 So, what we have done? We have looked at basically properties of various kinds of sets. We started looking at a set. The limits of sequences may not be inside. So, we looked at sets which are called closed sets. What are closed sets? We have the limits of sequences of elements of that set are also inside. So, that we called as a closed set. Then, we looked at a set may not be closed, but you can have something called the closure of a set. So, a is always subset of a closure and closure is the smallest closed set which includes a. That is interesting. I think I should say that something more about it. We proved a result that given a, a closure is the smallest closed set including, remember this we proved that. So, this was basically by proving that a closure is always a closed set, a is inside it and closure of closure is itself. So, that is a smallest. But now look at this given any set a, a given is contained in say a whole space. So, let us write say Rn. a may not be closed, but it is contained in Rn which is a closed set. The whole space is closed. So, let us collect together all sets C in Rn. C closed and C includes a. So, look at the collection of all closed sets in Rn which includes C. At least we have got one candidate. This set is non-empty because Rn is member of Rn belongs to you because Rn is closed and it includes a. Now, if I take the intersection of all sets C belonging to you, look at all the sets which are inside this collection and take the intersection. What can you say about this set? Intersection that will be a closed set because arbitrary intersection of closed sets is closed and each one of them includes a. So, a is subset of this and it is closed. My claim is this is smallest one because we have taken the intersection of all. So, in fact this is nothing but this is equal to this set is equal to a closure. So, this is another way of saying what is a closure. Look at the intersection of all closed sets which include it that will be smallest and that is closed. So, it has to be a closure. So, that is another way of defining a closure. You will find it somewhere. Now, can you say what is a interior? Given a set a, can a similar thing be said about a interior? Try to make a guess. Given any set a contained in Rn, what can you say about a interior? It is the largest. Interior is always an open thing inside but empty set is open. So, I should try to make it bigger. So, try if you like, try to prove it. A interior is the largest, you can deduce it from here also if you like, is the largest open subset of a. So, what we are saying, look at all open subsets which are inside a, take their union that must be a interior. So, various nice playing around with sets and definitions. So, I am leaving this also as an exercise because we do not really need it but it is good. It will help you to understand interior points and so on and open sets and so on. So, let us see what. Now, let us come to a very important concept. We started looking at sets whose limits may not be inside, limits of sequences but there can be sets which have no convergence of sequences at all. There could be sets which does not have any convergence sequence at all. But possibly there is a convergent subsequence of every sequence. There are no convergent sequences but every sequence has a convergent subsequence. There could be sets with that property. So, they turn out to be very important collection. So, we say this is such a set is called a compact set. A set is called compact. If every sequence, the sequence may not converge at all but it should have a subsequence converging where to a point inside the set. That is the condition. Are you understanding what I am saying? Yes, subsequence converges and the limit is inside that set. Look at the sequence minus 1 to the power n. That sequence is not convergent. Every sequence need not converge. It may not have any convergence subsequence at all also. Look at the sequence to n. Neither the sequence converges nor any subsequence converges. So, convergence of a sequence is independent of anything. But some sets have that property. There are sequences. Some of the sequences may converge. Some may not converge. Those who converge collect the limit points and put them together in a box. We call that a closure and so on. But now we are looking at those sets which have the property. Given any sequence, it should positively have a convergent subsequence, at least one. So, for example, let us look at examples. Look at convergent to a point inside the set. That is important. Keep that in mind. Look at the closed interval, say 0, 1. Take any sequence in that closed interval. Take any sequence in that closed interval. We proved that given any sequence, there must be a monotone increasing or monotone decreasing subsequence. Given any sequence in the closed boundary interval 0, 1, there is a subsequence of this sequence which is monotonically increasing or decreasing. It remains between 0 and 1. It is bounded. So, the closed boundary interval 0, 1 has the property that every sequence has a subsequence which is monotone and bounded. So, it must converge by completeness property. So, the closed bounded intervals have that property that we are looking at. Namely, every sequence has got a convergent subsequence converging to a point inside that set. So, any closed bounded interval a, b is a compact subset of the real line by this definition. So, I am giving examples now. So, example 0, 1 is compact. Is that okay? It is a compact set because given any sequence by our earlier theorems on sequences, it has a convergent monotone sequence and it remains between a and b. So, it is bounded. So, it must converge. So, every sequence has a convergent subsequence. Let us try to look at say a, b. Let us take any sequence inside. Let us try to copy the earlier proof where it goes wrong if at all or it works for this also. Take any sequence in the open a closed b. Then it has a monotone sequence, subsequence. It is bounded by a and b. So, it must converge. That is up to completeness. But the limit may or may not be inside open a and close b because earlier for the subsequence every term was between a and b. So, the limit could be equal to a or could be equal to b. But that remains inside that interval. So, the limit is inside here. The limit can become for example, a. If you look at the sequence a plus 1 over n, then that is going to converge to a. Any subsequence that also will converge to a because the sequence itself is convergent. But a is not inside the set. So, this is not a compact set because we have produced not compact. So, it looks like this closed bounded interval seems to be a prototype of compact sets. So, let us try to prove a result. We guess a result now that a set in R n is compact if and only if it is closed and bounded. So, we want to prove a theorem. So, a is compact if and only if a is closed and bounded. So, let us try to prove if we are able to prove then it is true if not then we have a contradiction. So, let us try to prove. So, let us say a is compact. So, this is given. So, let this is given to show a is closed and bounded. So, how do I prove a is closed definition? What is saying a is closed? If a sequence of elements of a converges somewhere then that must be inside the set a. So, let a n belongs to a, a n converges to a belonging to R. Then we want to show a belongs to small a, element a belongs to the set a. What is given to us? a is compact and the compactness is every sequence has a convergent subsequence converging in the set. So, a n is a sequence which is itself converging by the compactness property. It must have a subsequence which is converging but to a limit which is inside the set. But the sequence itself converges. So, every subsequence has to converge to the same limit and that is in a because it is closed. So, small a belongs to. So, let us write then by compactness as a subsequence. So, let us write a n k, a n k converging to something in a. It only says it has a convergent subsequence converging to something which is in a. But that something is a because a n converges to a and it is a subsequence because it is a subsequence implying that a belongs to a. So, we are using the fact compactness is there is a subsequence which is convergent. But the important thing is compactness is that the limit must be inside the set. On the other hand, sequence itself is convergent. So, every subsequence must converge to the same limit and the limit being a, a belongs to capital A. So, a is closed. So, next I should prove that a is bounded. If a is compact, it must be bounded. If not, if it is not bounded, what will happen? If it is not bounded, it is a subset of R n. When do you say a subset of R n is bounded? We are not defined it as such. For real line, we define bounded that it is between alpha and beta. When would you say a subset of R n is bounded? Because you can go any direction now. So, there is a ball which includes a. Is that a good enough definition? We will say a set is bounded if there is ball of some radius. You can take it center at origin or it does not matter which is inside. So, if you want very precise, you can say we will say set a is bounded if there is some radius R, say that the ball centered at origin of radius R includes the whole set a. Good enough. Is it okay? Let us take that as a definition of bounded. Now, I want to prove if a set is compact, it must be bounded. So, our definition of bounded is there is a ball which includes a. So, if it is not bounded, what will happen? Whole of a is not inside any ball. Whatever ball I take, there is something which is outside. So, what will be the distance of. So, if I take a ball of radius n, n natural number, then there is a point outside. So, that point outside, how much away it will be from origin, distance at least bigger than n. So, it is outside. It is a ball of radius. So, there is a ball of radius n at origin. So, this is n. If there is a point outside, what will be the distance of that point? It will be greater than n. So, saying that a is not bounded implies. So, I am saying this implies for every n, there exists some xn such that norm of xn is bigger than n. Is that okay? Just whatever we are discussing, we will just return it to you. So, can this sequence xn have a convergent sub-sequence? Because this is a sequence in A and A is given to be compact. So, by that compactness, it should imply this should have a convergent sub-sequence. But can this sequence xn have a convergent sub-sequence? It is going away and away. The points are going away and away from 0. You can say the sequence xn is unbounded. Norm of xn is bigger than n. So, as n becomes larger, actually this is becoming larger and larger. So, this does not have, this sequence cannot have a convergent sub-sequence. Is it okay? If you find visualizing in R n to be difficult, you can keep in mind your real line. You have got points on the line which are going away and away from 0. So, sequence is unbounded. Convergent implies that is a necessary condition. Every convergent sequence should be bounded. So, this is unbounded. So, cannot converge by that also if you like. So, saying this sequence cannot converge because it is unbounded. So, that means our assumption there is something outside. Every ball must be wrong. That means the whole of A must be inside one particular ball. So, implies A is bounded. So, what we have done? We have proved the theorem that a subset of R n is compact if and only if it is bounded and close. That a less than b that is closed interval that is compact, then it is closed. So, the whole space is not R or R n is not compact. Real line is not compact because it is close but it is not bounded. Same reason R n is not compact. What way it is useful? So, let us come to it is closed and bounded that we have proved.