 So, finally, we are headed towards proving Orison's matrization theorem. So, let us begin with some definitions. So, topological space x is said to be second countable if it has a basis B whose cardinality is countable. So, R n is second countable because we can take open subsets of the type balls A of radius epsilon, where epsilon belongs to rationals and this vector A belongs to q, right. So, this is this forms a basis and this is going to be a countable collection. So, therefore, R n is second countable and if x is second countable then it is subspace of x is second countable, ok. So, next let us define a regular space so topological space x is said to be regular if given a point x and a closed subspace A then there exists open sets u and v such that x is in u, A is contained in v and u intersection B is disjoint. So, earlier we had seen the example of normal topological spaces. So, regular is weaker than normal and obviously every normal topological space is regular. So, what we need is given a point x and this closed subspace A we should be able to find a neighborhood around x and a neighborhood around A which are disjoint, ok. Obviously, implies regular and for normal subspaces we had proved the following lemma. We have a similar we had proved a lemma similar to the following one. So, analog is a similar proof, ok let me just write the statement first. Let x be a regular space x in u where u is open. So, then we can find an open subset v such that x is in v which is contained in v closure which is contained in v. So, what do we have? We have a point x over here and we have an open subset u. So, then we can find a smaller neighborhood of x which is v such that the closure of v is completely contained inside and the proof we have proved a similar lemma for normal spaces, right. So, there instead of x we had a closed subset A. We had a closed subset A and an open subset open set u which contained A. So, imitate that proof. So, the main theorem we want to prove is the following. The main theorem of this lecture a regular and second countable space is normal, ok. So, let us prove this. So, let b denote a countable basis for x. So, let us say our space is x, right. So, x is second countable. So, that means by definition x has a basis whose cardinality is countable, right. So, let c and d be two disjoint closed subsets. So, we need to find open sets u and v such that c is contained in u, d is contained in v and u intersection v is empty, ok. So, first we use as follows fix for x in d not fix sorry using the lemma using the lemma we can find an open set v such that x belongs to v, v is contained in v closure and v closure is contained in x minus c, right. So, d is contained in x minus c since c and d are disjoint, right. So, a picture is something like this we have two disjoint open subset c and this is d, right. So, d is contained in the complement of c which is this, right. So, we just use any x in d and we can find a neighborhood v such that its closure is also contained in x minus c, this is using the lemma, ok. So, clearly we are allowed to shrink v if necessary, yeah. So, thus replacing v by a smaller basic open subset we may assume. So, we have a basis for our topology and that v is in, right x is in v. So, that means that there is a basic open subset which contains so, x is in v. So, there is a basic open subset which contains x and is completely contained inside v, yeah. So, therefore, we can replace our v by a smaller basic open subset which contains x and the same condition we will hold, ok. So, we can do this for each x in, right. So, for every x, so what are we done? So, for every x we get basic open set v sub x such that, so this is in v such that v sub x closure does not meet c, ok. So, since b is countable enumerate this collection v x, right. So, we can just number these, this is a subset of a countable set and therefore, this has countable cardinality. So, we can just say this v n is for n greater than equal to 1, right. So, now, define w n to be equal to the union i equal to 1 to n v i's, right. So, then what do we have? We have this w 1 is a subset of w 2 and so on. It is an increasing chain of open subsets to d is contained in the union of these w i's. This is because as every x in d is in some v i. The third condition is that w n closure is equal to union i equal to 1 to n v i closure, sorry w n closure intersection c, intersection c. Now, there is a finite union and therefore, the closure will go inside v i closure intersected c and each v i closure intersected c is empty. So, this is empty, right. So, we can do the same for points of c. So, in the above we were doing this for points of d. So, we can do this for points of c. We get once again this increasing chain of open subsets c is contained in the union of these and 3 u n closure intersected with d is empty, ok. So, note that this implies that is contained in x minus u n closure, right. So, this implies that when we intersect both sides with w we get w n intersected d is contained in w n minus u n closure. So, this implies that the union w n intersected d is contained in the union w n. Now, w n is open and u n closure is closed. So, when we look at w n minus u n closure this an open subset, right. And this union is equal to union n greater than equal to 1 w n intersected with d, but as d is contained in this union there is going to be equal to d, right. So, so thus we get d is contained in the union of this open sets. And similarly c will be contained in the union of u n minus w n closure. So, we claim that these are the required open subsets, yeah. So, we so we just have to show that the intersection is empty. So, we claim that the intersection is empty, right. If not there exists i comma j such that w i intersected with u j is now empty. So, let us assume that first assume that i is greater than equal to j w i w j is contained in w i and u j is contained in u i, right. So, if we take an x in the intersection this intersection, right. So, this implies that x is in x is in u j, but x does not belong to u i closure, right. But this implies that x is in u j, but and x does not belong to u i, yeah, which is a contradiction. So, similarly if j is greater than equal to i then we will have w i is contained in w j. So, which will imply that x belongs to w i and so this will imply that x does not belong to w j closure x that is x is in w i and x does not belong to w j which is a contradiction. So, thus the intersection. So, this completes proof of the theorem. So, we will end this lecture here.