 So, in this final lecture we will prove the Orison's metrization theorem. So, let us begin. So, let x be a regular and second countable to logical space right. So, recall that regular recall that regular meant that we can separate points and flow subsets by disjoint open sets and second countable meant that x has a basis whose cardinality is countable and in one of the previous lectures we have proved that this these two assumptions imply that x is normal right. So, then x is metrizable right. In other words we can put a metric on x such that the topology induced by the metric the topology induced by the metric. So, in other words this means that we can put a metric on x as a topology generated by the metric is the same as the topology on x that we started with. So, let us see how to prove this. So, the idea is to use what we proved in the previous lecture that Rn the countable product of R is metrizable right. As Rn is metrizable it suffices to produce a continuous map f from x to Rn such that 1 f is an inclusion and 2 when you look at f from x to f of x. So, this f is an inclusion. So, this is going to be a bijection right. So, this is an open map. So, this is a bijective continuous map and f is also open right. So, that will mean that this is a homeomorphism right. So, here f of x has the subspace topology. So, the subspace topology from Rn is induced by the metric because the product topology on Rn is induced by the metric as we saw and this will imply that the topology on f of x is given by a metric, but f from x to f of x is a homeomorphism that means that the topology on x is also coming from a metric. So, equivalently we have embedded this f is going to give an embedding of x into Rn and. So, let b be a countable basis for the topology on x this is because x is assumed to be second countable right. So, the subset up as v comma u such that v comma u are both basic open sets and v closure is containing u right. So, by Urson's lemma there exists ok there is a continuous function f sub v comma u from x to 0 1 such that f sub v comma u of v closure is equal to 1 and on x minus u is equal to 0. So, we have. So, the collection of such pairs where v closure is contained inside u that is a. So, first of all we have many such pairs right because given any point x we can first take a basic open set u inside that then as x is normal well x is regular and second countable and we have proved that it means x is normal we can find first one open subset v inside x as that v closure is contained inside u and inside this v we can take a basic open set w which contains x. So, then w closure is also going to be contained inside u right. So, there are many such open sets there are many such pairs right, but at most they are going to be countable because v is an countable set and u is an countable set and a product of countable sets is countable. So, we have a countable collection yeah. So, we put these together to get a continuous map right. So, using the product using the property of giving a map to the product. So, we get this map f from x to product over all these pairs u comma v are right this is a count. So, v comma u such that v closure is contained in u right and what is this map given a x it goes to this f v comma u. So, there is a countable product and there is a countable product. So, therefore, this is I mean we can enumerate such pairs and then this is going to be equal to r n right and therefore, this the product apology over here is coming from a metric because of our previous theorem ok. So, clearly ok this map is continuous we have already observed that right. So, let us prove this is an inclusion. So, if x is not equal to y and both these are points in x right. So, then there exists basic open sets v comma u such that x is in v is contained in v closure is contained in u and y is not in u. So, we can just take this x and this y we can first start with u over here and then we can find v. So, then f v comma u of x is equal to 1 and f v comma u of y is equal to 0. So, thus these two points f of x and f of x and f of y differ in the v u th coordinate which implies that f of x is not equal to f of y ok. This shows that this map is an inclusion right. So, this implies f is an inclusion next we need to show that x to f of x we have this f and this is an open map. So, what we have to show so that is given u in x which is open we need to show that f of u is open in f of x. Yeah f of x has a subspace topology right where f of x has a subspace topology. So, to do this it suffices to show that if x is a point in u then there is an open set y contained in this product such that f of x belongs to y intersected f of x this is contained in f of u right. So, our aim is to show that f of u is open. So, let us say this is our f of u. So, given any point f of x right we are going to show that there is this open subset in. So, y intersection f of x that is open in the subspace topology which is completely contained inside this f of u. So, that is what we have to show ok. So, for this find basic open sets w and w prime such that x belongs to w is contained in w closure is contained in w prime is contained in u right and consider the projection w comma w prime. So, from this product u v comma u's to the w w prime square right yeah. So, w w prime also occurs the pair w w prime also occurs in this collection of pairs because w closure is contained in w prime and these are basic open sets right. So, therefore, so this is so this map is ok. So, we had this is the projection map and let y be equal to pi w comma w prime inverse of 0 infinity y is open in this. So, by definition of f when we look at the composite x to f and we project to w w prime this composite is exactly f w w prime and as. So, as x belongs to w closure this implies f w w prime of x is equal to 1 which is in 0 infinity right, but this f is equal to pi w w prime of f of x right and this is in 0 infinity this implies f of x belongs to pi w w prime inverse of 0 infinity this is equal to y right. So, this implies that f of x is in y intersected f of x right. So, that proves this includes this part. So, now let us prove the this inclusion. So, suppose f of a is in y intersected with f of x right. So, this implies that so this implies that f of a is in pi w w prime inverse of 0 infinity which implies that pi w w prime compose f of a belongs to 0 infinity which is same as saying that f of w w prime of a is in. So, now, note that f w w prime of w closure is equal to 1 and f w w prime on x minus w prime is equal to 0 right. So, therefore, since f w w prime of a is in 0 infinity. So, this implies that f sorry this implies that a has to be in w prime w prime and w prime is contained in u. So, as a belongs to u this implies that f of a belongs to f of u. So, thus we have proved y intersection f of x is contained in f of u right. So, this proves that f of u is open. So, this completes. So, we will end this lecture here and this also brings our course to an end. Thank you very much.