 So, in the previous lecture we introduced compactness and we saw that the interval 0, 1 the closed interval 0, 1 is compact. So, just as in the case of connectedness we can ask what happens for products or subspaces. So, if x is compact what is compactness behave with taking products or taking subspaces. So, exactly as in the connected case. So, if x is connected and y is a subspace then y need not be connected. For instance we can take x equal to r and y we can test take this 2 points 0 and 1. So, clearly y is just the union of these 2 points. So, clearly y is not connected and in the same way. So, if x is compact y contain in x is a subspace then y need not be compact. So, we can take x equal to 0, 1 and y equal to the open interval 0, 1. So, there is a homomorphism from r to 0, 1 right and we saw that r is not compact. Thus the open interval 0, 1 since homomorphism obviously is going to preserve compactness. So, and we saw that r is not compact. We explicitly wrote down an open cover which does not have a finite sub cover. What this homomorphism is is left as an exercise ok. So, let us come to products and that is in the as in connectedness if x and y are connected to logical spaces we saw that their product is connected. So, in the same way we will prove that if x and y are compact to logical spaces then their product is also compact. So, the crucial ingredient for that we need is this lemma which is called the tube lemma. So, let y be a compact to logical space let x be any to logical space let w contain in x cross y. So, x cross y is obviously always given the product topology be an open set containing x cross y. So, then there is an open set u contain in x such that x is in u and u cross y is containing w. So, let us make a picture of this lemma and explain what it is saying and why it is called the tube lemma. So, this is our x and this is our y, y is compact and here we have this is x cross y and we have this open set w some open set which contains x cross y. Then what this lemma says is there is a small neighborhood u around x. So, is that this entire disk around x cross y is contained inside w it contains this entire tube w contains. So, this region is w and this is u cross y. So, let us see how to prove this lemma. So, given any point y in y there is a basic open set. So, given any point in y in y. So, we have the point x comma y which is in x cross y which is contained inside w and w is open in the product apology which implies that implies there exists open sets u y contained in x and v y contained in y such that this x cross x comma y this is in u y cross p y and this is completely contained inside. So, we are taking any point over here this point is x comma y. So, we are just saying that this is open neighborhood. So, this is u y this is v y. So, u y cross v y is contained. So, clearly and this happens for every y. So, thus we can write y as a union of y in y v sub y. So, since y is compact this has a finite sub color. So, we can write y as j equal to 1 to n v y j we can write this as a finite union. So, let u be equal to intersection of the u y j's. So, note that each of these u y j's it contains x right. So, therefore, u contains x each of these u y j's is a open set containing x we are taking a finite intersection. So, therefore, u is an open set such that an open set of x an open set in x the point x. So, then u cross v y i is contained in u y i cross v y i which is contained in w and this happens for all i. So, this implies that union i equal to 1 to n u cross v y i is contained in w, but this union is simply u cross union i equal to 1 to n v y i this is contained in w, but the union of the v i i is a simply y. So, this completes the procedure. So, using this lemma which is the crucial ingredient let us prove the theorem on x theorem let x compact to logical spaces then the product x times y is compact. So, let us prove this suppose we are given an open cover. So, then fix an x. So, then we have x cross y which is contained in x cross y. So, as x cross y. So, note that consider the map consider the map y to x cross y which sends y to x comma y right. So, this map is bijection factors through x cross y and this has this subspace to project. So, it is easy to check. So, the map from y to x cross y is obviously continuous because the image of y. So, let us call this f let us call this f sub 0 let us call this i that f sub 0 is a homomorphism. So, this shows x cross y is and now. So, we can write x cross y is equal to union i and i intersected w i right and since x cross y is compact this implies that this has a finite sub cover which implies that this x cross y is actually contained in some finite set w i j. So, by the previous lemma. So, let us this finite sub cover let us write this as union i sub x contained in i or rather i in i sub x i w i yeah. So, we. So, x cross y is contained in some finite sub cover we just collect the indices in that sub cover and put it into the set i sub x yeah. So, here is a finite sub. So, now by the previous lemma there is an open neighborhood u sub x of x in x such that. So, what are the previous lemma say? So, let us look at the previous lemma. So, y is compact and if x cross y is contained in w then there is a small neighborhood u of x such that u cross y is also contained in w. So, we will use that u sub x cross y is contained in union. Now, this can be done for every x. So, thus when we do this we get an open cover u sub x. Now, again as x is compact this has a finite sub cover. So, we can write x as u x sub j yeah. So, then x is equal to I am sorry x cross y u sub x x j cross y which is contained in j is from 1 to n each of these is contained in this indexing set w i right and now the index set is finite yeah. So, each as each i sub x j is a finite set subset of i this implies. So, let me just this is equal to union i in this index set is finite. So, thus x cross y. So, thus we have found a finite sub cover right. So, this sub cover is finite because this index set over here that is a finite index set. So, this completes the proof of this here. So, as a corollary of this we see that the closed interval 0 1 I mean when we take product of it n times is compact. So, next let us make some observations about compactness. So, proposition a closed subspace of a compact space. So, let us prove this. So, let y contain in x be closed and suppose. So, our aim is to show that y is compact. So, what we have to show is given any open cover for y it has a finite sub cover. So, given any open cover for y we will construct an open cover for x and from that we will deduce that y is compact. So, let y be equal to union i in i w i b and open cover for y right. So, since y has a subspace topology. So, thus there exists u i containing x open such that each w i is equal to y intersection u i right. So, this implies that y is contained in u i y is contained in this collection of open sets this open in x. So, suppose this is our x and let us say this is our y. So, first we are given an open cover for y this we extend to an open cover for x I am sorry we cover this like this. We find open subsets we find a collection of open subsets in x such that the union contains y right and now since y is closed we may write closed in x we have x minus y is open x right. So, this region is open. So, we can write we may write x as x minus y disjoint union y this containing x minus y union. So, this is open in x and each of these is open in x right. So, x is contained in this and all these are of course contained in x. So, this implies that we have gotten an open cover for x. So, since x is compact this has a finite sub cover. So, let us say the finite sub cover looks like this I mean this may or may not be present, but there is no harm in throwing it in throwing an extra open subset into it. So, now intersecting both sides with y we get y is equal to this is disjoint this empty y intersected with x minus y is empty we simply get this equal to w i j. So, thus the open cover we started with has a finite sub cover. So, this proves that. So, we will end this lecture here and in the next lecture we shall show that a closed subspace of R n is compact even only if it is closing down.