 in the previous lecture proved to Risson's lemma. And today we shall prove teacher's extension here. So, what does it say? So, let X be a normal topological space, A contained in X be a closed subspace. So, then there are two assertions. So, first let F from A to minus R R be a continuous function. Then we can extend to a continuous function. So, what does it mean? It means that there is a continuous function this which is this capital F such that F restricted to A is equal to this F V star. So, let us prove this. So, we will use Risson's lemma to construct a sequence continuous functions F n's from X to minus R R which are uniformly Cauchy. So, I just want to define what this means for every epsilon positive there is an n such that for all i comma j greater than equal to n and for all X in X we have absolute value of F i F j of X is strictly less than epsilon. So, this will in particular imply that for every X the sequence F and X is Cauchy is a Cauchy sequence. So, given such a uniformly Cauchy sequence that is try to show that we get a continuous function. So, so since so lemma given a uniformly Cauchy sequence of functions in X we get a function F from X to minus R R F is and this F is continuous. So, let us prove the continuity of F. So, proof of lemma. So, it is clear how we get this function F. So, for each X the sequence F n X n greater than equal to 1 is a Cauchy sequence in this interval minus R comma R and so it converges which we define. So, how to define F is clear. So, let us show that F is continuous. So, let us show F is continuous. So, we will sketch a proof of this. So, let us say F of X is somewhere over here and let us take epsilon neighborhood of F of X and we need to show that F inverse of the ball around F of X of radius epsilon is open in X. So, let us choose any choose Y in this open neighborhood. So, F of Y lies somewhere over here. So, now let us look at this equation. So, mod F of Y minus F of Y prime. So, this we can write as. So, let us write this as and this is less than equal to F of Y ok. So, since the F n's are uniformly Cauchy if n is very large then F of t minus F and t is let us say less than delta by 3 for all t in X right. So, this will imply that for n large this F of Y minus F of Y prime is going to be less than equal to. So, this quantity and this quantity are going to be less than delta by 3 right. So, this is going to be 2 delta by 3 plus mod of F n Y minus F n Y prime ok. But now as F n is continuous as F n is continuous there exists a small neighborhood around Y. So, let us say W such that for all Y prime in W F n Y minus F n Y prime is also strictly less than delta by 3 right. So, this will imply that for all Y prime in W we have absolute value of F of Y minus F of Y prime is less than equal to delta by 3. So, it is less than equal to delta sorry in fact, strictly less than delta by 3 right. So, now if we choose delta appropriately then this will imply that. So, if we choose delta is very small such that the delta neighborhood of F of Y is completely contained inside this epsilon neighborhood of X. So, this will imply that for all Y prime in W we have Y prime is contained in this B F of X. So, first is contained in F of Y comma delta. So, this is actually strictly less than delta which is contained in B F of X comma upside right. So, this implies that. So, this what we have proved is if Y belongs to F inverse of this open set this open set then there exists an open set W containing Y such that W is completely contained inside X upside yeah. So, this implies that F is continuous. So, this proves the lemma. So, now using this lemma we will prove teacher's extension theorem now. So, let us continue with the proof. So, it suffices to construct a sequence of functions f n's. So, first let us construct f 1 from X to minus r comma r. So, for this we divide the interval into three parts right. So, this is minus r this minus r by 3 this is 0 this is r by 3 and this is plus r ok. So, we call this I 1 r this part is called I 1 r this part is called I 2 r and this part is called I 3 r. So, I 1 r is defined to be minus r comma minus r by 3 defined to be minus r by 3 comma r by 3 and I 3 r. So, then this minus r comma r is the union of is I 1 right is the union of these three. So, as this I 1 r and I 3 r r disjoint it follows that f inverse of these two. So, it is joint closed subsets. So, recall that small f is a function from a to minus r comma r. So, by Orison's lemma there is a continuous function g from X to minus r by 3 comma r by 3 such that g of f inverse of this I 1 r is equal to minus r by 3 and g of f inverse of I 3 r is equal to r by 3. So, note that we had proved Orison's lemma for in statement of Orison's lemma we had a function f from X to 0 1, but obviously 0 1 is homeomorphic to minus r by 3 comma r by 3. So, we easily achieve this also. So, let us denote the restriction g to a by g sub a. So, we claim that the supremum f minus g sub a the infinity norm this by definition is supremum a in a f of a minus g a of a we claim this is less than equal to 2 r by 3. So, let us check this. So, suppose X is in f inverse of I 1 r right. So, then sorry suppose a then f of a belongs to minus r comma minus r by 3 and by definition g of a is equal to minus r by 3 right. So, g of a is this value minus r by 3 and f of a lies somewhere over here. So, obviously the difference between these two is less than equal to 2 r by 3. So, it is ok in this case right. Similarly, if X a is in f inverse of I 3 r then f of a belongs to r by 3 comma r and g of a is equal to r by 3 right. So, it is ok in this case also right. Finally, if a belongs to f inverse of I 2 r then f of a belongs to minus r by 3 comma r by 3 right and of course, g of a is also in this interval. So, we are ok in this case also right. So, this proves this claim that this happens ok. So, let. So, now, let us define f 1 we define from a to minus 2 r by 3 comma 2 r by 3 is defined as f 1 of a is equal to f of a minus g of a right. So, we have just checked that this f 1 is this makes sense it is a map from a to minus 2 r by 3 2 2 r by 3 ok. So, we repeat the above process. So, using f 1 we construct a g 2 right. So, if this was g 1 right. So, we repeat the above process with by replacing f with f 1 to get g 2 right. So, in other words we would have proved that this infinity norm of f 1 minus g 2 is going to be less than g 2 let us say g 2 restricted to a is going to be less than equal to 2 by 3 into 2 r by 3 right. So, this is equal to 2 square r by 3 square right. So, but when we substitute for f 1. So, this implies that the norm of f minus f 1 is this thing minus g 2 infinity is less than equal to. So, we go on when as we proceed like this. So, at each stage we get a g i from x 2 minus 2 i minus 1 r by 3 i comma 2 i minus 1 r by 3 i right. This interval such that when we look at this norm g 1 minus g 2 minus minus g n this norm is less than equal to 2 to the n r by 3 to the n ok. So, we construct the sequence of functions g i which have this property right and so, this is on a this is we are taking the restriction of all these g i's to a ok. So, let f n be equal to the function g 1 plus g 2 plus plus g n right. So, then f n is a function from x 2 minus r comma r right and it is easily checked that f i's are uniformly Cauchy 2 on a f i's converge to f point wise right. So, this obviously follows from this. So, using the lemma on uniformly Cauchy sequence we get that the f n's converge to a continuous function f continuous f n restricted to a converges to f, but it also converges to f restricted to a which implies that f restricted to a. So, this completes the proof. So, given a function given a continuous function f from a to minus r comma r we have extended this to a continuous function f from x 2 minus r comma r. So, let us prove from the second part of the theorem. So, let us prove the second part. So, we are given a continuous function f from a to r. So, fix a homomorphism phi from r to minus 1 comma 1 right and define f tilde to be the composite f and then we apply this homomorphism phi. So, then so, f tilde is defined to be phi composite right. So, as minus 1 comma 1 is contained in this compact interval. So, this implies that f tilde is a continuous map from a to minus 1 comma 1 and using the previous part we get a continuous function capital f tilde from minus 1 to 1 such that f tilde restricted to a is equal to this f tilde ok. So, now, we want to slightly modify this and get an extension of this f as a continuous function of x 2 r. So, how do we do that? So, let D the closed subset f tilde inverse of minus 1 union f tilde inverse of 1 right. So, as f tilde of a is equal to f tilde of a this contained in minus 1 comma 1 this implies that a does not meet D right. So, thus there is a continuous function h from x 2 0 1 such that h of a is equal to 1 and h of D is equal to 0 right. So, consider the function now we have 2 continuous functions f tilde and h. So, we take the product consider the function. So, the product is continuous because both them are continuous h times f tilde this is from x 2 minus 1 comma 1. So, we claim that the image is in the open interval right. So, we claim that is in the open interval minus 1 1. So, why is that? If not there exists some x such that h of x into f tilde of x is equal to plus or minus 1 right. Now, the only way this can happen is that. So, this implies since h of x lies between 0 and 1 and f tilde of x has absolute value less than equal to 1 the only way this is possible is that h of x is equal to 1 and f tilde of x is equal to plus or minus 1 right. But, if f tilde of x is plus or minus 1 then x is in D and h of D is 0 right which is a contradiction right as f tilde of x is equal to plus or minus 1 on D only on D h of D is equal to 0. So, therefore the image lands inside this open interval right. So, we let f be equal to h times f tilde right. So, then f is a continuous function from x 2 minus 1 comma 1 right and f restricted to a is therefore equal to h restricted to a times f tilde restricted to a, but this is equal to f tilde restricted to a which is equal to small f tilde right. So, thus so we have f from x 2 minus 1 comma 1 and we composite with phi inverse right. So, let us see what we get. So, phi inverse compose f is from x to r right and phi inverse compose f restricted to a is equal to phi inverse compose f restricted to a this is equal to phi inverse compose f tilde, but note that f tilde was equal to phi compose f this simply f. So, thus phi inverse compose f is the required extension. So, this completes a proof of D says extension theorem. So, we will end this lecture here.