 So, let us start by recalling what we were doing. We were looking at the closed subsets of R n. Then, we looked at what is called the closure of a set and we proved the result namely A bar is closed, A bar is the closure of the set A. It is a closed set and in fact we proved it is a smallest closed subset which includes the given set. Next we look at some more properties of special points of sets. So, let us start looking at what are called. So, we are given a set A contained in R n and we say a point x belonging to R n is a limit point of A. If you say it is a limit point of A, if for every epsilon bigger than 0, if you look at the ball centered at x of radius epsilon minus the point x that must intersect A. So, that is non-empty. So, whatever ball you give at the point x, it should intersect the set A at a point possibly other than x because we are not saying x belongs to A or not. So, if in case x belongs to A, then every point will be a limit point. So, we are not we are looking at points which are such that. So, this is what is called such a thing is called a deleted neighborhood of the point x. So, from the ball, we are removed the center. So, that is called a deleted neighborhood. So, a point is a limit point if every deleted neighborhood of that intersects the set A. Such points are also sometimes limit points. Also, some people call it accumulation points. Let us also define a point x which belongs to A is called an interior point. It is called an interior point of A if there exists some epsilon bigger than 0 such that the ball centered at x of radius epsilon is inside A. So, if you are looking a kind of picture, it says that if this is a set A, then a point is an interior point if there is some radius epsilon such that the ball open ball around it of radius epsilon is completely inside. And we say x is a limit point. So, this is x is a limit point. If I take any ball of radius epsilon, then it should intersect the set A at some point. So, that is called the limit point. Let us have a notation for A0, upper 0 is or denotes the set of all interior points of not limit points of interior points of A. So, here are some obvious properties. Let us look at those properties. So, interior points and limit points. So, a point is a limit point if the open ball, I have called it I here minus the center intersects A for every ball, every open ball, whatever the radius may be. And interior point if there is some ball which is inside a center at x, which is inside A. So, that is the interior, clear? The relation between these things, some properties of some sets. For example, if you take the open interval 0, 1, then every point is an interior point. Because if you take a point x, it will have some distance from 0, some distance from 1. Look at the smaller of whichever it is. So, that open interval will be inside the open interval 0 to 1. If you look at open 0 and close at 1, then 1 is not an interior point. Because at 1, if I take any ball, that will go outside. So, there is no ball centered at 1, which is completely inside. So, 1 is not an interior point, but 1 is a limit point. Because whatever ball I take at the point 1, it will intersect, it will have some point of 0, 1. Look at the set, for example, 1 by 2, 1 by 3, 1 by 4 and so on. So, what do you think is a limit point of this? 0 is a limit point. Because if I take any ball at center at 0, then 0 does not belong to A. So, whether removing from that ball, if I take an open ball centered at 0, any ball because this sequence 1 over n is going to converge to 0. So, after some stage, all the points will be inside that ball. So, whatever the radius of that ball may be, that is going to intersect A minus the center of force. So, that is a limit point. Can you say what are the interior points of this set? Interior point has to be a point of the set, first of all. So, can 1 by 2 be an interior point? No, obviously not, because I can find a whichever open ball I take around 1 by 2, it will go outside the set 1 by 2, 1 by 3 and so on. So, there is no interior point, interior is empty. So, similarly, you can show that set of rationals. There is no interior point for set of rationals. What about limit points? Limit point is a point in real line, so that whichever neighborhood I take of that point, it must intersect rationals and we know rationals are dense. So, that means, whichever ball I take, it will positively contain a rational. So, all real numbers are limit points for the set q. So, let us look at probably one of the last ones. If you take any open ball centered at in R n, I take an open ball, then every point is an interior point. If you pointed out yesterday also, given any point in the open ball, you can have a small radius. I left that as an exercise. So, do it in the problem sessions. That given any point in an open ball, for example, if I take an open ball like this, this centered at some point, if I take any other point, then I can find a small radius. What that radius should be? You have to figure it out. It is not very difficult to figure out what could be that radius. So, every point is an interior point for the ball. What are the limit points? What are the limit points for the open ball? Every point of the ball is a limit point anyway, because whatever ball I take, at any point, it will have points other than the center also inside A. What about the points on the geometric boundary? The distance equal to, if the ball is of radius R, so radius R, all points at a distance exactly equal to R. They are also points of their limit points, because if I take any ball at the boundary, that will come inside also. They have points inside. So, limit points are interior point and these are possibly limit points and of course, the ball inside also are limit points. Look at the complement of the closure. Given any set, I am looking at Rn minus A. What is that? That is the complement of the closure. The aim is, it is related to the interior of something. It is the interior of the complement. So, let us see why it is so. We are looking at the complement of, so let us look at, so let X belongs to Rn minus A closure. What is the meaning of this? It is in the complement, that means X does not belong to, it does not belong to A closure. It is in the complement of A closure. So, what does that meaning say, it does not belong to A closure. What was A closure? It was a set of all points such that if I take a ball, it should intersect A. Every ball should intersect A. That is the closure. Closure of a set were the points such that every ball centered at that point should intersect A. So, if it does not belong to closure, means what? Negation of that segment. That means there exists some epsilon bigger than 0 such that a ball centered at X of radius epsilon intersection A bar is intersection A is equal to empty set. That is by definition of A closure. But what does that mean? This intersection is empty. So, where is that ball? So, where is that ball? That is in the complement of A. So, what we have shown? If X is a point here, this is if and only if X is a point here. What is the meaning of this? That X, the ball at the point X, there is a ball. There exists some epsilon such that the ball is inside that set. That means it is an interior point for that set. So, it says X belongs to R n minus A. Radius set theory, nothing more than that. Definition of the closure. So, definition of the closure says X belongs to the complement of the closure. That means X does not belong to the closure. It does not belong to the closure means there is at least one ball centered at X which does not intersect A. So, that is what we have written. Intersect that A. That means it must be in the closure and that means this is an interior point. So, this says, so implies R n minus A bar is equal to the complement and take the interior. It is not doing much. We are just writing the definition and using the complement property. So, this is what we were saying here. The first one, the complement of the closure is the interior of the complement. If I take complement of this, what will I get? That means A closure is equal to complement of the right hand side. So, that is the second thing. Just by complements I get the second statement. So, a similar statement, if you take a complement interior where complement closure that is equal to R n complement of the interior. Basically same thing, definition. It belongs to closure. That means there is a ball which would intersect A, this set and hence go to the other side. So, just wrote it down and try yourself. I have given you the idea of the proof of the first one. All remaining are same actually. There is nothing much involved in it. It has definition of the closure and definition of the interior. Closure means every ball at that point must intersect that set. That is a closure. Interior means there is at least one ball at that point which is inside the set. So, that is the interior. So, using that all these are easy to verify. First one we have already verified. Otherwise, you can do complements and for example, from here to here, third to fourth, you take complement and you get that thing. Either by complements or by definition you can approve. Now, let us define. We will call a set to be an open set. Till now we have not defined what is called an open set. A set is called open if it is equal to its interior points. Given a set, set of interior points is a subset of it with some property. So, what we are saying? A set is open. The definition is if all its points are interior points, a is equal to a interior. So, if a is equal to a interior, look at this thing. What does it give you? What does third give you? If I replace a by a interior, then right hand side is complement of a. So, what is complement of a? It is equal to closure of that set. If a is equal to a interior, what does the third equation tell you? Third tells me that the complement of the interior is equal to its closure. That means the complement is a closed set because its closure is itself. Is it okay? In equation 3, if I say that a is equal to a interior, what does right hand side mean? Right hand side becomes complement of a. Complement of a is equal to the closure of the complement. Same set. Complement of a closure is equal to complement of a. What does that mean? Whenever for a set, a is equal to a closure, we prove that that is a closed set. A set is closed if and only if a is equal to a closure. So, a set is open if and only if its complement is closed. You get that property. From third you get that property that a set is open if and only if its complement is a closed set. Normally, sometimes it is taken. This has taken as a definition. One defines an open set and takes closed set to be the complement of a. Closed set is defined as the sets whose complements are open. We are doing other way around. We have defined what is a closed set first. How is a closed set defined? We defined what is a closure. The points which can be approximated by approach by limits of points in a. That was a closure. Then we said a is closed if a is equal to a closure. We also proved it is the smallest closed set which includes a. So, that is a closed set. So, we defined closed sets. Now, we are saying we define a set to be open if its interior is equal to itself. For example, open interval. What is the interior of the open interval? So, open interval that is why it is called open. That interval round bracket a, b is called open interval because it turns out to be an open set. Every point is an interior point. Similarly, if you look at the closed interval what we call as a closed interval a, b, its closure is equal to itself. That is closed. What is the complement of that closed interval? That is minus infinity to a open union. Can we say that is an open set? I know each part is open. So, we are saying then we need to analyze something like if a and b are open can I say a union b is open or not. So, we have defined open sets. Now, let us look at the properties of open sets like we looked at the properties of closed sets. So, let us look at properties of closed sets. So, we defined a is open. That was the definition we have said a is open if a is equal to a interior and consequences. So, if this is a definition, then a is open if and only if a complement is closed. So, let us look at some properties. Let us look at second if a and b are open. What about can we say that empty set is open set? Is there any problem if we say so probably somewhere before that let us just write empty set is open. Why is it open? You can look at complement is closed. If you want to say what is the interior of empty set is itself. There is nothing inside. So, interior is equal to itself. Whatever way you want to say it is open. So, let us look at the third that if a and b are open, then that implies a union b is open. So, what will be an argument for that? If x belongs to a union b, I want to show every point is a interior point. Or if you like you can use your a and b and go to complements and then complements back and so on. You can do that. Let us try definition straight. If x belongs to a union b, then either x belongs to a or x belongs to b. a is open. So, there will be an open ball at x included either in a or in b depending on whether a is open or b is open. So, that open ball will be inside a union b also. So, that no problem at all. So, if x belongs to a implies the interior is epsilon bigger than 0 such that the ball centered at x of radius epsilon is inside a. Implies the ball centered at x of radius epsilon is also in the a union b because it is in a. So, implies x belongs to a union b. So, implies x is interior b. So, every point x is interior point. So, implies a union b is open. No problem. What about if I add a third set c also? If something belongs to a union, it will belong to one of them. That is important. Either a or b we are saying. So, you can take arbitrary union also. It does not affect the proof at all. Take arbitrary collection of open sets a alpha. Then, their union is also a open set because if x belongs to a union a alpha, it belongs to one of them and that being open, there is a ball inside that particular one. Hence, in the union also. Keep in mind properties of closed sets. We said arbitrary intersection of closed sets was closed. Here, arbitrary union because they go complements. Set is open if and only if it is closed. So, keep that in mind. So, probably we should have another property about intersections. If a and b are open, can you say a intersection b is open? No problem. We can go by definition. If x belongs to a intersection b, then it belongs to both of them. That means there is open ball at x of some radius epsilon 1 contained in a, but it also belongs to b which is open. So, there is some open ball of some other radius epsilon 2 centered at x which is inside b. At same point, there are two radii, one bigger one and smaller. I can take the smaller one. That ball with the smaller radius will be in both a as well as in b. So, it will be inside a intersection b. So, every point in the intersection has an open ball contained in that set. So, intersection is an open set. So, here is if you like, here is a here is b, here is a point. So, there is one ball like this and other ball smaller. So, smaller one is inside the intersection anyway. No problem. Or if you like to go by a definition, we have already proved a set is closed if and only if its complement is open. So, if a and b are open, the complements are closed. Finite union of closed sets is closed. So, a complement union b complement is closed, but what is a complement union b complement? It is a intersection b complement. So, that is closed. So, a intersection b is open. If you want to go that, you can that route also you can go or you can just go by definition. Every point of the intersection is a interior point. So, hence it is a open set. Once again does not matter whether I take two sets or three sets or any finite number of them, same proof will work because the minimum of the two I was taking is radii. Any finite, I can take the minimum of that that will be inside any finite. So, into a finite intersection of open sets is also open. Question, can I say arbitrary intersection of open sets is open or not? So, think about this as a question. So, this goes to closed sets also. We have property of closed sets also looked at. So, I am leaving this as an exercise. So, this is an arbitrary intersection of open sets is open. If you think it is true, you have to give a proof. If you think it is not true, you need to produce example of a collection of sets, which are open, but their intersection is not open. So, these are various properties of open sets. So, let me just see if anything else is to be said. Then let us do it. So, close if and only if that we have done. So, that is the definition or we defined open sets and that is same as the complement is closed. So, that follows from the previous theorem. Here are the properties, which you have just now proved. Every open ball is an open set. That is like every open interval is open set. Every open ball, every point is an interior point. We discussed that A and B are open. Intersection is open. That you can extend it to finite number of sets. So, property is this. You can extend it to finite number of sets. Whether you can extend it to arbitrary intersection or not, that is a question left for you to analyze. Union is okay that we have proved. So, these are the properties of open sets. Here is something I will not prove that, but it is quite nice property to have it handy. It says if it characterizes open sets in the real line. It says a set is open in R, then it can be written as a disjoint union of open intervals. So, this every open set, it need not be an interval. Every open set can be written as a disjoint union of open intervals. That means open intervals are in some sense give you everything, all open sets, by taking countable, by taking union. Anyway, why this is important or useful let me just say supposing U contained in let us say Rn is open. Let us take any open set in Rn. Then for every x belonging to U, it is an interior point. So, there must be a ball around that point inside. So, implies there is some radius let us call it epsilon x bigger than 0 such that the ball centered at x of radius epsilon x is inside U. X belongs to it that is the center. Then what is U equal to? In terms of these balls, do you agree that I can say that this is same as x epsilon x union over all x belonging to U. All balls are inside U. So, the union is inside U, but all points are inside the balls. So, all U also is inside the union. So, both are equal. Is it okay for everybody that this implies this obviously because if I take union, union of singleton x that is the whole space U whole set U. So, U is inside the union of balls which is inside U again. So, everything must be a equality. So, U is equal to. So, this says an open set can be written as a union of open balls. Interpret this every open set is a union of open balls. So, if you interpret in the real line that says every open interval for example, every open set is a union of open intervals. Balls are open intervals. Now, given two intervals, you can always make them disjoint kind of by taking intersections. So, that is the basic idea. So, we will not prove that. So, it says not only that. So, in the real line, you can write not only as a disjoint union of interval, as a union of open intervals, you can write as a disjoint union of open intervals. So, that is a very useful thing. If you do not, this indexing set I is not saying it is finite or countable in finite or what? Some collection. Actually, one can show if you forgo disjointness, then you can write every open set is a countable union of open intervals. Then you make it countable. That property, you will not be using it, but that is the beginning of something called a subject called metric spaces and you say real line is where everything comes from open intervals, from countable union. So, you say it is a second countable space and such kind of thing. So, that is something separate, but this is useful sometimes or it is a beautiful result in itself saying that every open set can be decomposed into a union of disjoint union of intervals. So, just keep that result in mind. Sometimes you may use it in your courses somewhere else.