 So, in the last few lectures we discussed the notion of compactness, we discussed of compactness. However, as we saw there is an important there are lots of important spaces which are not compact. However, note that several important spaces which are not compact for example, R n. These places are however, locally compact, locally compact which is the next property we shall study ok. So, let us define locally local compactness ok. So, let x be a topological space. So, we say x is locally compact if every point x in x has an open neighborhood u such that u closure is compact right. And example it is clear that is locally compact because we can take any point and we can take a disk of let us say it is one around it. So, the closure will be this closed disk and this is obviously closed and bounded in R n and therefore, it is compact right. And so, one of the main propositions a very useful proposition we are going to prove about locally compact spaces is the following. So, the following proposition is very useful x is locally compact if and only if for any pair x comma w where w is an open subset containing x we can find an open subset v such that x is in v which is contained v closure and v closure is contained in w v closure is compact. So, in other words given any x and given any open subset w which contains x we can find a neighborhood v of x such that v closure is compact. So, this is v closure ok. So, let us prove this it is clear that if this condition holds for any pair x comma w we can find v such that x belongs to v v closure is contained in w and v closure is compact then x is locally compact right. We simply apply this condition we simply apply this condition to the pair right. So, when we apply this condition to this pair we will get that there is an open subset u which contains x and u closure is obviously, going to be contained that inside it x and u closure is compact. So, this proves that. So, that is precisely the meaning of being locally compact. So, let us prove the converse let us assume x is locally compact ok. So, now recall the following assertion. So, which we have already seen before. So, let y contained in x be a compact subspace ok this is more general. So, consider the following more general assertion. So, let y contained in x be a compact subspace x be a point in x minus y then there are open sets subsets c and d y is contained in c x is contained in d and the intersection c and d are disjoint. So, let us see how to prove this. So, this is our x right and this is our y and this is our x right. So, for each point y in y we can find a neighborhood since x is host of we can find a neighborhood u sub y of x and v sub y of y such that u sub y containing x v sub y containing y such that u sub y intersected v sub y is empty right. So, cover. So, we can cover y as we can write it as a union of all these v sub y's right. As y is compact this implies there is a finite sub cover and then we let u be equal to the intersection of these u y sub j's. Each u y j contains x. So, this implies that x belongs to u, u is a finite intersection of open sets and therefore, it is open and right. So, it is not we can cover. So, y is contained in this sorry y is going to be contained in this and since y is compact and from what we had seen in one of the earlier lectures if we can put y into a collection of open subsets then we can find a finite sub collection of that collection such that y is contained in that. So, x belongs to u let us look at. So, let d be equal to u and let c be equal to this union j equal to 1 to n v y j's right. So, this v y j's could be something like this. So, then c contains y and d intersected c is equal to u intersected union j equal to 1 to n v y j this is equal to union j equal to 1 to n u intersected v y j, but this contained in union j equal to 1 to n u y j intersected v y j which is empty. So, this proves our general assertion which is something and we had seen the same proof before earlier as well. So, we will use this general fact ok. So, suppose now we return to the proof of our assertion. So, we are assuming that x is locally compact and we want to show that x has this property. So, suppose we are given where w is open and x is contained in w. As x is locally compact this implies there exists a neighborhood open subset such that u closure is compact. So, this is our x this is w and let us say u closure is this. So, consider the closed subset y defined as u closure minus w. So, u closure I will write it as minus w, but this same as u closure intersected with x minus w. W is open. So, x minus w is closed. So, this is and we are intersecting two closed subsets. So, that is a u closure is compact and compact subspace of a host of spaces closed and therefore, the intersection of these two is going to be closed. So, this closed subset it looks something like this right. So, from u closure we are removing w right. So, this region is y and x does not belong to y right. So, thus a closed subspace of a compact space since y is a closed subspace of u closure which is compact this implies that y is compact. So, thus using this general result that we proved yeah. So, thus there is there are open sets such that y is contained in c, x is contained in d and c intersection d is empty right. So, c could be some open set like this. So, c contains y right and d could be some open set like this. So, we can replace d by an open subset which is smaller than d as long as it contains x yeah. So, we may replace d intersection u right. So, u contains x. So, therefore, d intersection u also contains x and assume d is contained inside u. So, so, our set d is going to be contained inside u right. So, this implies that d closure is also going to be contained inside u closure and as u closure is compact this implies d closure is also compact because it is a closed subspace of a compact space yeah. So, now no point of y the closure because if y is a point in y then there is a neighborhood of y namely c such that c intersection d is empty right. So, this implies that y does not belong to the closure which implies the closure intersected y is. So, as the closure is contained inside u closure and the closure intersection y is empty this implies the closure is contained in u closure minus y and as y is equal to u closure minus w this implies that u closure minus y is equal to u closure intersection w which is equal which is contained in w right. So, thus we have found open set d such that x is in d, d closure is contained in w, d closure is compact ok. So, this completes. So, this proposition is very useful and this this set theoretic check is very easy and I will leave it to you as a simple exercise. So, ok. So, next we are going to talk about so compactifying locally compact spaces. So, note the space the open interval 0 1 is locally compact. So, that is clear because I mean if we take this interval 0 1 then given any point x we can find a small closed interval around x and we know that this closed interval is yeah homomorphic to 0 1 which we proved is compact right. Now, notice that it is contained. So, this interval 0 1 is contained as a dense open subset it is obviously contained in the interval 0 1 and also in s 1 right. So, to see this note that 0 1 is homomorphic to r and using the stereographic projection yeah r is homomorphic to s 1 minus this north pole. So, r is homomorphic to s 1 minus this north pole n. So, this implies that and this is embedded inside s 1. So, although we can compactify a locally compact topological space in many ways there is one compactification which is special among all these and which is now which is what we are going to explain now yeah. So, we want to compactify. So, thus they can be compactify a locally compact topological space. However, one particular compactification is very special we shall next describe that ok. So, we will do that in the next lecture. So, we will end this lecture here.