 So, in the previous lecture we gave a convenient, we gave a criterion for a metric space of when a metric space is compact. So, this lecture we will begin with a lemma which is very useful and it is called the Lebesgue number lemma. But before that we need some preliminaries, so x be a metric space, z contain x be a subset. So, given the subset z we can define. So, we define function dz from x to r by dz of x is defined to be infimum of z in z distance of x from z. So, what does this mean? So, these are x and this is my z let us say. So, this is x. So, we take the distance of x from all points in z and take the infimum right. So, the infimum may or may not be attained at a point. For instance, if we take the real line and we take the open interval 0 1, we take this to be z right and we take x to be minus 1. So, then the infimum, so the distance between d of x comma a is equal to a plus 1 right. If a is a point here, then this distance is a plus 1 right. So, the infimum is never attained because so the infimum I am sorry. So, the infimum is 1, but it is not attained for any a in the subset z right. So, the infimum may or may not be attained. So, this is a function which we define on x. Next let us check that this is continuous. So, we claim that d z is a continuous function. So, for this note that if distance between x and y is strictly less than delta, then for any z in z, we have the distance of x from z is less than equal to x and y and z which is strictly less than delta plus distance of y from z. So, when we take infimum this implies taking infimum d z of x is less than equal to delta plus d z of y which implies d z x minus d z y is less than equal to delta right. So, similarly replacing switching the roles of x and y, x and y we get d z y minus d z x is less than equal to delta which implies the absolute value of d z x minus d z y is less than equal to delta right. So, thus for any epsilon positive we let delta to be equal to epsilon by 2 right. So, then d of x comma y is less than equal to delta will imply that d z is strictly less than d z x. So, this shows that d z is continuous. So, using this distance function let us prove this Lebesgue number lemma. So, let x be a compact matrix space suppose we have given an open cover u i's. So, then there exists delta positive such that for any for every x p delta x is contained in u i for some i. So, now this is something which we have already proved while we are proving a previous theorem because since x is compact every sequence has a convergent subsequence and this is precisely the claim one which we proved in the while we are proving the previous theorem. However, using the this distance function we let us see a clean proof of this nice proof of this. So, proof. So, as x is compact there is a finite sub cover and we may assume x is equal to union i equal to 1 to n u i's. So, let us see i be x minus u i these are close of spaces. So, define f from x to reals greater than equal to 0 by f of x is defined to be summation i equal to 1 to n d c i of x. So, given any point x we take the distance of x from each d c i and we just add all these ok. So, then f is continuous as all the d c i are continuous right. We are simply taking a finite sum of continuous functions and that is going to be continuous ok. So, if this function takes the value 0 then we have summation d c i of x is equal to 0 yeah, but each of these d c i's is greater than equal to 0 this implies d c i x is equal to 0 for all i, but what does that mean? So, note that recall that d c i x is equal to infimum of z in c i distance of x from z right and this is equal to 0 ok. So, this implies that there is a sequence z n in c i such that the distance of x from z n is tending to 0, but this implies that z n is converging to x right. However, we also know that c i is closed right as c i is closed in x this implies that x belongs to c i which implies that the distance of c ok. This implies that x belongs to c i that is all that we need sorry for all i. This implies that x belongs to intersection of c i intersection of c i is equal to intersection of x minus u i which is equal to x minus union u i, but this is the empty set as the union is equal to x right. So, thus we get a contribution. So, thus f of x is always positive ok. So, as x is compact this implies f of x is compact. This implies f of x is closed as 0 does not belong to f of x. So, since f of x is closed and 0 is not in is in the complement of f of x which is open this means there is a neighborhood 0 delta or delta neighborhood of 0 which does not meet f of x. This implies that f of x is greater than equal to delta for all x in x. So, this implies that the summation i equal to 1 to n d c i x is greater than equal to delta by n for at least right because if it was strictly less than delta by n for all i that would mean that the sum would be strictly less than delta ok, but this the. So, what does this mean? So, we have our c i is this and our x is this. So, distance of x from c i is greater than equal to. So, this distance is greater than equal to delta by n right, but then the complement of c i is u i. So, this implies that delta by n x is containing u i. So, this proves. So, we will end this lecture here.