 Welcome to point set topology part 1, module 38 today. So we are going to study compactness and linear look properties in the larger topic of smallness properties of topological spaces. You have to start with the definition, take a topological space x and a subset a. Recall that a family UI of subsets of x is called a cover for a, if a is contained the union of UIs. So this family will be called an open cover if in addition each member UI is an open subset of x. By a sub cover of a given family, the family must be covered, sub cover means what? For a given set a, we mean a sub family of UI which is a cover for a, it is a sub family and it is a cover so that will be a sub cover. Now we make a proper definition, subset a is called compact if for every open cover of a, we have a finite sub cover for a. Likewise we say a is Lindelof if every open cover for a has a countable sub cover. Remember countable includes finite therefore by this definition compact will be automatically Lindelof but Lindelof may not be compact because countable includes counter infinite also. So compactness Lindelof-ness but Lindelof-ness may not imply compact. However because of the similarity these two concepts can be studied together to some extent. Lots of more properties will be there for compact spaces which are not true for Lindelof spaces in general. Most of the properties for Lindelof space will be automatically true for compact space, compact space or Lindelof space. Ectopological space is called compact or Lindelof if it is compact or Lindelof as a subset of a given space. So if the whole space if x is the whole space a itself is equal to x then what is the meaning of that? a being a compact subset is simply a compact space that is what it is. To immediate that a subset is compact if you know only as a subspace as a space as a take subspace topology etc. if a space is compact or Lindelof. These two concepts are the same because the subspace topology if UIs are open subsets in a given a UI intersection a those will be open subsets inside a that will form a cover whenever that sub cover is there from a intersection UIs the corresponding UIs if you take that will be a sub cover for the original thing and so on. So going to the subspace and back to the space is immediate so compact space is simply as being a compact subspace of a given space that is no problem. Also it is immediate from the definition that a closed subset of a compact space see every time I am saying something of a compact space it is true for Lindelof I am writing it in the bracket here. So here a closed subset of a Lindelof space is limited low a closed subset of a compact space is compact. Similarly it follows that a finite union of compact spaces is compact when you take Lindelof of course finite you know Lindelof space compact you can say more namely any countable union of countable union of Lindelof space is also Lindelof. Now let us do this little more carefully why take all open subsets let us see whether we can do it some economically start the topology space suppose b is a base for the topology then x is compact or Lindelof respected if we do not live every cover of x by members of b admits a finite sub cover admits a countable sub cover respected instead of every open cover we are having now covers which are coming from members are coming from a given base so family UI is a sub family of b that much I can get you if that is true for all sub families of b itself then it will be true for all open covers so this is the lemma here let us see how this is true clearly if you have members of b they are also open they must admit finite that is all right but why that is sufficient is what we have to see to start with the condition that uj be an open cover for x I have to produce a finite sub cover under the assumption that for a basic open subset this set this is true okay for each point x in x this x will be in one of the uj's that j will depend upon x so I write the uj x select one x is inside ujx for ujx being jx being 1000 these j's okay but by the definition of a base what happens there will be a member of this curly b namely bx such that x is in bx and bx is contained u of bx then you vary x all these bx's form an open cover for x but that are members of this curly b only the basis okay that will admit a finite sub cover this bx1 bx2 bxn's so these will cover the whole of x but now each bxi I can replace by uxj's ujx's so they will cover the entire of x so I have got a finite sub cover exactly if you if you allow this one to be countable here instead of ironed 1 to n I want to infinity 1 to infinity a countable cover then you get little of property okay now another important thing is to investigate what happens under continuous maps okay the image of a compact receptacle in the low space under a continuous map is compact respectively in the low so I will now on won't say this in the low part at all this will be automatic and it will be written down here okay okay so the proof will be also exactly similar and so on so let us look at how why this is compact and so on you start with a contained zx, p contained inside y, x and y are topological spaces and f from a to b is a continuous function such that f of h b that is subjective continuous function okay I have to show that if a is compact then b is compact okay assuming that a is compact I am sure b is compact okay b is a compact subject of y so take an open cover of y take an open cover of b open subsets of y f inverse of those will open subsets of x and they will cover a because f of a is inside b f of a is actually equal to b right so start with the uj ui i inside j open cover of b then each f inverse of ui intersection b is open in a and a equal to i i range on f inverse of ui intersection b okay i f inverse of ui itself I will take then intersection with a is the same thing as take a f inverse of ui intersection b because f is not defined on the whole of x I do not say f inverse of bi I have to take a intersection here f may not be defined on the whole of x that is not necessary f is defined only on a that is why I am writing this otherwise it is just f inverse of ui okay so this is an open cover now f in a is compact so I will get a finite sub cover okay therefore the finite sub cover so i is a sub finite subset of this is that a is inside union of bi's i belong to i what are these bi's bi's or bi these f inverse of ui's okay I have put f inverse of ui intersection b as bi intersection a bi's because these are open subsets of a means what they are bi intersection where bi's are open zx okay if f for defined on the whole of x I could have taken f inverse of ui as bi here I have to separately say what these bi's are okay so this will admit a finite sub cover here a intersection bi it follows that b will be inside the union of now corresponding ui's here where this i will run only over i okay after this bi's are just damaged here to show that this is an open subset in here so an immediate consequence is that its compactness and end end over topological properties what does that mean that means if you have homeomorphism from x to y if x is compact then y is compact and conversely so that is the meaning of topological property you don't need homeomorphism you just need continuous surjective man a is compact f of a is compact a is length of f of a is length of so that will show that compactness and end end ofness are topological properties okay so notice that it just means that the compactness of subset is a topological space does not depend where a is taken as a subset okay a may be a topological space it may subspace of some y it may subspace of some x it may subspace of some the moment a is compact in one of them it will be always compact as a space itself and then it will be compact in every other every other say so it is independent of being a subset of something okay the same remark holds for little of property connectivity path connectivity etc if you think they are topola once the something of topological properties the topological space where it is embedded is immaterial so now we have proved a theorem about quotient maps with strong property then if each fiber is connected and the base space is connected then the total space is connected x to y is a surjective map all right y is compact y is connected when this x is connected so there was a converse theorem this 3.13 similar to that we may expect if y is compact and all the fibers here f in most of y's are compact will x be compact okay the answer is no even under a stronger assumption that f is an open mapping open surjective is quite strong you can assume it is closed mapping also closed surject even then this will not be true namely a very simple example is here take the open interval minus 1 to closed one it is half closed minus 1 open plus 1 closed take this open interval to 0 1 closed interval what is the map eta x equal to x square x going to x square that's surjective okay is it an open mapping if you have an open interval here what will be the image of this open interval open interval open interval or half open on this side one closed what will be the image of those will that be open subsets of of 0 1 you see you can't take uh some epsilon some r here less than 1 then the open interval you have to be open subset after you open interval itself okay but if you take one close one also then you can take the closed interval also that will be still an open subset of the half closed interval the but image will be a half closed interval here half open interval here so that will be open subset of of 0 1 all right so eta x is both an open mapping as well as a closed mapping it is surjective inverse image of every point here is just one point or two points depending upon if it's zero inverse image is zero any other point there will be two points here okay there are two square outs so fibers are all compact the image is a closed interval that's compact but we know that half closed intervals like this are non-compact so very easy example okay in fact you can produce build on this one you can produce many such examples indeed in application this is what causes a problem so you have to be careful of exactly precise in this one what you have actually done is you look at the full thing here you take minus 1 to plus 1 close interval that will be compact that will include all the solutions what you have done is certain solutions are missing here right so that is the one which causes problems okay this precisely the way how the the expected answer here y is compact fibers are compact y x may not be compact okay so keep this example in mind so one of the consequence of this example is that we cannot do compactness of product of two compact spaces in a similar way as we have done it connectedness remember connectedness x cross y x and y connected x cross y connected which can be proved in different ways simpler way is to take the quotient map from x cross y to y apply that theorem y is connected the fibers are x cross singleton wise right for each y so they are all connected because they are a homomorphic place therefore the the entire thing with the total space with x cross y that will be connected that kind of proof is not possible here is what this example says okay so we have to do a little more work here and that gives us a beautiful theorem here called valance theorem which will be used elsewhere also okay so what does go through this valance theorem x be a compact space then for every topological space y and a point y inside y and an open subset v of x cross y such that this v is a neighborhood of is capital x cross singleton y suppose if you have this situation so x is a sub x cross singleton y is a subspace right of x cross capital y the v is a subspace which is a neighborhood of is open open subset neighborhood x cross then you have some kind of a uniform neighborhood is what it is provided this x is compact compact so what is that uniformity there exists open neighborhood n which is I am writing n y depends upon y inside y such that this entire x cross n itself will be contained inside every neighborhood has a unique single neighborhood of this point because that whole x cross n will be contained inside v this is the kind of uniformity uniform neighborhood comes it is like this n little y is uniform epsilon if you write epsilon y here that will depend upon the y point y here now this capital n does not depend upon any of them for all x cross y this is this is the same thing that is what it says okay so the proof will actually give you better view of why I call that uniformity and so on for each x comma y as x varies x cross y inside contained inside v by definition of product apology you will get open subsets ux and vx if you have matrix basis you could have get epsilon same epsilon x epsilon y and so on so now I have to write no metric here some open subsets ux vx and x cross y x as well as y sorry not x cross y respectively such that if you take the product x y will be x y will be neighborhood of ux cross vx will be neighborhood of x comma y and that neighborhood is contained inside the open subset v that is the meaning of an open subset in the product space right so these are basic open subsets here by compactness of x okay what happens we will get finitely many x i's such that x is contained inside i read 1 to n ux i's because as x varies over x ux will cover the whole of x so they will get a finite now put n equal to this will depend upon the second coordinate y for each of them the y is fixed here that is why I am not writing uxy vxy and so on I am just writing ux vx n equal to n y which is intersection of the corresponding vx1 vx2 vxn you depending upon ux1 ux2 vxn here right so take the intersection finite intersection of open subsets will be an open subset all of them see vx is for all vx they are neighborhoods of y okay so intersection of all of them finite many of them that n is an neighborhood of y so we claim that x cross n is inside v now so this is a typical way you take the union here you take the corresponding intersection there okay so why this is contained inside v look at any x comma z inside x cross n z is in the intersection means z is in which vx i's okay so choose an i such that x is in one of the ux i's then ux i cross vx i is inside v so x comma z is inside v that is so okay so vales theorem says that if you have a slice it is a compact space inside a x cross y take a slice and x cross little somewhere a neighborhood of that you can take uniformly okay one single neighborhood of y cross the whole of x will be contained in that neighborhood so this will be very useful in proving many other things also so we are going to use that one over here in the proof of product of two compact spaces compact now you notice that I have stopped telling about lindeloff if you take lindeloff what you get you will get some somewhere countable infinite things may be get intersection of infinitely many opens of such may not be open so that is where these results will not be true for lindeloff spaces okay so we are we are not having those results for lindeloff spaces okay so x and y are compact then x cross y is compact so how does it prove that start with an open cover of x cross y for each y inside y singleton x cross y is compact right and this ui's will cover that also therefore you will get a finite subset j y that depend upon little y contained inside j subset x cross y is contained in the open cover which is sub sub family or depending on j y only y side is at j y union of all those ui's so let us call that open set v y it depends upon y now this is true for every y x cross y since that vi if i range y that will cover the whole of x so before that before that since x cross y is contained in that vi i am going to use valet's theorem here now by valet's theorem there exists an open subset n y of y such that x cross y is contained inside x cross n y contained inside v y i don't have to take this funny v y but i can take now n y and forget about x so these n y's as y varies over capital y they will cover capital y that is compact so i will get a finite even y 1 y 2 y k such that y is contained inside union of n y i i 1 2 k okay now take i equal to the corresponding j y 1 j y 2 j y k take the union of those so that is a subset of capital j okay each j y i is finite set and this is the finite union so the actual i is a finite subset of capital j the claim is that if you take now only u i's inside i this finite set that will cover the whole of x cross y and that is very straight forward start with any x comma y in x cross y this y will be in one of the n y i's because y is contained in n y i's here then x cross y will be contained in x cross n y y but that is contained inside v y j corresponding y i j y i whatever y n y j i okay v y j okay but v y j see v y j's union now is u i right where i is in j y j here okay one single set here j y j and take those unions so that is just a union of under j y i's okay so that is definitely contained in all the u i's where i rings over capital i where capital i is actually union of all j y i's okay so i have shown that x cross y is contained inside this union y is arbitrary so the whole x cross y is contained inside that that's all so product of two compacts space is compact is what we have proved immediately it follows that finite product of compact space is compact so if you have finite product like this then it's compact okay and if x is compact each xi is compact because they are we can take the projection maps so you have a product space like this is compact you spend on this each factor is compact okay so for it immediately you want to prove this for arbitrary products that is for techno's theorem okay so there is two for infinite product as well goes under the name techno's theorem but for that you will have to wait okay but let us do a little more of compact space here techno's theorem is a bit special thing we will do it's on its own at its own time after one or two more modules it x be a compact product space f1 contain is f2 contain is monotonically decreasing it's a nest of a of sequence of non-empty closed subsets okay remember such a thing we had in cantor's intersection theorem wherein we had a complete metric space then the diameters were going to 0 etc right so here there is no metric no diameter and so on all that I assume is that instead of metric space complete matrices a compact logical space then the intersection of all these f is is non-empty so here the proof is surprisingly much much simpler than cantor's introduction theorem even that was not very difficult anyway all that I have to do is apply de Morgan's law f i's are closed compliment of f i's are open f i's are decreasing compliments of f i's are increasing union okay intersection is non-empty is what we want to show is the same thing as that union is not the whole of x is the same thing intersection is non-empty by de Morgan's law if the union is the whole space what you have got is an increasing sequence of open subsets such that the union is the whole space x is compact therefore it has a finite sub cover an increasing union of open subsets the finite sub cover means at one of the the maximum one of maximum one you take that itself will be x what does that mean one of the ui's itself is x which means when you go to compliment those one of these f i's must be empty but I have taken an unempty close subsets okay so that is the full proof I have just given you a prior de Morgan's law rest of them you must be able to do anyway now I have told you all right so next time we will do some more things about compact metric spaces okay thank you