 We looked at last time, the real number system, we looked at what we called as the nested interval property. Some more concepts about special subsets of real line, they are also common to higher dimension spaces. So, let me introduce briefly R2, R3 and in general, Rn. I think most of you are familiar with R2 and R3, but the same thing works for Rn also and we will come back to it later on also. So, the basically idea is that say for example, R2 or Rn is a sort of all n tuples where each component or each coordinate is in real line. So, you can think it as a vector with n components x1, x2, xn. There is on Rn, there is addition. So, what is addition of two vectors? That is a component wise addition. Scalar multiplication, you can multiply each component by the same scalar. So, there is scalar multiplication. There is no multiplication of vectors as such in R2 or Rn. Like in real line there was, there is notion of a dot product and cross product will not be using them and there is no order on R2, R3 or Rn. You cannot compare, you cannot define an order between vectors, one vector bigger than the other vector. So, there are problems about that. However, what you can do is like absolute value, you can define what is called the magnitude of a vector. So, for a vector x with components x1, x2, xn, we define what is called the magnitude of the vector or the norm of the vector, that is, sigma absolute value xi square raised to power 1 by 2 i equal to 1 to n. In R2, this is just the Pythagoras distance, right, right angle triangle x1, y1. And this has properties which are very similar to the absolute value function. So, properties, this is always bigger than or equal to 0, equal to 0 if and only if each xi is equal to 0. So, that is corresponding to the absolute value. The second property is, as far as scalar multiplication is concerned, alpha times x is equal to mod alpha times norm of x. So, that is scalar multiplication. How does this magnitude behave? And the third is the triangle inequality property, namely absolute value of x plus y is less than or equal to 0. So, basically it behaves very much like the notion of absolute value for real numbers. So, as a result, it gives rise to a notion of comparison between or the distance between two points. One point is close to another. So, we define norm of x minus y, right. So, you can call it as the, you can define this as the, you can call this as the distance between x. So, basically the idea is that the notion of limits of sequences, right, and notion of convergence of a sequence depended upon the notion of distance, right. We said a sequence xn of real numbers converges to a point x if xn is coming closer to x. And that closeness was measured in the term of the distance. Now, we have a notion of distance available in R n. So, we can consider notion of sequences in R n and notion of convergence of sequences in R n. So, let us just write because that will be useful. So, let us write definition of a sequence and ordered collection points xn and bigger than or equal to 1 is called. So, we will put a underscore to indicate this is a vector, ok. It is called a sequence n. As usual written as, we write it simply as xn, same notation n bigger than or equal to 1. So, let us say definition 1, definition 2. We can define now convergence a sequence xn is said to converge to say a vector a belonging to R n if, what should be definition? The distance between xn and a becomes smaller and smaller. That is same as saying norm of xn minus a goes to 0 as n goes to infinity. Now, keep in mind this quantity norm of xn minus a is a number, is a real number. So, for every n you are getting a non-negative sequence of real numbers, is a sequence of numbers which are non-negative and real. So, we can ask whether the sequence converges to 0 or not. So, this part, this is precisely convergence of real numbers. So, we are using convergence of real numbers to define convergence of vectors in R n. So, keep that in mind. Now, here is a small observation, we cannot write it as a note. So, supposing our vector, let us note something namely x minus y, if I look at the distance between two vectors x and y, this is always bigger than x has got components, y has got components. So, let us say x is x1, x2 and so on, xn and y has components y1, y2, yn. Then this is always bigger than or equal to xi minus ya. Is that okay? Because what is the distance norm of x minus y, that is xi minus ya absolute value square raise to power 1 by 2. So, that summation is 1 to n. So, it is always bigger than each term and this is, is it okay? So, this implies that if xn converges to a, then I am using n. So, I should probably, because I am using n for R n also. So, let me write xk, if xk converges to a as k goes to infinity, then this is same. This implies, then this implies that look at x vector k. Let me write this as i. What is that? That is the ith component of the vector xk will converge to the ith component ai for every i between 1 and n. Is that okay? This inequality just now, we are saying that if a vector is convergent, this sandwich theorem mod of xi minus yi is less than this. So, if this becomes smaller, then this becomes smaller. Sandwich theorem, is it okay for everybody? If not, get clear. So, let me write here, because a is bigger than xk ith component minus ai. So, each component distance is dominated by the distance between the vectors. So, sandwich theorem implies that if that goes to 0, then this goes to 0. So, what we are saying is, if a sequence of vectors is convergent, then each component converges to the corresponding component. Now, let us look at the converse of this. Suppose the ith component converges to the ith component for every i. Can I say that the vectors also converge? Clearly yes, because what is this distance between xn and a? That is summation of the distance is summation here now, if it is quality. Each one is going to 0. So, sum goes to 0 by the limit theorems for real numbers. Is that okay? Because norm of xn minus a is equal to summation i equal to 1 to n norm of x, I am writing k here. So, xk i minus ai or not norm is absolute value because the real numbers square raise to power 1 by 2. That is the definition of distance. Here, each term is going to 0. So, square goes to 0 by limit theorems. The summation goes to 0, non-negative square root goes to 0. So, by the limit theorems for sequences of real numbers, this for each i goes to 0 implies the distance between the vectors also goes to 0. So, what we are saying is, a sequence of vectors is convergent if and only if each component converges to the corresponding component. So, let us write it as a theorem. If you like, n converges to a if and only if each component sequence is convergent to where? To the corresponding component. So, as far as convergence of sequences of vectors is concerned, it is same as analyzing convergence of each component. So, no, not a big deal. So, that is okay. So, now, we will be using this concept to define something. So, let us start with, let us take a set A contained in Rn. A is a subset in Rn. So, given a sequence in A, a sequence so that each element is in A, supposing it is convergent, then the limit of that sequence may or may not belong to A. So, we are trying to now specialize special subsets of Rn. So, we say definition. A subset A in Rn is said to be closed. It is called a closed set if for every sequence xn belonging to A, xn converging to A implies A belongs to A. So, what we are saying is a set which includes all limits of sequences of its elements. We will say it is a nothing goes out kind of thing. So, it is called a closed set. The normal colleture is very clear. So, let us look at some examples. So, let us look at A belonging to Rn and the set A is a singleton A. Is this set closed? Is the set consistent of a single point closed? Obviously, because the only possible sequences that we can consider of elements of the set are the constant sequence and constant sequence converges to the constant and that belongs to A. So, A is closed. Let us look at when n is equal to 1, some special cases. So, A is equal to that is a subset of R. An open interval. The name itself says open interval. So, it should not be closed. But anyway, that is not the reason. So, if you want to prove this set is not closed, then what should I prove? There is a sequence of elements of the set. It converges somewhere, but the limit is not inside that set. So, precisely if this is my A and this is my B, I can take a sequence sort of coming from the right side to A. The limit will be A, which is not inside the set. So, what could be such a sequence? So, you can look at A plus 1 by n, but n suitably large enough so that you are inside. So, here is A plus or A plus 1 may be outside. B may be bigger than. But anyway, let us from some stage onwards, A plus 1 over n will be inside the interval A B for some n large enough. So, this will converge to A, which is not part of A. So, this is A is not closed. For the same reason, you can take as A B. It does not matter. 1.1 counter example is good enough or you can take A B. These are all or you can even take A 2 plus infinity and minus infinity to A are not closed. All these are not closed subsets. Obviously, if I include here, can I say A B is closed? Why is that closed? It is not because a sequence converging to A, that means A is inside it. To say it is closed, I have to show for every sequence in the set. If it converges, that point limit must be inside the set. So, let X n belong to A and bigger than or equal to 1 and X n converges to some x. Then what can I say about these X n's? They are inside the set A and A is the interval A B. That means this is true for every n. Is it clear? Because it belongs to A. A is the interval A B, which is closed interval. So, A less than or equal to and that implies, if it converges, X n converges, what can you say about the limit? If X n's are bigger than or equal to A, the limit cannot go below A. We have seen that. So, A less than or equal to X less than or equal to B. But if they were open, if it was a strict inequality, then the limit could become equal to A. That we have seen in the example. So, then that will not be true. So, less than or equal to is okay. The limit can become equal to, even though each A n is strictly bigger than A, the limit can become equal to A and that is okay for us, because we want A less than or equal to X less than or equal to. So, hence, A which is equal to A B is closed. Let us, similar thing probably will be nice to, okay. Let us define something which is going to be useful for every X belonging to R n and epsilon n number B X epsilon. I am going to define a set. This is all vectors Y belonging to R n such that the norm of X minus Y is less than epsilon. All points, all vectors in Y in R n such that it is distance from X. X is a point which is fixed. So, because we are looking at Pythagorean distance or Pythagorean way of looking at. So, if I take it R2, it is very easy to visualize. Here is the point X. A point Y will be inside if its distance is less than at the most epsilon. So, it will be, right? It will be all points inside what we visualize as the ball. So, you can think it as, so this is a visualization of this thing, right? Sometimes it is good. But if I look at a point, geometric kind of thing on the boundary, then its distance will be equal to epsilon. So, that will not be a part of the set. So, that is why it is called an open ball centered at X radius. Geometrically, it looks like that. So, we call it as an open ball centered. Let me also define, because we are defining this thing, let me also define B bar X epsilon to be the set of all Y to the distance X minus Y is less than or equal to, then we are also including the geometric boundary kind of thing also, equal to also are included. So, let us look at, can I say this is close? So, which of them are close? Can I say the ball X epsilon is close, is a close set? No. So, what that means, you have to construct an example of a sequence. So, I leave it as an exercise. It is not difficult to, just intuitively is very clear, right? You can go towards boundary kind of thing, take a sequence going towards boundary. Like we did for the open interval a, b, something similar you can do. Slightly, we have to do slightly more kind of, be slightly more careful about, because there are many possible directions there. In real line, only one left or right, here are all possible directions. So, but you can go any direction. So, what about this thing? Why this is a close set? Now, I have to prove that for every sequence, like in the interval close interval, for the same proof works basically. If I take any sequence X n in this close ball and converging to a, which is, that means what? The distance between X n and a will become, X n goes to a. So, what is the distance of X n from X? Because if I want to show it is close, I should show that the distance of the limit from X is less than or equal to epsilon, right? But the distance of each point X n from X, because X n is inside is less than or equal to. So, can I say that, so here is something. I am giving a hint. It is true. If X n belong to ball X epsilon and X n converges to a, then X minus a is less than or equal to epsilon. So, that is what is to be done, right? And what is given? Because X n belong given norm of X n minus X is less than or equal to epsilon for, sorry, X n, yes, is less than or equal to for every n, right? Because X n belong to, X n belong to the ball, close ball. So, less than or equal to. We want to show that this implies, so that is the exercise, okay? Just if I y norm of X n minus X implies that, is it okay? Idea is similar to that of the close interval a, b. If X n is in the close interval a, b, then X n minus a, right? They remain, this is the order here. Here is the order of the distance actually. So, try that. So, that will also prove, modulo this part of exercise that this also is a. So, by the way, this kind of things, okay? Let us give it a name because they are going to be useful. For every point Y belonging to ball centered at any X and radius epsilon is called, okay? I should write it properly. Every ball, every open ball is called a neighborhood of Y belonging to B at epsilon. So, what we are saying is, if it is an open ball, if it is an open ball of radius epsilon centered anywhere, it does not matter. Then for every point Y inside that ball, the ball is called a neighborhood of the point Y, right? The ball is called a. So, that mean for this Y could be anywhere. So, here is a point Y or it could be anywhere Y here. For all elements of that open ball, this open ball is called a neighborhood of that point. So, obviously, it is a neighborhood of the center also anyway, right? So, it is something similar to, let us look at what happens in the real line. So, in the real line, if you have got a, so here is point X. What is an open ball of radius epsilon? So, it will be X minus epsilon and that is open interval, simply, right? So, what we are saying is for every point inside, this is a neighborhood, okay? It is a neighborhood of the point Y or this is ball, okay? So, I think I will elaborate it slightly more later on. So, it is a concept of neighborhood for everybody and open ball is neighborhood of every point inside it. That is the definition, okay? So, now, let us look at closed sets. So, we have looked at many examples of closed sets. So, we have seen that there are sets which are not closed and there are some sets which are closed, right? So, can we do one thing? Take a subset of Rn, some of the limits may go out. Let us throw them in and make a bigger set. So, what do you expect of the bigger set? The bigger set should have the property that it contains limits of all sequences in that, right?