 So, recall that in the previous lecture we had proved this nice result. So, in the previous lecture we proved that S o n is connected. So, before we proceed let me make the following remark. So, if we replace so using the same using the same methods or using similar ideas we can also show S u n are connected. So, recall what is S u n? S u n is all those complex matrices complex n cross n matrices such that a star a is equal to identity and determinant of a is equal to 1 and u n is all those a in m n c which are really the first condition the unitary matrices. So, just as in case of S o n we define this map to S n minus 1. So, for u n we define this map to S 2 n minus 1. So, here a matrix a gets sent to the first column. Now, the inner product a e 1 comma a e 1 is equal to so in the case of u n we use the standard metric sorry standard inner product C n which is given by v comma w is defined to be w star v and recall that w star is equal to w transpose conjugate. So, we have a column vector and we take its transpose and conjugate all the entries. So, this with this definition this inner product is going to become star which is equal to e 1 star a star a e 1 and since a star a is equal to identity because a is in u n this is going to be e 1 star e 1 which is equal to 1 right. But also notice that the columns the column of a the first column consists of complex numbers z 1 z 2 up to z n right. If each of the z i if I write it as z j if I write it as x j plus i y j right where x i comma x j comma y j are real numbers then this implies that since the inner product is 1 this implies that summation or z i square is equal to 1 which implies that summation x ok x j square plus y j square j equal to 1 to n is equal to 1 which implies that we actually get an element of a e 1 can be viewed. So, let us call this map ok if we call this map psi or this map phi. So, over here we proved that in this situation we proved that phi of a is equal to phi of b if and only if b inverse a belongs to is a matrix of this type this something in S 1 minus 1 right. So, here we will have to prove that psi of a is equal to psi of b if and only if b inverse a is equal to is in 1 0 0 and then the base case. So, the base case for the induction which is you can easily check this for S 1 which is connected and similarly the base case for. So, in the same way we can define a map from S u n. So, we again let me call this psi 1 S 2 n minus 1 once again this map is a goes to a e 1 right and here the base case for the induction will be S u 1 which is simply 1 which is connected. So, the details does not exercise. So, having said this let us begin with today's lecture. So, today we will start our discussion on compact metric spaces. So, basically we are going to. So, the main theorem of this lecture is the following let x be a metric space then x is compact if and only if every sequence in x has a convergent subsequence. So, let us prove this first let us assume that first assume x is compact and suppose we are given a sequence n is greater than equal to 1. So, we are given the sequence of points and we want to show that S has a convergent subsequence. So, let this be a sequence in x. So, if S is finite. So, if the cardinality of S is finite. So, then there is some or yeah. So, there is some x in x n in S such that x is equal to x n j for infinitely many n j's for infinitely many indices right. So, and then. So, we can take this we can take this subsequence right. So, this subsequence in this subsequence every element all the elements are the same and therefore obviously it converges. So, let us assume. So, let us assume cardinality of S is infinite right. So, we may replace S by a subsequence of S. So, since we may replace S by a subsequence S. So, we may assume all elements of S are distinct. So, the cardinality of S is infinite. So, therefore, we can find a subset which is infinite and which contains and each member of that subsequence is all the members of that subsequence are different. So, if you can find a subsequence of this subsequence which converges then we would have found a subsequence of S which converges. So, we can assume this. So, let y be the closure of S in x. So, if so, S is a set of points like this and if y minus S is non-empty that means there is some point x naught in y in y minus S right there is and as x naught belongs to S closure this implies when we take any neighborhood of x naught b 1 upon n x naught intersection S is non-empty right and this implies that there exists a sequence x n S such that. So, we are done in this case. So, we are done. So, if y minus S is non-empty then we. So, let me say there is a sequence y m because such that y m converges to ok. So, then we have found a subsequence in S which converges. So, let us assume that y minus S is empty. So, note that S is always contained in S closure which is equal to y and y minus S is empty implies that y is equal to S is equal to S closure. If now if x n. So, if x n is any point in S there is subsequence in S converging to x n then we are done right. So, for instance we may have we can take a sequence like this and this sequence may converge to this point x n over here yeah. So, let us assume that. So, thus we may assume for every is no subsequence in S which converges to in other words what this means is for every x for every x n there is a small neighborhood. So, for every x n in other words for every x n there is delta n of x n such that B delta n of x n intersected S is just this x n ok. So, now as y is S closure and this is equal to S this implies S is the compact is a close subspace of a compact space compact space right recall that we had assumed x is compact here. So, thus S is compact close subspace of a compact space is compact right. However, we may write S as the union over n greater than equal to 1 B delta n x n intersected S right this is x n right this simply equal to union n greater than equal to 1 x n x n right yeah this shows that S has an open cover which has no finite sub cover which is a contradiction. So, this completes the proof of one part. So, this shows that. So, now let us prove the converse. So, to prove the converse we prove the converse. So, let us assume that. So, suppose every sequence in x has a convergence of sequence. So, we shall show x is compact. So, for this let us begin with. So, our aim is to construct a finite sub cover of this ok. So, we make a claim there is a delta greater than 0 which works for all x such that when we take this open set B delta x right. So, we are given space x and this has an open cover. So, no matter which x we choose when we take this ball of radius delta around x it will be contained in one of the UIs yeah it is completely contained in one of the UIs. So, let us prove this. So, if claim 1 is not true then for each. So, no matter how small we take delta there will be an x such that the ball of radius delta around x is not going to be contained in any of the UIs right. So, for each n greater than or equal to 1 there is an x n such that we take B 1 upon n x n is not contained in UI for any I. So, let us take S to be the sequence of x n. So, then there is by our assumption there is a convergence sequence a convergence of sequence x n j and let us say this converges to some x naught in x. So, there is some U naught such that x naught is in U naught in our open cover and for this U naught yeah there is an epsilon positive such that this ball of radius epsilon around x naught is contained in U naught right. So, R U naught could be this x naught could be somewhere here this could be R U naught right and this ball of radius epsilon is completely contained inside U naught yeah. So, for j for j very large we will since R x n since R x n j is converged to x naught x n j is going to go belong to B epsilon by 2 x naught right. So, we can take the ball of radius epsilon by 2 around x naught right. So, we can take this to be x n j. So, if so we can choose. So, choose j large so that 1 upon n j is strictly less than epsilon by 2 then it is easily checked. So, in fact so if we take the ball so x n j is here or x n j is here let us say this x n j. So, if I take a ball of radius epsilon by 2 around x n j right. So, then this ball is also going to be completely contained inside the epsilon ball around x naught. So, then it is easily checked. So, before that so then the ball of radius 1 upon n j around x n j is contained in the ball of radius epsilon by 2 around x n j and this ball is going to be completely contained inside the ball of radius epsilon around x naught which is containing. But this contradicts the assumption 1 upon n j x n j is not contained in u i for any i. So, recall that we had constructed this sequence x n's by requiring that they satisfy this condition, but we have now contradicted this condition ok. So, this proves now let us prove claim 2. In fact, claim 2 is the main assertion right. So, there is a finite sub cover of x ok. So, let us prove claim 2. So, by claim 1 there is a delta such that for every x B delta x is contained in u i for some i. So, thus we have B delta by 2 x is contained in B delta x is contained in u i. So, let us start a process of covering x as follows. So, choose any y 1 in x and let x 1 be equal to B delta by 2 y 1 ok. So, our x is like this and we can just pick any y 1 and we take this ball of radius delta by 2 around y 1. So, we choose y 1 and we take this ball of radius delta by 2 around y 1 right. So, assume that. So, now choose y 2 from x minus x 1 right. So, we can choose any y 2 over here which is not in this ball and this y 2 and we take the ball of radius delta by 2 around y 2 x 2 B equal to B delta by 2 y 1 union B delta by 2 y 2 right. So, assume that we have defined x n defined x n ok and define x n plus 1 as follows. So, we choose any y n plus 1 in x minus x n and let x n plus 1 be equal to B delta by 2 y n ah sorry we should y n plus 1 B delta by 2 minus 1 union x n right. So, we have these balls sorry. So, we have these balls and this is our y n plus 1 right. So, then we have x 1 is containing x 2 is containing x 3 and so on right. So, we claim that ah this this process has to stop finitely many steps ok y what happens if it does not stop if not then we get a sequence of points y n's right such that the distance between y i and y j is greater than equal to ah delta by 2 for all i naught equal to j right. This is how the y n's have been constructed. We take the open balls ah of radius delta by 2 around y 1 and we chose y 2 outside that ball which means that the distance of y 1 from y 2 is greater than equal to delta by 2. Now, we take open balls of radius delta by 2 around y 1 and y 2 took the union and y 3 was selected outside this. That means, that the distance of y 3 from y 1 and y 3 from y 2 is also greater than equal to delta by 2 right and this sequence cannot have a convergence of sequence right. So, thus ah this process stops. So, this implies that r x is contained in the finite union of B delta by 2 y i's i equal to 1 to some n right and each of these B delta i's is contained in some u j equal to 1 to n u i j. So, this implies that there is a finite sub cover. So, this completes proof of the theorem. So, we will end this lecture here.