conclusion. I think the confusion was in considering the complement of A not as X/A = (4, infinity) but as all real numbers outside of (3,4]
Am really enjoying this lecture. I can't believe after 5 years I can still enjoy maths like this.The professor i very clear, he is more of a teacher than a lecturer.
The important subtlety in the last example is that A would not be an open substet X if X were to be the whole real line R. It is sometimes confusing because we tend to forget that our set X actually has a hole in it. in another word {3< x
Really love the video! Now one of my roadblocks is solved. I remember watching a Sith lord of the Dark Math teacing differential geometry on YT, he also said a set can be both open and closed, but only gave a very short explanation that I wasn't able to grasp. At that moment I thought I didn't deserve to wield the Dark Side of Math. Now the Bright Side hath shewed me the truth, and I believe I belong here, for now. 9:10 A is also closed, is it because that the complement of A is open?
Thank You for great lectures! I have a question about: 5:05 " A is open A ∩ ∂A = ∅ " 5:22 " A is closed A ∪ ∂A = A " I think there should be implications only in one direction: A is open
@@brightsideofmaths If A is both open and closed, it means that simultaneously "A ∩ ∂A = ∅" and "A ∪ ∂A = A", which is possible only if ∂A = ∅. Am I correct?
I’m really excited to see more functional analysis videos. I have an exam coming and you’re really helping me out. Could you maybe make a video about operators (why boundedness implies continuity, Neumann series and investable operators). Also something about properties of infinite dimensional spaces like the theorem of Arzelà-Ascoli. The most difficult topic, so I find, is however the Sobolev spaces (although it might be too early). You’re a great teacher so please continue with this playlist haha.
Incredible well explained! Love it! I'm thinking about what you said, points in A that are boundary points but are inside A, I always think of boundary points as points that are not inside A. I'm going to think about it
Could I ask a question? On Example (a), you have proven that 'For x=3' B_1(x) belongs to A and that A is open. Epsilon was assumed(?) 1, so it turned out that it doesn't contain 4 in set X. How does it make sense that the result from assumed epsilon could define the general characteristic of a subset A? +) I could not understand that openness and closedness is not an opposite pair of notion. Into which video could I check?
@@brightsideofmaths That makes sense! Also, did you choose an epsilon value less than one for the case where x=3 because, if you put x = 3 into the original epsilon equation you used for when x≠3, you'd get min (2,0) x 1/2 But epsilon needs to be above 0 so you pick 2, and half it for one? Is that why you chose one? or did you just randomly pick it?
Thank you for these lectures! I have a question about last example (a): if we consider X:=R instead of X presented in the lecture then A won't be open, right?
I'm a tiny bit confused by the last statement. Wouldn't C_bar equal (1,2) because 2 has been shown to be a boundary and not included in (1,2]. The compliment of C would be (2,3],(4,inf)?
I have a question with your example at the end with the subset A=(1,3] : you mention this set is closed but the intersection of the ball of center 3 and radius 1/2 (or less) with the complement of A is empty, so doesn't that contradict the first definition ? Interestingly however it is true that the complement of A is open, so aren't these two definitions inconsistent ?
Hiiiiii I didn't understand the explanation at 7:50 if you take the point x = 3 and you take an epsilon ball around the point it will always exceed so the space is not open? And so if we follow this logic then by the same process then the complementary is also an open and therefore I am lost ^^
Just bare in mind that the same set can be open and closed at the same time (such sets are sometimes called "clopen"). It's a bit counterintuitive, but there is no contradiction here.
I think it's because there's no X defined between 3 and 4, so even though the ball on 3 will always go to 3.1 or 3.001 or 3.00001, we ignore it because it's outside A, but not in X. Because X is only defined again at values greater than 4. It's like there's nothing between 3 and 4, so we don't count the ball there. Nothing exists there, the universe we are in is defined by X only. I'm not sure about this explanation but is what I concluded trying to understand.
ZirTaaah The whole set and the empty set are trivially clopen. Let R^n be a metric space than the empty set is open because you can choose an arbitrary episilon. The complement is R^n itself and this is open too. So the empty set has to be clopen. You can argue the same way for R^n.
When we have a topological space we call elements of the topology open sets as well? Is it the same notion of openness as shown here? I mean I don't really see how openness makes sense when we aren't considering R^d or some subset of R^d. Also, do we need a metric to define open sets or are there other ways?
You can induce a topology from any metric by defining the open sets as shown in this video. But not every topology has a corresponding metric. Therefore topologies are a more general construct, but not as useful for more specific applications. The good thing about metrics is, that they already have a notion of "distance" which you need for analysis. Edit: As far as I know topological (which basically means ad hoc) and metric (by the given definition) spaces are the two only general ways to define open sets, although you can fand equivalent expressions, which may require special circumstances. Like you can derive a topology from any metric, you can also derive a metric from any norm (by saying d(x,y) = ||x - y||). With this in mind you can define metrics on function spaces (for example with the sup norm for bounded functions) and consequently have open sets. In doing so you can mathematically make sense of "openness" on this abstract infinite-dimensional vector space though I wouldn't know how to greatly visualise this.
@@sven_lu_ Thank you for replying, your comment did provide some insight. I'm guessing the term "open set" originally came from using metrics to induce topologies and then stuck around as the general name of elements of the topology.
@@brightsideofmaths The definition of a boundary point for A reads: a point around which there can not be an epsilon ball small enough, such that the epsilon ball only contains points of A. A simply does not have boundary points. So the closure of A would be A again. Is this proper?
Very nicely presented lectures. My only comment comes where the use of the phrases "inside of A" and "outside of A" where boundary points for A is concerned--- Incorrect is the statement a boundary point of A can be "inside of A". Generally, we associate "inside of a set A" with the topological "Interior of A" and we similarly associate "outside of a set A" with topological "Exterior of A". Thus, a point x is a boundary point of A if every open ball about x meets both the exterior and the interior of A in points other than x. In my opinion, a boundary point x of A may set theoretically "belong to A" or not "belong to A" would be proper.
In the first example, If A is a semi closed interval then at the point x=3 the open ball would not be a subset of A as it would contain points outside of A, points greater than 3 so how can it be an open set? As open set needs to have open ball of every point inside it?
I really enjoy the way you teach math. Now I have strong interest in learning functional analysis more seriously. Could you recommend some textbook ? I hope those books are not just presenting definitions or theorems, but also giving some interesting examples and incentives of how those math object are discovered or developed. Thank you!
I can really recommend Rudin's book about functional analysis. However, there are a ton of good books and free lecture notes about the topic. Maybe you just try some of them.
I cant uderstand why. Not all open balls will have others point. just for big ones. If the natural line is the metric for example. for each x, only epsilon bigger than one will have other point. i cant understand why it is boundary. To be boundary dosent need that for all balls need to have other point outside of the subset and also inside of the subset ?
I felt so confused about the example (1,3] in this video. This video says (1,3] is open and also closed. But my lecture notes told me such interval (a,b] is neither closed nor open. Could someone explain this to me? Thank you!!
In the example (1,3], if x=3, B_1(x) does not belong to (1,3] totally, the part (3,4) is outside (1,3], so I couldn't understand why you said this example is open.
@@Sarah-bn7nj The sourrunding space is important. "Open" is always to be understood with respect to a metric space (or in general: a topological space).
Hi, I have a question. In your example, A:=(1,3], it is both open and closed. Does it have a boundary? If it doesn't have a boundary, can we still say it's closed?
@@brightsideofmaths According to my understanding, given X:=(1,3]U(4, ∞), A:=(1,3] is closed with no boundaries. Is that correct? It's a bit counter-intuitive though...
Using the definition of Closed set that its complements is open, in this case, Ac is open right?[ (4,inf)]. But using the definition of closure, that a closed set has the interior and boundary points, i.e. the closure is equal to the set. I cant see the A having its boundary point, because when I apply the definition of boundary point on 3 it dosent work, because not for all open balls i will have point outside and inside of A, just for balls bigger than one. Could you help me with it? Thanks !
Is it correct that C would not be closed in the example if the first value of the interval is any bigger than 1? So e.g C=(1.001,2] is not closed and the closure would be [1.001,2] with the new boundary point being 1.001? Thanks so much, very insightful!
I'm not sure I'm thinking about this correctly, but in the definition of a closed set, must it be assumed that X has no boundary? Otherwise it seems that both A and X could be closed, due to each of them having a boundary.
In your last example, if epsilon is 1.5, then the open ball (1.5, 4.5) belongs to set X but does not belong to set A. So A is not open? Or did I misunderstand something?
so how do we decide on an epsilon? Couldn't you just pick any eps so small that it eventually works? Or is it just that, that eps is defined as some arbitrary very small number?
I dont quite get how our A = (1; 3] u (4, +inf) is a closed set? Is it buecause the contradiction of A is a subset of X equall to (4; +inf) which is clearly open and hence by definition A must be closed?
About example a). By definition a set A is open if for EACH x from A we can find a ball. In the example (1, 3] the 3 IS included. So why I can't chose 3 and try to find a ball? Isn't 3 is boundary point? Isn't A is closed?
I really love your lectures!, I have a question on this one, so If I have disk on R^2, the disk is an open set but also a closed set because I can extract a ring from it as a boundary, creating another open set and a boundary, and os on, am I right?
@@brightsideofmaths so if X is all the points like (a,b) where "a" and "b" are real numbers, and d is the euclidean distance. Not sure If expressed myself correctly, I'm learning math as a hobby. Thanks in advance for your response.
@@AndresFH7233 If X = R^2 and d the normal euclidean distance, the *subset* A given by your disc can't be open or closed at the same time. For example, it is open if the circle line is excluded from A.
Can an epsilon ball be formed around 3 as the center, when at 3 as the center will the ball still be in the interval (1,3]. Will the ball not get out of the interval hence not open?
I think that the answer is no because outside of 3 towards 4 is an interval which is not part of the metric space X in which A is a subset. If A was a subset of R instead then it would be closed. At least that's what I got from.the last example.
I have one question , in the book "Introduction to topology by Simmons" it is mentioned that the interval X = (0, 1] is an open set, if it is given that X in itself forms a metric space (X,d). So if we consider X as a subset of Real line, when real line forms a metric space with "usual" metric, it is NOT an open set. But if it is given that X itseld forms a metric space, it is open So is the concept og being open or not more abstact than we define? i mean removing bounday points? I mean , as far as Iearned, X here is defined open as there is no point even talking about points like "0" as metric space (X,d) is not even defined for them... What is this sorcery ... please explain
@@brightsideofmaths So, indeed if X is defined as a metric space, no matter if it is (0,1] it can be open? W.r.t to some metric d....hmmm....now I understand why the definition of open set with respect to a space, its complement etc is so important than simple an Open Ball one. There is a theorem as well, that whole set of a metric space, along with empty are open sets.
@@brightsideofmaths agree, but anyone after seeing a semi closed interval would think it is not open. Infacf most text books and videos teach students that open sets are nothing but generalizations of open intervals on real line....
@@brightsideofmaths that is actually not difficult to comprehend, arbitrary unions and finite intersections are closed....I mean that is why empty set has to be open...otherwise finite intersection would not even work 🙄
Hello friend, (A) intersection (boundary of A) is empty iff open and (A) union (boundary of A) = A if closed. Is this always true or only for the standard topology ?
I have a question if you don't mind. Is the example you gave with the set (1,3]. Since you said that A is an open set, then 3 cannot be a boundary point, as all boundary points have to lie outside of A for A to be considered open. You also said that the set A is closed. Since A has no boundary points, but it is closed, then a set can be closed without any boundary points? Is my reasoning correct? This seems to go against intuition.
Isn't the definition of a closed set , "a set that contains all of it's limit points" ? where by limit point we mean a point such that any open ball of that point when intersected with the set minus the point itself , it will be non empty ? So interior points are limit points too , but if all points are interior points then the set is open etc etc
@@brightsideofmaths thank you.1 more question. There is a theorem in Rudins Principles of mathematilca analysis that states "a set is open iff its complement is closed" here we said that A is open , and its complement is also open. Does that mean that both of them are also closed?
@@brightsideofmaths I am glad that I have looked through this comments. You are totally right in saying, that the set is opened and closed. But I think it would have been a little better to start with an example from the real line and then go to these special cases. But it is very good that you discuss these special cases, these namely make the difference. By the way, very well explained. Thanks for your great effort!
@@lvlanson Thank you for you comment! I have a whole series about real analysis, where one focuses on the real number line. Here, we have to go to the abstract case immediately. Therefore, I started with these special cases :)
@@brightsideofmaths This sounds absolutely reasonable. Maybe it would be good to add like a link into the video linking to the real analysis part concerning this video. This might help to get a better grasp on it. Anyhow you do it, it is well done. These are only some points to grind out the fractions of perfection.
See here is something I never really believed, and I guess this reveals my insecurity about the real numbers being densely packed; we can just arbitrarily choose any ol' epsilon we want in order to guarantee that a set is open. . . seems fishy to me.
@@brightsideofmaths too true, I find myself far too often debating believability of claims rather than attempting to see how the axioms support the ideas.
The example of openness, closeness and boundaries towards the end of the video is an absolute gem... shows that blind intuition (drawing analogies mainly from the continuous realm of reals) can deceive us more often than not...
my intuition says is inline with the examples, idk its like theres a void so i can feel that it was open
For all the people who are confused why in the last example (1,3] is open.... thats because the epsilon-ball B(x) = { y element X | d(x,y) < eps) and when you consider the point at x=3, there are no points complementary to A between 3 and 4, is just not part of the metric space X by definition, so the eps-ball definition gives you a thumbs up and this is why its an open set
But the ball isn't included in the interval at point 3
ah i saw X as A
It's not quite right
@@kisatsronal256
I'm having fun with these short lectures! I'm super curious to see how far this will go into the topic. Congratulations on your ability to explain these concepts so clearly!
These concepts are a nightmare when self studying. You made them a little easier to understand. Thanks man !
0:30 if you fix a point x in a matrix space, you can look all the other points that have the same distance from the point x
0:44 the notion of the ball we want to generalize in a metric space
1:20 the epsilon ball is never empty
1:40 open set
3:15 boundary point
4:58 open set A is the set where all the boundary points of A do not belong to A
5:10 closed set define using the above logic
5:29 the definition of closed set
5:52 closure
6:33 example : X is a subset of real number
6:55 considering subset of X
I am afraid the statement at 4:58 is not true. The example explains exactly this. The boundary points can be part of the open set as long as the ball at the boundary dA does not include elements of the complement of A, A^c. A set can be both open and closed.
But boundary points are clearly defined :)
I'm a Korean student, and recently started studying college level mathematics for double-major. This is the part I've been having a hard time with these days, and this video was very helpful. Your other videos are very cool, too. Thanks alot! :D
I'm just curious; what are you majoring in, and how have your studies been going?
The last example of closure was very enlightening to me! Thank you and, please, continue doing examples!
If you want to work just in the real numbers, my real analysis series might be helpful. See here to learn about open sets: ruclips.net/video/Wqo4Svs4erw/видео.html
Dang that example with X := (1,3]U(4,∞) and A := (1,3] really threw me for a loop. If I understand correctly, the point x = 3 is *not* a boundary point because its epsilon-ball when d < 1 lies entirely in the region that is less than 3, since the region (3,4] doesn't exist in the set X. Hence the epsilon-ball is entirely contained in A, showing that x = 3 isn't on a boundary but rather is in the interior of A. Now that I (think I) understand that concept, it is *very* cool!
Man, these videos are ballin!
Yes, totally correct! This example shows why "boundary", "open", "closed" and so on are always to be considered in the respective context. Here, the whole metric space has to be known.
@@brightsideofmaths This lecture was very illuminating. Sometimes intuition can lead to things that turn out to be very un-boundary-sonable. 😎
3 not interior point
@brightsideofmaths
Would this be better stated as A is open with respect to X? Because with respect to the real numbers R, the set A is neither open nor closed.
The most important lessons of this great video are the followings:
1) closeness and openness are not mutually exclusive concepts
2) closeness and openness must be checked independently
3) the same subset A can have different properties depending on the structure of the set X it belongs to.
6:30 In the example, A is a subset of X and 3 is an element of A, so 3 is either a boundary point or not of A. If 3 is a boundary point of A then any open ball centered at 3 will contain elements of A and elements of X/A. In particular, a ball of radius 1 centered at 3 will only contain A and (3,4]. Since (3,4] is not in X/A = (4, infinity), then 3 is not a boundary point of A.
Since none of the points of A are boundary points of A, we can say that A is open.
Also, since the complement of A is open, then A is closed.
I think that you could use a ball centered at 3 with any radius < 1 and come to the same conclusion.
I think the confusion was in considering the complement of A not as X/A = (4, infinity) but as all real numbers outside of (3,4]
spent about an hour watching this 10 min videos. But worth spending time. all concepts clear.
Nice :) Thank you very much! You can download additional material with the link in the description.
Am really enjoying this lecture. I can't believe after 5 years I can still enjoy maths like this.The professor i very clear, he is more of a teacher than a lecturer.
I am a Russian student and I found your videos really helpful and useful. Thank you !
Thank you very much :)
How have your studies been going?
The important subtlety in the last example is that A would not be an open substet X if X were to be the whole real line R. It is sometimes confusing because we tend to forget that our set X actually has a hole in it. in another word {3< x
I love your lectures on measure theory and analysis. This also helps me to understand the concepts clearer :-)
Thank you!!
You're welcome! And thank you :)
Very clean presentation. Thank you!
Really love the video! Now one of my roadblocks is solved. I remember watching a Sith lord of the Dark Math teacing differential geometry on YT, he also said a set can be both open and closed, but only gave a very short explanation that I wasn't able to grasp. At that moment I thought I didn't deserve to wield the Dark Side of Math. Now the Bright Side hath shewed me the truth, and I believe I belong here, for now.
9:10 A is also closed, is it because that the complement of A is open?
The last example is very helpful, thank you
You are welcome!
Thank You for great lectures!
I have a question about:
5:05 " A is open A ∩ ∂A = ∅ "
5:22 " A is closed A ∪ ∂A = A "
I think there should be implications only in one direction:
A is open
Thanks for the question. You are missing the definition of the boundary points ∂A :)
@@brightsideofmaths If A is both open and closed, it means that simultaneously "A ∩ ∂A = ∅" and "A ∪ ∂A = A", which is possible only if ∂A = ∅. Am I correct?
this is super helpful, really neat video and clear explaination
Glad it was helpful! You are welcome :)
I’m really excited to see more functional analysis videos. I have an exam coming and you’re really helping me out. Could you maybe make a video about operators (why boundedness implies continuity, Neumann series and investable operators). Also something about properties of infinite dimensional spaces like the theorem of Arzelà-Ascoli. The most difficult topic, so I find, is however the Sobolev spaces (although it might be too early). You’re a great teacher so please continue with this playlist haha.
How did your exam go?
Incredible well explained! Love it! I'm thinking about what you said, points in A that are boundary points but are inside A, I always think of boundary points as points that are not inside A. I'm going to think about it
Could I ask a question?
On Example (a), you have proven that 'For x=3' B_1(x) belongs to A and that A is open.
Epsilon was assumed(?) 1, so it turned out that it doesn't contain 4 in set X.
How does it make sense that the result from assumed epsilon could define the general characteristic of a subset A?
+)
I could not understand that openness and closedness is not an opposite pair of notion. Into which video could I check?
Exactly
For around 8:20 why is it (2,3]? wouldn't it be [2,3] because a distance of less than 1 off of 3, would be 2 and include 2?
The distance should be strictly less than 1. Therefore, 2 is not included.
@@brightsideofmaths That makes sense! Also, did you choose an epsilon value less than one for the case where x=3 because, if you put x = 3 into the original epsilon equation you used for when x≠3, you'd get min (2,0) x 1/2 But epsilon needs to be above 0 so you pick 2, and half it for one? Is that why you chose one? or did you just randomly pick it?
Hope you continue uploading this videos as soon as possible c:
Thank you for these lectures! I have a question about last example (a): if we consider X:=R instead of X presented in the lecture then A won't be open, right?
Thank and you are very welcome! Of course, you are right: A is not open in R.
I'm a tiny bit confused by the last statement. Wouldn't C_bar equal (1,2) because 2 has been shown to be a boundary and not included in (1,2]. The compliment of C would be (2,3],(4,inf)?
A boundary can lie in the set C or not! Be careful :)
@ 7:56, while defining B1(x), why are you considering all the points 'y' in 'X', shouldn't it be all the points 'y' in 'A'?
B1(x) is the open ball in X
I love this channel!
I have a question with your example at the end with the subset A=(1,3] : you mention this set is closed but the intersection of the ball of center 3 and radius 1/2 (or less) with the complement of A is empty, so doesn't that contradict the first definition ? Interestingly however it is true that the complement of A is open, so aren't these two definitions inconsistent ?
Ah my bad, that's the definition of boundary points
Such amazing and detailed videos
I enjoy this lecture. thanks!
Hiiiiii I didn't understand the explanation at 7:50 if you take the point x = 3 and you take an epsilon ball around the point it will always exceed so the space is not open? And so if we follow this logic then by the same process then the complementary is also an open and therefore I am lost ^^
Just bare in mind that the same set can be open and closed at the same time (such sets are sometimes called "clopen"). It's a bit counterintuitive, but there is no contradiction here.
I think it's because there's no X defined between 3 and 4, so even though the ball on 3 will always go to 3.1 or 3.001 or 3.00001, we ignore it because it's outside A, but not in X. Because X is only defined again at values greater than 4.
It's like there's nothing between 3 and 4, so we don't count the ball there. Nothing exists there, the universe we are in is defined by X only.
I'm not sure about this explanation but is what I concluded trying to understand.
Douglas I understand what you mean that s mean this set is not open on R but He is on X ^^ ty !
Evgeniy Kaptsov yeees i never seen set like that because I always work in Rn i think , ty :p
ZirTaaah The whole set and the empty set are trivially clopen. Let R^n be a metric space than the empty set is open because you can choose an arbitrary episilon. The complement is R^n itself and this is open too. So the empty set has to be clopen. You can argue the same way for R^n.
When we have a topological space we call elements of the topology open sets as well? Is it the same notion of openness as shown here? I mean I don't really see how openness makes sense when we aren't considering R^d or some subset of R^d.
Also, do we need a metric to define open sets or are there other ways?
You can induce a topology from any metric by defining the open sets as shown in this video. But not every topology has a corresponding metric. Therefore topologies are a more general construct, but not as useful for more specific applications. The good thing about metrics is, that they already have a notion of "distance" which you need for analysis.
Edit:
As far as I know topological (which basically means ad hoc) and metric (by the given definition) spaces are the two only general ways to define open sets, although you can fand equivalent expressions, which may require special circumstances.
Like you can derive a topology from any metric, you can also derive a metric from any norm (by saying d(x,y) = ||x - y||). With this in mind you can define metrics on function spaces (for example with the sup norm for bounded functions) and consequently have open sets. In doing so you can mathematically make sense of "openness" on this abstract infinite-dimensional vector space though I wouldn't know how to greatly visualise this.
@@sven_lu_ Thank you for replying, your comment did provide some insight. I'm guessing the term "open set" originally came from using metrics to induce topologies and then stuck around as the general name of elements of the topology.
Thanks a lot! Great, very clear explanations
I understood how A = (1,3] is open but how is it closed? Shouldn't a closed set contain all its boundary points and 'A' doesn't contain '1'
The important thing is the definition of "boundary point" in the general context.
@@brightsideofmaths The definition of a boundary point for A reads: a point around which there can not be an epsilon ball small enough, such that the epsilon ball only contains points of A. A simply does not have boundary points. So the closure of A would be A again. Is this proper?
Closure of A = A union boundary of A
So this works :) @@janniserhard2810
@@brightsideofmaths thanks for the answer :)
Great video! Would you have any example of subsets that are neither open or closed? It would help give me more intuition. Thank you!
Consider the reals with usual metric.
Consider A=(2,4] and try to see that A is neither open nor closed.
Very nicely presented lectures. My only comment comes where the use of the phrases "inside of A" and "outside of A" where boundary points for A is concerned--- Incorrect is the statement a boundary point of A can be "inside of A". Generally, we associate "inside of a set A" with the topological "Interior of A" and we similarly associate "outside of a set A" with topological "Exterior of A". Thus, a point x is a boundary point of A if every open ball about x meets both the exterior and the interior of A in points other than x. In my opinion, a boundary point x of A may set theoretically "belong to A" or not "belong to A" would be proper.
Totally agree! I try to be careful with the words in future :)
In the first example, If A is a semi closed interval then at the point x=3 the open ball would not be a subset of A as it would contain points outside of A, points greater than 3 so how can it be an open set? As open set needs to have open ball of every point inside it?
There are no points between 3 and 4 in our space X :)
@@brightsideofmaths right i got it thanks ♥️
I really enjoy the way you teach math. Now I have strong interest in learning functional analysis more seriously. Could you recommend some textbook ? I hope those books are not just presenting definitions or theorems, but also giving some interesting examples and incentives of how those math object are discovered or developed. Thank you!
I can really recommend Rudin's book about functional analysis. However, there are a ton of good books and free lecture notes about the topic. Maybe you just try some of them.
@@brightsideofmaths Thank you professor. You make a lot difference to my math world.
@@Peter-xc1zo I am happy to help :)
What does the boundary represent in a discrete set? Is the boundary the set of number in the set ..? Many thanks
For a discrete set it means the same by definition. And yes, you get that all the points of the set are also in the boundary.
I cant uderstand why. Not all open balls will have others point. just for big ones. If the natural line is the metric for example. for each x, only epsilon bigger than one will have other point. i cant understand why it is boundary. To be boundary dosent need that for all balls need to have other point outside of the subset and also inside of the subset ?
was the radius for the epsilon-ball in the last example where x = 3 chosen to be equal to 1 for any particular reason? Why not radius 1/2 or 1/10 ?
I just want to keep it simple :)
Of course, also smaller numbers will do the trick.
I have a question, hope you help me soonest
Why 1 is not a boundary point of C (the last example)?
1 simply does not exist in our universe (in this example). The metric space X does not know anything about 1.
Oh thanks, I'm clear!
I felt so confused about the example (1,3] in this video. This video says (1,3] is open and also closed. But my lecture notes told me such interval (a,b] is neither closed nor open. Could someone explain this to me? Thank you!!
In the example (1,3], if x=3, B_1(x) does not belong to (1,3] totally, the part (3,4) is outside (1,3], so I couldn't understand why you said this example is open.
@@Sarah-bn7nj The sourrunding space is important. "Open" is always to be understood with respect to a metric space (or in general: a topological space).
Hi, I have a question. In your example, A:=(1,3], it is both open and closed. Does it have a boundary? If it doesn't have a boundary, can we still say it's closed?
Maybe you know the definition of boundary points?
@@brightsideofmaths According to my understanding, given X:=(1,3]U(4, ∞), A:=(1,3] is closed with no boundaries. Is that correct? It's a bit counter-intuitive though...
Hello at the example A which is an interval from (1, 3] when x is 3 it is can not be open, why you said it is open?
Openness depends on the metric space. That is important to know!
Very helpful, thanks
Glad it was helpful!
Using the definition of Closed set that its complements is open, in this case, Ac is open right?[ (4,inf)]. But using the definition of closure, that a closed set has the interior and boundary points, i.e. the closure is equal to the set. I cant see the A having its boundary point, because when I apply the definition of boundary point on 3 it dosent work, because not for all open balls i will have point outside and inside of A, just for balls bigger than one. Could you help me with it? Thanks !
Is it correct that C would not be closed in the example if the first value of the interval is any bigger than 1? So e.g C=(1.001,2] is not closed and the closure would be [1.001,2] with the new boundary point being 1.001? Thanks so much, very insightful!
Yes, you are correct :)
@@brightsideofmaths Danke!!
Examples are damn good!!
thank you so much
I'm not sure I'm thinking about this correctly, but in the definition of a closed set, must it be assumed that X has no boundary? Otherwise it seems that both A and X could be closed, due to each of them having a boundary.
In your last example, if epsilon is 1.5, then the open ball (1.5, 4.5) belongs to set X but does not belong to set A. So A is not open? Or did I misunderstand something?
If the set (0,1) is open,
Does there exist a positive real number d such that:
[0+d, 1-d] = (0,1)
I appreciate any answers to this question.
no
so how do we decide on an epsilon? Couldn't you just pick any eps so small that it eventually works? Or is it just that, that eps is defined as some arbitrary very small number?
You ask about the definition of openness?
@@brightsideofmaths ah yes, sure!
@@Epiphonication Okay! You just have to find one epsilon. That's all :)
(Of course, there are cases where no epsilon would work)
@@brightsideofmaths aaah alright, I see. Thank you! Your videos help me a lot!
I dont quite get how our A = (1; 3] u (4, +inf) is a closed set? Is it buecause the contradiction of A is a subset of X equall to (4; +inf) which is clearly open and hence by definition A must be closed?
Yes, complements of open sets are closed.
@@brightsideofmaths ahh yes, a complement not a contradicion. I confused words. Thank you so much for the answer!
About example a). By definition a set A is open if for EACH x from A we can find a ball. In the example (1, 3] the 3 IS included. So why I can't chose 3 and try to find a ball? Isn't 3 is boundary point? Isn't A is closed?
I think I understood. A is a subset of X, and ball is subset of X. Everything that outside of X is in vacuum and not counted. It's so difficult :'/
I really love your lectures!, I have a question on this one, so If I have disk on R^2, the disk is an open set but also a closed set because I can extract a ring from it as a boundary, creating another open set and a boundary, and os on, am I right?
Thank you! The answer for your questions depends completely on the chosen metric space. What is X is your case? And what is d?
@@brightsideofmaths so if X is all the points like (a,b) where "a" and "b" are real numbers, and d is the euclidean distance. Not sure If expressed myself correctly, I'm learning math as a hobby.
Thanks in advance for your response.
@@AndresFH7233 If X = R^2 and d the normal euclidean distance, the *subset* A given by your disc can't be open or closed at the same time. For example, it is open if the circle line is excluded from A.
@@brightsideofmaths do you mean that the perimeter of the circle is excluded from the set A am I right?
Can an epsilon ball be formed around 3 as the center, when at 3 as the center will the ball still be in the interval (1,3]. Will the ball not get out of the interval hence not open?
I think that the answer is no because outside of 3 towards 4 is an interval which is not part of the metric space X in which A is a subset. If A was a subset of R instead then it would be closed. At least that's what I got from.the last example.
I have one question , in the book "Introduction to topology by Simmons" it is mentioned that the interval X = (0, 1] is an open set, if it is given that X in itself forms a metric space (X,d). So if we consider X as a subset of Real line, when real line forms a metric space with "usual" metric, it is NOT an open set.
But if it is given that X itseld forms a metric space, it is open
So is the concept og being open or not more abstact than we define? i mean removing bounday points?
I mean , as far as Iearned, X here is defined open as there is no point even talking about points like "0" as metric space (X,d) is not even defined for them...
What is this sorcery ... please explain
"Open" is always only meaningful with respect to a space. "(0,1] is open" does not mean anything if the context is not clear.
@@brightsideofmaths
So, indeed if X is defined as a metric space, no matter if it is (0,1] it can be open? W.r.t to some metric d....hmmm....now I understand why the definition of open set with respect to a space, its complement etc is so important than simple an Open Ball one.
There is a theorem as well, that whole set of a metric space, along with empty are open sets.
@@brightsideofmaths agree, but anyone after seeing a semi closed interval would think it is not open.
Infacf most text books and videos teach students that open sets are nothing but generalizations of open intervals on real line....
@@paulhowrang Open intervals are already very special open sets. Any union of open intervals is an open set :)
@@brightsideofmaths that is actually not difficult to comprehend, arbitrary unions and finite intersections are closed....I mean that is why empty set has to be open...otherwise finite intersection would not even work 🙄
Hello friend, (A) intersection (boundary of A) is empty iff open and (A) union (boundary of A) = A if closed. Is this always true or only for the standard topology ?
Standard topology in which space?
@@brightsideofmaths I mean is this true for for any topological space (X, T)
In every topological space A union boundary of A is closed.
NIce, Make more lecture on geometry, Riemannian geometry, fourier series computation
So yours next video will be ??
@@47lokeshkumar74 don't you know it's more elegant to write "please" in your sentence?
@@guilhemescudero9114 yes please sir
great videos. can you make videos on measure theory
Yes I can! But do you know my measure theory series? You find it here: tbsom.de/s/mt
I have a question if you don't mind. Is the example you gave with the set (1,3]. Since you said that A is an open set, then 3 cannot be a boundary point, as all boundary points have to lie outside of A for A to be considered open. You also said that the set A is closed. Since A has no boundary points, but it is closed, then a set can be closed without any boundary points? Is my reasoning correct? This seems to go against intuition.
Yes, the set can be closed without having any boundary points.
@@brightsideofmaths Thanks. Is it correct to say that: a set can only be open and closed at the same time when it has no boundary points?
@@usefulknowledge6074 If a set is open and closed at the same time, the boundary of the set is empty. :)
Isn't the definition of a closed set , "a set that contains all of it's limit points" ? where by limit point we mean a point such that any open ball of that point when intersected with the set minus the point itself , it will be non empty ? So interior points are limit points too , but if all points are interior points then the set is open etc etc
Yeah, this is also possible but I didn't want to go into the details of "limit points" at the beginning.
@@brightsideofmaths thank you.1 more question. There is a theorem in Rudins Principles of mathematilca analysis that states "a set is open iff its complement is closed" here we said that A is open , and its complement is also open. Does that mean that both of them are also closed?
@@DeadPool-jt1ci That is exactly our definition, isn't it? A set is closed if the complement is open :)
Thankyou sir for the video
Only the empty set and R are both open and closed, so I think the last part of the video is wrong...
Can you explain?
We are working in a different metric space :)
That would be great, If I fully could understand English :) but thank you very much !
Great value
Can you explain why A is closed please?
The complement of A in X, (4, inf) is open, so A is closed. Alternatively, the boundery of A is empty so 𝛿A∪A = A.
@@kristianthulin thanks.
(1,3] is neither open nor close
with respect to which metric space?
@@brightsideofmaths I am glad that I have looked through this comments. You are totally right in saying, that the set is opened and closed. But I think it would have been a little better to start with an example from the real line and then go to these special cases. But it is very good that you discuss these special cases, these namely make the difference. By the way, very well explained. Thanks for your great effort!
@@lvlanson Thank you for you comment! I have a whole series about real analysis, where one focuses on the real number line. Here, we have to go to the abstract case immediately. Therefore, I started with these special cases :)
@@brightsideofmaths This sounds absolutely reasonable. Maybe it would be good to add like a link into the video linking to the real analysis part concerning this video. This might help to get a better grasp on it. Anyhow you do it, it is well done. These are only some points to grind out the fractions of perfection.
@@lvlanson Thank you. I am always happy for suggestions to improve future videos!
I don't think when x=3, x is an interior point that belongs to A. B/C there exists NO SUCH B(3,ϵ) s.t. B(3,ϵ) ⊂ A
See here is something I never really believed, and I guess this reveals my insecurity about the real numbers being densely packed; we can just arbitrarily choose any ol' epsilon we want in order to guarantee that a set is open. . . seems fishy to me.
It's not about believing. It's about your axioms for the real numbers.
@@brightsideofmaths too true, I find myself far too often debating believability of claims rather than attempting to see how the axioms support the ideas.
I need help with my functional analysis exercises...Can I have your email?
See description :)
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The Bright Side Of Mathematics what kind of topics i must to prepare for my LA II Klausur? 😁
@@easystudy2925 I don't know.
The Bright Side Of Mathematics schade😢
@@easystudy2925 Probably, you should prepare all the topics you learned in your lecture.