 So, welcome to this lecture. So, in the last few lectures we studied the topological property of being connected, connected and path connected and saw several examples. So, two examples which we did not see were. So, the problems of whether S o n, u n are connected or path connected will be pending will be taken up later. So, we need to develop a few more tools before we can answer the question of whether S o n and u n are path connected. So, however for the next few lectures we will study the property of being compact. So, let us begin with the definition of a host of topological space. So, let X be a topological space. So, we say that X is host of if for points x 1 distinct points x 1 and x 2 in X. So, we have our X, we have x 1 here and we have x 2 here. So, there should be two small open sets which contain x 1 and x 2 which are designed right. So, we can take any two distinct points x 1 and x 2 and there exists open sets u 1 and u 2 such that x 1 belongs to u 1, x 2 belongs to u 2 and u 1 intersection u 2 is empty and u 1 and u 2 are disjoint. If this happens for all pairs of distinct points x 1 and x 2 then we say X is host of ok. So, here are some easy exercises if X and Y are host of the product of topology is host of. So, let us see how to do this. So, let us say we have two points x 1 y 1 and x 2 y 2. So, now since x 1 y 1 is not equal to x 2 y 2. So, this implies either y 1 is not equal to y 2 or x 1 is not equal to x 2 ok. So, let us simply assume that. So, let us assume that y 1 is not equal to y 2. So, then our y 1 is here I am sorry y 1 is here and y 2 is here. So, since y is host of this implies there exists open sets such that y 1 belongs to v 1, y 2 belongs to v 2 and v 1 intersection v 2 is empty right. So, this is our v 2 this could be our v 1 right. So, then we know that. So, then x cross v 1 and x cross v 2 are open sets in the product of topology x cross y right. So, this is x cross v 2 this is x cross v 1 right. So, clearly x 1 comma y 1 is in x cross v 1 x 2 comma y 2 is in x cross v 2 and the intersection right. So, this shows that x cross y is host of ok. So, in the same way we can easily check that if x i for i and i is a family of topological spaces of host of topological spaces then the product. So, obviously every time we talk of the product as I said we are giving in the product topology is host of ok and imitate the proof of the proof of 1. The same proof as in the previous case will work and I will leave it as an exercise. So, finally, the third easy example or easy exercise is x is host of y containing x is a subspace then y is host of right. So, let us quickly see this. So, suppose these are x and let us say y is this. So, if you take any y 1 and y 2 then since x is host of there is a open set u u 1 containing y 1 u 2 containing y 2 such that u 1 intersection u 2 is empty right. So, then clearly u 1 intersection y contains y 1 u 2 intersection y contains y 2 and these are disjoint open subsets thus y is host of ok. So, the above exercises give us plenty of examples of host of topological spaces. So, we are mostly interested in host of topological spaces and we will almost deal exclusively with host of topological spaces for the rest of this course. So, for the rest of this course interested only in host of topological spaces. So, if nothing is mentioned then it is safe to assume that the topological space we are working with is host of ok. So, we have defined what host of means. So, now we are able to define compact compactness. So, let x be a host of topological space. So, we shall say that x is compact if the following happens. So, every open cover u i. So, what do we mean by an open cover? It means each u i we are given a family of open sets u i the union of all of them is equal to x. So, we are given any open cover and given any open cover has a finite sub cover that is there exist finitely many indices i 2 up to i n such that x can actually be covered by these finitely many open subsets u i j sorry. So, let us see an example of a topological space which is not host of I am sorry which is not compact. So, example r is not compact. So, to say that r is not compact it is enough to produce an open cover for r which has no finite sub cover. So, that is easy. So, what we do is we take any n n plus 1 we take n minus 1 by 4 and n plus 1 plus 1 by 4. So, then we can write r as the union over n in integers n minus 1 by 4 and n plus 1 plus 1 by 4. So, clearly if we take any finite subset. So, clearly any finite sub collection in this union it will not cover r any finite sub collection of intervals of the type n minus 1 by 4 comma n plus 1 plus 1 by 4 will not cover r. So, thus this cover has no finite sub cover thus r is not compact exactly. So, r is not compact. So, now we want to construct some very basic examples of compact spaces and prove some very basic results about compact spaces which is exactly the same as we did when we talked about connectedness. So, the first result we are going to prove is the interval 0 1 is compact. So, the proof is very similar to how we prove that 0 1 is connected. So, let us prove this. So, suppose we are given an open cover. So, our aim is to find a finite sub cover of this. So, we need to show that this has a finite sub cover. So, as before we let s be equal to those x in 0 1 such that this when we look at this interval the close interval 0 x is covered by finitely many u s. So, clearly 0 belongs to s as. So, for 0 to be belong to s what we need is 0 comma 0 this is just the single set 0 is covered by 1 u i. So, we can just take any u i naught which contains 0 or any u j let us say any u j which covers which contains 0 and then obviously, this interval which is just this 0 is contained in this u j. So, when I say covered by finitely many u i's I mean that this interval 0 x is contained in finitely many u i's union of finite union of u i's ok. So, let x naught be equal to supremum of x in s x. So, we claim that x naught is also in s. So, let us prove this. So, there is a sequence x n's converging to x naught with x n's in s right. So, let x naught be in some open set. So, we have 0 we have 1 and let us say our x naught is here. So, x naught is going to be in some open set u i naught. So, there is an epsilon, epsilon positive such that epsilon x naught intersected with 0 1 is containing u i naught ok. So, and since these x n's are converging to x naught all, but finitely many x n's are in this interval. So, all these x n's eventually all of them will land inside this small open neighborhood around x naught ok. So, we choose some x m in p epsilon x naught intersected with 0 1. So, this is 0 this is x naught this is our interval and let us say x m is here let us say this is 1. So, then as x m is in s this implies 0 comma x m is covered by or let us say is contained in a finite union of u i's finite union of u i's right. And moreover we also know that this interval moreover this interval x m comma x naught is contained in p epsilon x naught intersected 0 1 which is contained in this u i naught. So, this implies that 0 comma x naught which is equal to 0 comma x m union x m comma x naught which is contained in 0 comma x m union u i naught right. Now this interval 0 comma x m is contained in a finite union and therefore, 0 comma x naught is contained in that finite union along with u i naught which is therefore, also going to be finite. So, therefore, we conclude that 0 comma x naught is contained in a finite union of the u i's yeah. So, thus x naught is in s right. So, this proves our claim the claim that x naught is in s yeah. So, now next week claim that next week claim x naught is equal to 1 yeah. So, once again if x naught is strictly less than 1 then that would mean that there is some. So, we can just make a picture here once again and this is our epsilon and this is our x m that would mean that there is some epsilon positive such that x naught comma x naught plus epsilon is containing u i naught right. So, again so, this implies that 0 comma x naught plus epsilon which we can write as 0 comma x m union x this x naught plus epsilon is contained in a finite union of the u i right. So, this implies that x naught plus epsilon belongs to s which contradicts t assumption that x naught versus supremum. So, thus x naught has to be 1 x naught cannot be strictly less than 1 right. So, the since and since x naught is equal to 1 this implies that this 0 comma 1 we have already seen that x naught is in s. So, therefore, this implies that 0 comma 1 is contained in a finite union. So, thus there is a finite sub cover is contained in j equal to 1 to n some u ij this implies that. So, therefore, this open cover. So, this this proves that the open cover we started with has a finite sub cover thus this interval is compact. So, this proves this complete theory. So, we will end this lecture here.