 So, let us define such a thing. So, definition, A subset A in Rn is said to be closed. No, no, we have already defined closed. So, let us, for a subset A closed, we have already defined. Let us define A bar to be the set of all points x belonging to Rn, such that there exist a sequence xn in A, xn converging to x. So, we are given the set A. Look at all possible sequences of elements of A and their limits. So, all those points in Rn which are possibly limit of a sequence in the set A, collect them together in a box and call that box as A closure. A bar called, so let us give it a name called closure of, so it is called closure of the set. So, we are given a set A. Look at all possible sequences of elements of A. Look at their limits. So, all of them put together, all the limits put together, put them in a box, call that as set A. We want to study the properties of this set A. For example, keep in mind, say open interval A, B. Keep in mind the open interval A, B. If I take a sequence inside the open interval A, B, the possibilities are the limit. What can happen to the limit? It can become equal to A. It can become equal to B. So, positively it will be between A and B. It can be possibly A or B. Is it okay? So, it seems the closure of a set includes the set. Closure of a set is something bigger than the given set. It is enlargement of the given set. And that is so because if I give a point in the set A, then I can take the constant sequence. So, it is converging to that point. So, that point must belong to A closure. Is it okay? So, property is one, an obvious property. A is contained in A closure for every A contained in R. Is that okay for everybody? It is something bigger. We are enlarging. For the points in A, we can look at the constant sequences. They are the limits anyway of the constant sequences. So, they belong to A closure. So, what is A closure? It is all those points which can be put as limits of sequences in A, which can be approached as limits. So, every point of A can be approached by a constant sequence anyway. So, what do you expect now as far as the closeness of the set A is concerned? Can you say A is closed? Because we are just trying to put all the limits inside now. So, claim A closure is a closed set. Before that, let me just write something which will be useful. X belongs to A closure if and only if every neighborhood. So, I will be shortening neighborhood for NBD instead of writing NE, IG, HBR. So, every neighborhood, what shall we call? Say B of X, B intersection A is non-empty. So, let us observe this. What we are claiming is we are trying to give a characterization of the points in A closure, the set A closure in terms of a property of the neighborhoods of that point. So, let us see why is that? So, let us write a proof. So, let it belong to A closure and B be a neighborhood. So, to show what is to be shown, this neighborhood must intersect A. Now, if B is a neighborhood of X, what is X in A closure? That means, there is a sequence of points in A closure. So, there is a sequence by the definition. There must be a sequence of points in A which is converging to X. But if there is a sequence X and converges to X, what happens to the tail of the sequence? It must come closer to the point X. So, it will come inside that neighborhood, right, after some stage onwards. Is that okay? So, let us write that B a neighborhood of X belonging to A closure implies there is a sequence X n belonging to A such that X n converges to A. Sorry, X n converges to point is X, not A. It is the X implies. This is actually true if and only if. When do you say X n converges to X? This is for every epsilon, there exists some n not such that norm of X n minus X is must become smaller than epsilon for every n bigger than n not. So, let me, if you like, let me draw a picture. So, here is a neighborhood B and here is probably the point X and I have got a sequence X 1, X 2 which is converging to X. So, Michael, so this says this less than, so can you say that, see this says the distance of, so here is a, right, for every epsilon, is it clear? This says for every epsilon there is a, what is the meaning of this? That means X n belongs to, so that is same as saying X n belongs to ball centered at X of radius epsilon, right? So, keep in mind the geometric and the analytical thing. The distance between X n and X not and X smaller than epsilon means it belongs to the ball. Now, the only question is I am given a neighborhood B, right? So, what is the neighborhood? It is some open ball centered at some point. We do not know where. So, can I say that for every neighborhood of a point, there is a small ball inside the neighborhood. That is the only thing required, right? Is that okay? Given a neighborhood, given this neighborhood, okay, and a point inside X, can I say there is a epsilon such that the ball centered, see this ball, bigger ball may not be centered at X. We do not know where it is centered at. It may be centered at some other point, say X naught. But what I am saying for every point inside it, I can have a small epsilon, say that there is an open ball which is inside the neighborhood. Is that okay for everybody? Yes, requires a bit of writing down. So, let me write this also as exercise, write down, because geometrically very obvious. If the point X lies inside, then its distance from the center will be strictly less than, it will be strictly less than, right? So, you can always go very close so that all the points are inside. So, that is saying, implying that the ball intersection A is non-empty. Is that okay? Because all the points X n's will be inside the ball now, after some stage onwards. After some stage onwards, all the points X n's will be inside the ball of radius epsilon at X. But that is inside the bigger ball B, bigger neighborhood. So, there will be points X n's of A which are inside B. That is same as saying the intersection is not empty. Basically, saying that convergence means it is coming closer, right? So, it must come inside a ball centered at that point, very, very small. So, epsilon is arbitrary, but this is happening for every epsilon. So, you can always choose epsilon suitable enough so that the ball centered at X of radius epsilon is inside B. For that ball, everything is okay. Yes? Is it clear to everybody? So, what we are saying is a point X belongs to a closure implies every neighborhood of that point X must intersect A. Let us look at the converse part of it. Suppose, every neighborhood B of A intersects claim every neighborhood B of, sorry, of the point X of X intersects A, claim X belongs to a closure. If every neighborhood of the point X intersects the set A, we want to claim that point X is in the closure of A. So, what is to be shown? To show X belongs to A closure, what is to be shown? What is the definition of A closure? I have to produce a sequence in A, right, which converges to X. So, there exists some X n belonging to A such that X n converges to X. And when will X n converge to X? Now, you see, keep target in mind what you want to show. I want X n in A of course, right, but that X n will converge to A only when the distance between X n and A goes to 0. So, basic requirement is norm of the distance or the magnitude of X n and A, X n and X. So, that is, should go to 0, right. And saying it goes to 0 means what? After some stage, it becomes less than some epsilon, given any epsilon, right. So, let us write. So, given, but I have to choose X n. I have do not know what is X n. I have to find X n such that this happens, right. But what is given to me is for every, this becomes, that means, so let us write what does means, that means, this will become less than epsilon, that means the point X n should be in the ball centered at X of radius epsilon. See, what is given to me? Every ball intersects. What is given to me is every ball or every neighborhood of the point of the point X intersects. So, somehow I have to bring everything to the neighborhoods, right. Then only I can use the given thing. So, what I do is, because there is a sequence X n converging to X, right. This is, this we have to produce and this will happen if this goes to 0 and this can happen if I take this less than 1 over n. If you can choose X n such that the distance between X n and X n less than 1 by n, then I will be through, right. And that is same as saying X n should be in a ball centered at X of radius 1 over n. So, I should specialize that neighborhoods now. So, consider neighborhood ball centered at X of radius 1 over n, n bigger than or equal to 1. We are given that every neighborhood must intersect A. So, in particular, by given hypothesis, the ball centered at X of radius 1 over n, this is a neighborhood of the point X. This is a special neighborhood. We are given for every neighborhood something is happening. So, let me take this special neighborhood. This intersection A must be non-empty and this is motivated by the fact I want a point X n at a distance at the most 1 over n. Then I will have X n converging automatically, right. So, this is non-empty. So, choose X n belonging to ball X 1 over n intersection A. This is non-empty. So, choose any point inside that. So, then X n belongs to A and norm of X n minus X goes to 0 as because this is less than 1 over n. So, let me probably write, make it more specific. It is less than 1 over n goes to 0 as n goes to infinity. So, essentially, there is nothing very great about the proof, but it is intuitively very clear that given a point X, let us take the neighborhood of radius 1 over n. This must intersect A. So, there must be a point X n inside it and make it smaller and smaller. Shrink it so that the sequence X n comes closer and closer to X. This is happening because we are given that for every neighborhood something is happening. So, we are specializing. So, what we have shown is the following that proves that a point is in the closure if and only if every neighborhood of that point intersects a set. Now, let us look at the next property that A closure is a closed set. A may not be closed, but A closure is a closed set. So, what is to be shown? What is given and what is to be shown? It is always very clear. Things become clear when you write down what is given and what is to be. So, proof given A closure, I want to show it. We are not given anything. We are just given A is a set, A closure. So, let, if I want to show it is a closed set, what is to be shown? So, let us write to show. See, there is English A closure is a closed set. What is the mathematics behind it? Every point, what is to be shown to show for every z belonging to A closure, there is a sequence. There is X n belonging to A such that X n converges to z. No, no, no. Is that the property to be shown? Or is that the definition of A closure? What is meaning of z? z belongs to A closure. So, let us go back and just recall, because since looking at various, z belongs to A closure. Means what? There is a sequence that is given to me. So, let us, so this is given. So, let us write, this is not, so let, so let us write, let, so that means, so let z belong to A closure. To show, what to show? A is closed. So, what is the meaning of A is closed? Now look, what is A? A is closed. I want to show that. That means, if a sequence X n belonging to A and X n converges to A, then A must belong to, so what is to be shown here? Yes? What is to be shown? z belongs to, that is given. So, we are given z belongs to A bar. Now, that is not to be shown, right? It is not given. So, let us write. Since, I want to prove A bar is closed. So, that means what? If I just take a sequence in A bar and it converges to some point, that point must belong to A bar. A bar is closed. That means, every sequence in that set, if it converges, the limit must be inside the set. So, let us write, what is A bar is closed? So, that means, to show, if a sequence in A bar is convergent, then the limit is inside. For every sequence in the set, if it converges, the limit must be inside. That is what it says. So, how do you prove it? Let us take a sequence x n belongs to A closure. x n converges to z to show z belongs to A closure. Is that okay? That is to be shown, right? Now, x n belongs to A closure. I want to show this means this. So, that is, every neighborhood that is same as saying every neighborhood of B of z intersects. So, this is same, right? We want to show z belongs to A closure that is same as saying every neighborhood of z should intersect A. So, you want to show the point is in A closure. So, let us fix a neighborhood B of z. Now, what is given to us? x n belonging to A bar converges to z and B is a neighborhood of z. So, what should happen? x n is converging to z. So, x n must come inside B. So, implies because x n converges to z, x n belongs to B, x n belongs to B for every n greater than some n naught. So, x n belongs to B, right? Is that okay? Because x n is converging to z and B is. So, that implies what? x n belongs to B and what is B? Where is x n? x n is in A closure and z belongs to B. So, what does this imply? See here is a very small thing. The idea to be x n belongs to B also x n belongs to A bar. So, that means what? If x n belongs to B, B is also a neighborhood of x n. If x n belongs to B, then B is a neighborhood of x n and x n belongs to A bar. That means what? Implies B intersection A is non-empty. If x n belongs to A bar and x n also belongs to B, right? So, B can also be treated as a neighborhood of x. Now, it was a neighborhood of z, but x belongs. So, it can be treated both as a neighborhood of x as well as a neighborhood of z. So, if I treat it as a neighborhood of x, then B n belongs to A bar, right? So, x n belongs to A bar and x n belongs to B. So, B is a neighborhood of x n. So, it must intersect A because x n belongs to A bar by the definition of A bar, right? Very small point. That is all. x n belongs to B and z also belongs to B. So, B can be treated as a neighborhood of both of x n as well as z. If you treat it as a neighborhood of x n, x n belongs to A closure. That implies x n belongs to A closure implies it must be non-empty, right? So, this gives us hence z belongs to A bar. So, we have proved A bar is a closed set, okay? A bar is a closed set. That we have proved just now, right? A bar is a closed set.