 So, in this lecture, we will show or in this lecture and the next one there is a large class of topological spaces which admit many continuous maps or yeah or many which admit many continuous functions. So, for that we can recall that a host of space is one in which we can separate points by open sets yeah. So, what does this mean? So, we called a space X to be host of if given any two points on X we can find two open subsets U and V. So, U contains X and V contains Y and U intersection V is empty right. So, similar to this we have the following definition. Topological space X is said to be normal if for any two disjoint close subsets A and B contain in X there are open subsets U and V. So, here is our X we take two let us say disjoint subsets A and B this A this is B and these are closed right. So, then there should be open subsets U and an open subset V A is contained in U B is contained in V and U intersection V is empty ok. If this happens for any two pair of disjoint close subsets then we say that A is normal. So, let us see this proposition which says that there are lots of normal spaces. So, every metric space is normal. So, let us see a proof of this. So, let X be a metric space A and B be close subsets. So, recall the functions D sub A from X to R and D sub B. In fact, we had for any subset Z of X we had defined this map D sub Z from X to R as follows. The distance of X from Z is defined to be the infimum over all Z in Z out this distance of X from Z right. So, if our X is this like this X is here and let us say this is our Z right. So, we want to compute the smallest distance yeah ok. So, we take points in Z compute the distance of X from Z and we take the infimum in that collection yeah. So, that is the distance of that is defined to be the distance of X from Z yeah. So, we had seen that this function is continuous for any subset. So, let us prove this easy proposition or rather lemma if A is closed if Z is closed. So, let us prove this easy lemma first if Z is closed then D sub Z the distance of X from Z is equal to 0 if and only if X belongs to Z ok. So, proof. So, first assume that this distance is 0. So, this implies that the infimum of Z in Z is equal to 0 this implies that there is a sequence Z n such that the distance of X from Z n is converging to 0 right. This implies that Z n converges to X this implies that X belongs to Z closure, but as Z is closed is equal to Z. So, this implies X belongs to Z and obviously if X belongs to Z the distance of X from Z is 0 right because X appears in this collection. So, DXX is 0. So, having proved this lemma right. So, we will use this lemma right. So, now let us return to a proposition. So, we have we are given 2 disjoint closed subsets A and B and we have to find open subsets disjoint open subsets one which contains A and the other which contains B right. So, let us say this is our A and let us say this is our B. So, for each A epsilon A we define to be the distance of A from B divided by 4. So, we can take any A over here. So, let us just take A over here and let us compute the distance of A from B right and we take this epsilon A to be one-fourth of that distance right ok and define ok. So, similarly for B in B let epsilon B be the distance from A of B divided by 4. So, define open sets. So, we define U to be equal to the union over A in A epsilon balls around A of radius epsilon A right. And so, we are taking this ball of radius epsilon A yeah and taking the union over all these A's right and similarly V to be union B in B the ball of radius epsilon B. So, ok note that epsilon A is positive or else as B is closed by the lemma B is in B sorry A is in B which is a contradiction as A and B are disjoint and similarly epsilon B is also positive right. So, in other words so thus we get open sets open sets such that 2 contains A and V contains B right. So, we claim that U intersection V is empty right if not. So, our U is going to be something like this. So, union of V is going to be something like this right and U is going to be a union of something like this. If not there exists Y in U intersection B right. There exists A in A and B in B such that Y is in B A epsilon A intersected B B epsilon B. So, now note that the distance of B from A is less than equal to the distance of this yeah which we are trying to inequality is less than equal to d A Y plus d Y B right. Now as Y is over here right this quantity is less than epsilon A. So, this is strictly less than epsilon A plus epsilon B right. So, this is strictly less than epsilon A as Y is here and this is strictly less than epsilon B as Y is here. So, we may assume that without any loss of generality that epsilon A is less than equal to epsilon B and then this is going to be less than equal to strictly less than less than equal to. So, this will imply that the distance of B from A is less than equal to 2 epsilon B, but now remember what was epsilon B epsilon B was equal to d distance of B from A by 4 right. So, this implies that d A B is equal to 0 yeah and as A is closed this implies that B belongs to A yeah which is a contradiction. So, thus U intersection B is empty. So, this completes the proposition ok. So, this shows that there are lots of normal spaces. In fact, any metric space is normal right and let us prove another lemma. So, we will prove a few results which are in preparation towards proving Urison's lemma. So, let X be a normal space A B a closed subset. So, let W. So, we have a closed subset A and we have an open subset W right. So, then this lemma says that there is a V. So, this is R W and this is R V an open subset such that A is contained in W right. Then there is an open subset V such that A is contained in V is contained in V closure is contained in W. So, let us prove this lemma the sets A and X minus A X minus W. So, W is open. So, X minus W is closed. So, X minus W is this region is the region outside W including this boundary this joint closed subsets. So, thus there are open subsets such that U contains X minus W and V contains A. This is using normally. So, every time we have disjoint closed subsets we can separate these using disjoint open subsets. So, that is the definition of normality. So, now so we claim that V closure intersected X minus W is empty. So, why is that because if not so let us just make a picture. So, U is this region. So, this region is U if not there exists this T in X minus W such that T belongs to V closure right. But this is not possible but this is not possible as U is an open set containing T such that U intersection V is empty right. What does it mean for T to be in the closure of V? It means that every open subset containing T has to meet V that is not possible right. So, therefore right. So, this implies that V closure is does not mean X minus W which implies that V closure is completely contained. So, thus A is contained in V is contained in V closure is. So, this proves the lemma and the final proposition let X be a metric space let A and B be disjoint closed subsets. Then there exists a continuous function f from X to 0 1 such that f of A is 0 and f of B is equal to 1. So, we have our X we have two disjoint closed subsets A and B. So, we are claiming that there is a map to 0 1 which takes A to 0 and it takes B to 1. So, let us prove this. So, for metric spaces this is very easy to prove. So, take f of X to be equal to d A X divided by X plus d B X ok. So, now let us look at the. So, the functions d A and d B are continuous this implies that d A plus d B is continuous right and since A intersection B is empty right d A plus d B for any X for any X in X. When we look at d A X plus d B X this has to be positive because if it is 0 then it will mean that if not then we will have d A X is equal to d B X is equal to 0 which will imply that X belongs to A intersection B as A and B are closed. So, thus this function d A plus d B it takes positive values. So, this implies that this 1 upon X goes to d A X plus d B X is continuous right and since the product of continuous functions is continuous this implies that f of X is continuous right. So, now if X belongs to A then clearly f of X is equal to 0 right and if X belongs to B then f of X is equal to the distance of but this is 0 because X belongs to B. So, this is equal to 1. So, this completes the provisional proposition. So, in the next lecture we will prove Orison's Lemma. So, Orison's Lemma proves the same proposition, but it removes the hypothesis that X is a metric space and it replaces metric space by being normal. So, we will end this lecture here.