 So, let us begin this lecture. So, in this lecture we will prove Orison's lemma which is rather amazing result as you will see that x B are normal to logical space. So, recall what normal meant? It just means that given any two disjoint closed subsets we can separate them by disjoint open subsets that A and B be disjoint closed subsets there is a continuous function from x to 0 1 such that f of A is equal to 1 is equal to 0 and f of B is equal to 1. So, we saw that this happens in the case of metric spaces when x is a metric space and there we saw such an f we constructed such an f very easily and for a general normal topological space it is far from clear why there would exist any non-constant continuous function. So, let us before we begin the proof let us make a small observation. So, suppose we have such a continuous function f from x to 0 1 with f of A is equal to 0 and f of B is equal to 1. So, then for each q in 0 1 intersected rationals we get the set u sub q this is defined to be f inverse of 0 comma q this is an open subset of x and this is an open subset of x which is contained in x minus B because it cannot contain any point of B because all points of B go to 1. So, if it contained any point of B that would mean that the image of this u of q would contain 1, but that is not possible because the image of u of q is contained in this interval 0 comma r and also note that if q 1 is strictly less than q 2 and these are rational numbers in this interval this half open interval then obviously u q 1 is contained in u q 2. Moreover we will also have f of u q 1 closure which is contained in f of u q 1 closure. So, f of u q 1 is contained in 0 q 1 this implies f of u q 1 closure is contained in 0 q 1 which is contained in 0 q 2 which will imply that u q 1 closure is contained in u q 2. So, thus so what we conclude is if q 1 is strictly less than q 2 and both q 1 q 2 are in this interval then we have u q 1 is contained in u q 1 closure is contained in u q 2. So, motivated by these by this observation to prove the theorem we begin by constructing such sets. So, so let us this was a remark and now let us begin the proof of Horizont's lemma. So, since a and b are disjoint close subsets. So, this implies that a is contained in x minus b and we define this to be u 1 this is an open subset. So, using the lemma proved in the previous class proved in the previous lecture. So, let us just recall that lemma that lemma said the following x is normal anthropological space a contained in w a is closed w is open implies there exists v open and a is contained in v is contained in b closure is contained in w right. So, using the lemma proved in previous class we get that a is contained we can find out set u 0. So, set a is contained in u 0 is contained in u 0 closure is contained in u 1. So, now, a step 1. So, as q intersected 0 1 is countable we may index its elements using the set of natural numbers q is equal to q i's for i greater than or equal to 1 and further we may assume q 1 is equal to 0 and q 2 is equal to 1. So, we have defined u q 1 which is u which is u 0 and u q 2 which is u 1. So, inductively we will define u q j for each j greater than or equal to 3. So, let us do this. So, let us assume that we have defined u q i for q i in q 1 q 2 or 2 let us say q r. So, now we have this is 0 which is our q 1 this is 1 which is our q 2 and let us say q 3 is here, q 4 is here, q 5 is here, let us say q r is here and let us say there is some q i here. So, when we take our q r plus 1 it is going to q r plus 1 is different from all these q i's. So, it is going to lie between two of these. So, there is a unique. So, there are unique a comma j such that q r plus 1 lies between q i and q j. So, take u q r plus 1 to be an open subset which satisfies u q i is containing u q i closure is containing u q r plus 1 is containing u q r plus 1 closure is containing u q j. So, we can do this using the lemma we can do this using the lemma. So, I should have said here that assume that we have defined u q i such that if q i is strictly less than q j then u q i closure is containing u q j. So, because we have inductively, so this is true for r equal to 2 and so we proceed like this. So, note that for we have this inclusion because of our hypothesis and then using the lemma we can find a q r plus 1 which satisfies this. So, proceeding in this way in this way we would have found sets open subsets open sets u sub a for a in 0 1 intersectional rationalism. Such that if a is strictly less than b then u sub a closure is containing u ok. So, this we inductively construct a sequence of open subsets ok. So, this completes step 1. Step 2 is for p strictly less than 0 define u p to be and for p strictly less than 0 define p strictly greater than 1 define u p. So, p strictly greater than 0 and p in q for p strictly greater than 1 and p in q. So, now we have defined for all rationalism open sets u sub a such that if a is strictly less than b then a closure is containing u. So, next let us go to step 3. So, given x in x we note by s sub x to be those rationales such that x is in u. So, note that if p is in s x then for all q greater than equal to p we have q is also in s x. So, why is that? Because as x belongs to u p which is containing u p closure ok. So, let me just say for all p q strictly greater than q obviously holds for p which is contained in u q. So, we will use this observation later. So, let us just keep this in mind here. So, now using this we define a map f from x to r f of x is defined to be infimum p in s x. So, we just take the infimum of all elements in s of x right and define that to be s x and define that to be f of x. So, by so define a map infimum of the elements in s x or in other words the infimum of elements in s x. So, we need to check the following. So, this is our candidate for the function and we need to check the following 1 f of x is contained in 0 1 2 f of a is equal to 0 3 f of b is equal to 1 and 4 f is continuous. So, let us check these 1 by 1. So, let us prove 1. So, let x in x if a is in q and a is strictly less than 0, then recall that we define that u sub a is empty right. So, thus a does not belong to s of x. So, in other words are is 0 no rational which is strictly smaller than 0 is in s of x right. So, this implies that infimum of this is has to be greater than equal to 0 right. So, this implies that f of x is greater than equal to 0 for all x in x. So, similarly if a is in q and a is strictly greater than 1 then u sub a is equal to x and so every such a is in s x right. So, we have 1 over here and no matter which rational we take that is in s x therefore, when we take the infimum when we take infimum every rational which is larger than 1 is in s x. So, when we take the infimum this implies that the infimum is less than equal to 1. So, this shows that f of x belongs to 0 1. So, this proves 1 right. So, next let us prove 2. So, to prove 2 we need to show that f of a is 0. So, if x is in a then x belongs to u 0 right. So, remember we had this is our a and this is b the complement of b was defined to be u sub 1 and inside that we chose u sub 0 which contains a. So, that its closure is also contain inside u sub 1 right. Then if x is in a. So, we have x belongs to u 0. So, this implies that 0 belongs to s x right. So, this implies that now we know that the infimum cannot be less than equal to 0 right. So, thus f of x is less than equal to 0 right because f of x is the infimum of all elements in s x and 0 is in s x therefore, the infimum is less than equal to 0. So, this forces that f of x is equal to 0 right. So, this proves 2. Let us look at 3 if x belongs to b then we claim s of x is contained in 1 comma infinity intersected with q right. So, in other words s of x does not contain any rational is less than equal to 1. So, why is that? If if not then there exists a rational p such that p is less than equal to 1 and p belongs to s of x right. So, if p is equal to 1 is equal to 1 then we get 1 belongs to s of x which implies x belongs to u 1, but u 1 is by definition x minus p right which is not possible as x belongs to b right. So, thus p is strictly less than 1. So, in this case we have u p closure is contained in u 1 right and so, again again if x belongs to u p then x belongs to u p closure which is contained in u 1. So, which is not possible for the same reason. So, thus s of x is contained in 1 infinity intersected with q right and therefore, the infimum is going to be greater than equal to 1. So, this implies that the infimum of s of x is greater than equal to 1 which implies that f of x is greater than equal to 1, but f of x lies between 0 and 1. So, this implies that f of x is equal to 1 for x in b. So, we have proved the first 3 conditions. So, finally we just have to show that f is continuous we show that f is continuous. So, it suffices to show that for any two rationals c comma d with c strictly less than d the set f inverse c comma d is open in x right because intervals of this type c comma d with c strictly less than d and both of them rational form a basis for the topology on the interval 0. So, therefore, we will show that for any two rationals c and d this set is open right. So, suppose x belongs to f inverse of c comma d ok. So, then choose the rationals p and q such that c is strictly less than p is strictly less than f of x is strictly less than q is strictly less than d right. So, we will show that. So, we will show two things 1 x is in u q minus u p closure and 2 u q minus u p closure is contained in f inverse c right. So, let us try to prove these two things ok. So, as f of x is equal to infimum of s sub x which is strictly less than q this implies that there exist q 1 in s of x with q 1 strictly less than q which implies that x belongs to u q 1 which is contained in u q. So, this implies that x belongs to u q. Similarly, as p is strictly less than f of x which is equal to infimum over all elements of s x this implies that there exist a ration p 1 such that p is strictly less than p 1 which is strictly less than infimum of elements in s x. So, this implies that p 1 does not belong to s x this implies that x does not belong to u p 1, but we have u p closure is containing u p 1 this implies x does not belong to u p closure right. So, thus x is in u q minus u p closure ok and this is an open subset which contains x right and next we will show that this open subset is completely contained inside this f inverse c d. So, for every point that would mean that for every point in f inverse c d we have found an open neighborhood of that point in x which is completely contained inside f inverse c d which will mean that f inverse c d is open. So, next let us prove that u p minus u q bar is contained in f inverse c d. So, let us take if y belongs to u q minus u p bar this implies that y belongs to u q which implies that q belongs to s of y which implies that. So, q is in s of y therefore, when we take f of x which is equal to infimum of elements in s of y this is going to be less than equal to q right. So, similarly y does not belong to u p closure which implies y does not belong to u p which implies p does not belong to s of y. So, what this means is that. So, we have p here right. So, if s of y contained anything which is in any q which lies over here then recall that ok. So, let me call this p prime if s of y contained some p prime which is strictly less than p then recall what we have proved here we have seen here right. So, since s of y would contain since s of y contains p prime it would contain all rationals which are larger than p prime in particular it will also contain p. So, using this property of s of y. So, therefore, it will follow that s of y it does not contain any p prime which is smaller than p that means s of y is completely contained in this region right which means that infimum of s of y which is equal to f of y is going to be greater than equal to p right. So, thus we have proved that if y belongs to u q minus u p closure this implies f of y sorry this should have mean f of y f of y lies between p and q right. So, this implies that u q minus u p closure is contained in f inverse of c comma d. So, this proves that f inverse of c comma d is open. So, this proves that f is continuous which completes a proof of this theorem and what is the corollary which is really the moral of Orison's lemma is that on a normal topological space continuous functions separate disjoint closed subsets. So, in this sense there are lots of continuous functions on a normal topological space. So, we will end this lecture here.