 So, let us begin with the main theorem which we are going to prove. So, let x be a locally compact, topological space which is not compact. So, then there is a compactification. So, then there is another topological space which we call x hat along with an inclusion i from x to x hat such that the following conditions hold. First is x hat minus i of x is just one point which we denote p 0. Second is i is continuous. Third is the image of x under i is and fourth is the image of x is dense. So, let us prove the existence of such a topological space x hat. As I said we take x hat. So, the first condition also already tells us that. So, we want i to be an inclusion and x hat minus the image is just one point. So, therefore, we have to take x hat is just x disjoint union one point. So, next we have to define a topology on x hat. So, define a collection subsets of x hat satisfying any one of the following two conditions. So, a is if u is an open subset of x and u is open and b is if p naught belongs to u and x hat minus u note that x hat since p naught is in u x hat minus u is a subset of x right which is contained in x the subspace. So, we take we let tau be the collection of subsets which satisfy these conditions and we need to check that tau. So, let us check that tau satisfies the conditions for being. So, let us check the first condition. So, we need to show it. So, we need to show that the empty set and the full set x are in tau. So, the empty set satisfies a and the full set x hat satisfies b. So, thus phi and x hat are in tau. So, next let u i be an arbitrary collection. So, we need to show that v which is the union of these is in tau. So, if all the u i satisfy condition a then clearly v which is the union also satisfies. So, which would mean that each of the u i is contained inside x and each of these is open which means the arbitrary union is going to be open. So, v also satisfies condition a. So, let us assume that one of these. So, assume let us say u i naught does not is does not satisfy a so it satisfies b for some i naught. So, then let us look at so consider the set. So, then v contains p naught and therefore, to show that v is in this collection we can only show that v satisfies b. So, consider the set x minus x hat minus v. So, we have first look at the set x hat minus v this is equal to x hat minus union i in i u i which is equal to the intersection i in i x hat minus u i. So, as u i naught contains p naught this implies x hat minus u i naught is completely contained inside x. So, thus we can write x hat minus v is equal to intersection i in i x hat minus u i minus intersected with x. So, if u j satisfies a. So, then this clearly implies that u j is contained in x and therefore, x hat minus u j intersected x is simply equal to x minus u j. This is a easy check and similarly. So, therefore, we can divide this into two parts intersection i in i u i satisfies x hat minus u i intersected with x. Then this whole thing intersected with intersection i in i u i satisfies b x hat minus. So, this is equal to because of what we saw here this is equal to intersection i in i u i satisfies a x minus u i intersected with x minus u i intersected with u i satisfies b x hat minus u i. So, this collection is not empty as u i naught is here. Now, all these are closed of x and this is a compact each of these is a compact subspace. So, this implies that x hat minus v is a close subspace of a compact subspace of x which implies that x hat minus v is compact is a compact subspace. So, this implies that v satisfies b which implies that v is in tau. So, therefore, tau satisfies the second condition for being a topology. So, finally, let us check that tau satisfies the third condition for being a topology. So, for that let u 1, u 2 up to u n we finitely many elements of tau and we need to check that v which is the intersection of the u i's is in tau. So, first assume that consider the case where all u i satisfy b. So, then v contains p naught. So, b was p naught is in each of the u i's and x minus u i is compact. So, then v contains. So, x hat minus v is equal to a finite union of x hat minus u i each of these is compact and clearly a finite union clearly a finite union of compact subspaces is compact. So, this implies that v satisfies b which implies that v is in tau. So, now let us consider the second case which is so assume that one of the u i's satisfies. So, then v does not contain p naught and we need to show that v satisfies a. That is the only way we can show that v is in tau. So, notice that as v does not contain p naught it is already contained in x. Note that v is contained in x as p naught is not in v. So, thus we can write v as v intersected x which is equal to intersection i equal to 1 to n u i intersected x. So, if u j satisfies b right. So, it is easy to check. So, this is the second case which is always true u j is equal to x hat minus x hat minus u j right. So, intersecting both sides this is always true this is a very simple aesthetic statement since u j is contained in x hat. So, intersecting both sides with x. So, this implies that u j intersected x is equal to x minus x hat minus u j right, but x hat minus u j is a compact subspace of x as u j satisfies b. This implies a close subspace of x right. This implies x minus x hat minus u j is an open subspace. So, thus we get that v is equal to intersection i equal to 1 to n u i intersected x. So, we can break this intersection into two parts u i satisfies a u i intersected x intersected u i satisfies b u i intersected x right. And in both cases each of these members is open in and since a finite intersection of open sets is open this implies v is open. Therefore, we have checked that tau satisfies all three conditions thus tau defines the topology. Next let us prove that. So, I should have said host of which is what we are going to check next. So, we have defined a topology on x hat. So, let us check that this topology is host of right. So, let us prove this host of. So, if x comma y are in x. So, we have to take any two points in x hat and we have to construct neighborhoods of these points which are disjoint. So, if x and y are in x right then we are done as x is host of. So, we can since x is host of there are two neighborhoods there are two open subsets u and v inside x. So, is that u contains x and v contains y and their intersection is empty. And since both these u and v are in tau by condition 1. So, therefore, in this case we are done right. So, let us consider the next case if x belongs to x and y is equal to p. This is the only other case we need to consider. So, then let w be an open subset small x in x such that the closure of w in x is compound right. Such a w exists because we are assuming that x is locally compound right. So, then x hat minus w closure satisfies condition is an and so is an open subset of right. So, clearly w and x hat minus w closure are disjoint. So, is an open subset of x hat containing p naught. So, thus we have found the open neighborhood of x hat an open subset of x hat containing p naught another open subset of x hat which contains w such that their intersection is empty. So, this shows that ok. So, next let us prove that x hat is compact. So, suppose we are given. So, x hat is equal to union u s. So, there is a sub sub sub sub sub sub u i u i naught such that p naught belongs to this u i naught ok. So, thus x hat minus u i naught is contained in x and is compact ok. So, now, so x hat minus u i naught this is contained in this union u i and this implies that x hat minus u i naught intersected x is contained in this i and i u i intersected x right. So, now, exactly as we have found saw over here right if u i satisfies b then u i intersected x is an open subspace of x right. We saw this above. So, this is if u i satisfies a then obviously u i intersected intersected x is equal to u i and it is an open subset of x. And on the other hand if u i satisfies b then as we saw above u i intersected x is an open subset of x right. So, but this is equal to x hat minus u i naught since x hat minus u i naught is contained in x yeah. So, therefore, this open cover since x hat minus u i naught is compact and it is a subspace of x and we have found an open cover of this inside x. So, this open cover has a finite sub cover. So, thus x hat minus u i naught is contained in finite union u i j this implies x hat is contained in u i j union right. So, this proves that next we have the obvious inclusion denote the obvious inclusion right. So, if u contain in x hat is open and u is of type a u satisfies a then clearly i inverse of u is simply equal to u which is open. On the other hand if u satisfies b then as we saw above u intersection x x not x j is open in x yeah. So, this implies the i inverse of u is u which is exactly u intersection x is open. Thus in both cases the inverse image of an open subset is open. So, this implies i is continuous as we already seen that x hat is host of and in a host of topological space a single point is always close is a close subset because the complement is an open subset yeah. So, this implies that i of x which is equal to x hat minus this close subset is an open subset. And the last thing we need to show that finally, we need to show i of x is dense in. So, for this u be an open set in x hat which contains. So, we need to show that u intersection x is non-empty yeah. So, if u intersection x is empty right. So, this will imply that p not. So, as this will imply yeah as x is equal to x hat is equal to x disjoint u in p not this will imply that u is equal to p not right that is the singleton set p not is open in x hat right. So, thus this will imply that x which is equal x hat minus this open subset containing x hat is a close subspace which will imply that x is compact. So, but this contradicts our assumption. So, thus u intersection x is non-empty which implies that i of x is dense in x hat. So, this completes the book with you. So, we will end this lecture here.