 with the topological aspect of topological manifolds. So today we shall study paracompactness to begin with. Essentially we are more interested in the partition affinity part of paracompactness rather than all those topological conditions of locally finite open refinement and so on and so on. So we will directly prove the existence of partition affinity together with horse dwarfness, the paracompactness is equivalent to existence of partition affinity. This is a general aspect. So we will not lose anything. And if you have seen this proof in the study of your differential topology like every open subset of Rn, every subset of Rn is paracompact. The proof is much simpler here. There what you do doing is not only you will be studying paracompact but you will be studying smooth partition affinity. So proof is if not difficult it is even slightly less easier you may say than in the case of a smooth partition affinity. But what I wanted to say is proof more or less repeats the steps that you have been going through in the case of a subset of Rn is paracompact. So let us do those things, rituals here. Start with the topological manifold. The statement of this lemma is that there exists a nested sequence of open subsets Wj in X such that each Wi closure is compact, Wi closure will be contained inside Wi plus 1. Wi plus 1 closure will be contained at Wi plus 2 and so on. And X itself is the union of Wis. This itself, this lemma itself would have been totally easy in the case of Rn because you can then you take just ball centered at the origin ok ball with radius 1 ball with radius 2 and so on you are done. But suppose you want to prove this one for any open subset of Rn then whatever we are doing here is the same proof there also ok. There are many ways of doing it not just one but in this case one of the proofs which you work very nicely for topological manifolds. So I am going to prove that proof here ok. So what we are using here use a second countability ok. It follows that there is a countable family of homeomorphisms Fi from Ui to Rn where Uis are open in X and X is union of Fi inverse of open balls ok open balls. The domain of Fi is may be larger ok and the image of this one I will assume that they are larger than the single open ball of radius 1 B10 denotes the open ball of radius 1 inside Rn ok. There is always such an atlas out of which I can take a countable cover because every second countable space has a countables whenever you have an open cover it admits a countable sub cover that is lindel of property ok. Now let us have this notation B10 is what is the open ball of radius 1 at the origin. So more generally BRX is set of all Y in R is set norm of Y minus X is less than R open ball of radius R centered at X ok. So put Vi equal to Fi inverse of B10 these Vi's cover the whole of it. Uis may be larger now Vi's are Fi inverse of B10 ok. Each Vi is an open subset of it. The closure of Vi bar and then Vi bar it will be Fi inverse of B1 bar it is a closed ball ok. So that is compact. So inverse image under this homomorphism is compact ok. Homomorphism closure of the open ball B10 ok. So start with W1 equal to B1. B1 bar is compact ok. Inductively suppose we have defined Wk to satisfy the property 1 and 2 ok. Suppose you have constructed W1 W2 Wk with that property you select finitely many members of Vi that covers Wk bar because Wk bar is covered by these Vi's and Wk bar is compact therefore there are finitely many Vi's which Vi's I do not know to collect them. Let Wk plus 1 be the union of these members and Wk plus 1. These members Vi's cover Wk bar but Wk plus 1 may not be there. So put Wk plus 1 also. In other words to begin with we had V1 when we second try we will be definitely have V2 also at the K plus 1 stage we definitely have Vk plus 1 along with there may be many others Vi's which might have come. If Vk plus 1 has already come there is no need to put it extra. So there is no harm in putting it again that is all. So check that the property 3 holds namely now exceeds the union of all the Wi's. Obviously Wk bar is contained inside this Wk plus 1 because it is a union of all Vi's that itself covers it one more Vk plus 1 ok. Being a finite union of all these and each of them meaning compact the closure will be compact. So property 2 is obvious. Property 3 is because at each stage we have been careful enough to include V1, V2, Vn, V and so on. The union of Vi's is the whole of X right. Therefore property 3 also. So that is the limit ok. Having done that we will attack partition of unity. So let us go through the definition here. I have given full definition here ok. Let X be any subspace of a manifold wire. So we are going to prove that every subspace is para-compact which is very stronger result than saying that the space is para-compact ok. X be any subspace of a manifold wire and U alpha be an open covering for X. Then there exists a countable family theta j of continuous real varied functions on Y with compact supports such that all these this is part of the definition of para-partition affinity ok. They are all taking values between 0 and 1 in the closed interval ok. And they are all defined on the whole of Y. The X is subspace of Y. The domain of these theta j's is the whole of Y ok. For each X in X there exists a neighborhood NX of X such that only finitely many members theta j's are non-zero on NX. So this is called locally finiteness at each point of X. So I am not claiming the local finiteness on Y. That is very important. Note this carefully ok. The third condition is for each j support of theta j into section X is contained in one of the members of U alphas. So now relating with the given covering. Otherwise there was nothing to. So theta j support of theta j intersection with that is contained inside one of the U alpha name U alpha j ok. The fourth condition is some total of theta j is one which makes its partition affinity. The third condition is support of theta j contained. This one means this partition affinity subordinate to the given covering ok. I am just recalling this one. I presume that you know paracompassiness. I presume that you know partition affinity. Because at least in part one I have done all this. Taking U equal to U alpha ok. It is an open covering right. If you take U equal to U alpha union of U alpha that will contain X ok. Instead of doing these things on X I can do it on U on the whole of and then restrict it to X. Because functions are going to be defined on the whole of Y ok. So it is enough to prove the theorem for open subsets of Y ok. So that is the simplification first thing. Therefore you may assume X itself is open environment. By previous lemma X is the increasing union of a countable family of open subsets Ki. I written one with the closure of each Ki being compact. So I am using lemma here ok. Every open subset of a manifold is a manifold. So manifold can be written like this is the previous lemma. So I can write X as a increasing union of open subsets. Each of them such that their closures are compact and contained inside the next one. Just for completion of this just for argument sake put k naught equal to empty set. This will be needed for us. You can also put k minus 1 also empty set ok. We shall first construct a family B ij double indexed ok of open subsets of X with homeomorphisms B ij to the unit ball ok dn such that if we put half B ij this is just definition notation half B ij is phi ij inverse of half the ball ok half the radius is half ok not you have not cut it laterally ok radius of this one is half of the ball psi ij inverse take that part then these half B ij's themselves cover X ok. These are open subsets at each point. So it is a covering alright and is a locally finite open refinement of u alpha. So this is what we are going to do we have not yet done it. This is we shall first construct a family ok. Locally refinement means what? Definement means what? Each B ij's contained inside u alpha one of the u alpha's. Local finite means what? This condition namely at each point there is a neighborhood nth which will intersect only finitely many members of these B ij's. So once you have done that inductively suppose B ij's have been constructed for i less than or equal to k for b1 for w1 you can always construct. Actually that is why I have put k0 equal to empty set you can start with the induction hypothesis here ok it is no problem. B ij have been constructed for i less than or equal to k so that half of them cover the kk bar up to k you have done ok. For each point X in the complement of kk inside kk plus 1 bar see kk is an open subset kk plus 1 bar is closed subset. So this complement is a closed subset of a compacted this compacted you can so you can choose a neighborhood wx contained in some member of u alpha because u alpha's cover them and not intersecting kk minus 1 bar ok. So kk contains kk minus 1 bar so I have thrown away kk itself kk minus 1 is far away and that is a closed subset there will be a distance and so on in the Euclidean space alright but you do not need all that you have to just see that this is it is a compact subset and let us say closed subset away from that right. So you can always do an open subset which is smaller than that one here which one ok. You can further assume that there is a homomorphism psi x from wx to b10 ok you can take always a smaller open subset and then declare that that open subset is homomorphic to whole days such that psi of x goes to x goes to 0 that is psi x is a coordinate chart at that point. Now psi x inverse of half ball is an open cover for this kk plus 1 minus kk for each point I have done that and this is a compact set. So I get finitely many x1, x2, xr such that this complement is contained in the union of psi x or half balls ok. This is what I am going to do in the inductive step. So if you have doubt about what where the inductive step so put k equal to 0 this is nothing but this is just first stage I am doing kk plus 1 ok. All these not intersecting etcetera is emptyly you know vacuously true that is all. So I have taken kk plus 1 bar is a compact set I have taken covering of that one that is the starting point. Logically I can should have started with k not equal to empty set and then there is no problem. So this is the inductive step. Now I will put the new family kk plus 1 comma j as psi xj inverse of these things I do not know how many of them so psi xj of v1 bar. So these are the extra balls inverse image of these balls I am adding to the original list. I am calling them as bk plus 1j. It follows that kk plus 1 bar is contained inside this family now because whatever is not covered by not covered kk those things they have been they are all here now namely this new one. So they are they will cover kk plus 1 bar. Everything is in kk is already covered by induction hypothesis. So this should be covering for kk plus 1 bar. So for each compact thing we have selected these opens of sets which are smaller than one of the members of this one namely they are what they are refinements of the original covering and now as kk range is overall these will be a covering for whole of x now because union of kk plus 1s are all kks are all is the whole of x. The only difference is in the lemma we have w say here they are kis that is the difference. So inductively the construction of bij is over. Now this is a open refinement of the family u alpha that is what I have told you because each time you are selecting they are contained inside u alpha right contained in some member of u alpha here and it is a covering for the whole of x. Now given x belong to x suppose x will be in one of the kks because x is union of all kks as kk is increasing family okay in one of the kks okay take nx itself kk itself that is an open subset I want to find a neighborhood with some property. So take that neighborhood to be kk itself then this nx does not meet any of the bij's if i is bigger than k plus 2. So in the k plus 1 in the k plus 1 stage remember all these things did not intersect kk minus 1 okay at kk plus 2 they will not intersect kk so they will not intersect nx. So how many of them will you intersect nx only up to k plus 1 for each k there are only finitely many j's so there are only finitely many these balls which will intersect nx where nx is kk itself okay so this proves the local finiteness all right finally let beta from b10 to i be this function in the case of rn you can actually get smooth maps and so on so I am not interested in smooth maps so I am just writing minimum of the distance between x and x and sn minus 1 okay and minimum distance x and x and minus 1 and half okay you take that and multiply by 2 so that this will be always between 0 and 1 and it is a continuous function okay you could have taken lots of suspensions anyway so I have taken one continuous function. So the idea is if you take the half the ball what is the value take half the ball here the sn minus 1 is the boundary of dn then distance between x and the boundary will be bigger than distance between x and sn minus 1 will be bigger than half right so minimum is half half multiplied by 2 is 1 so on the half ball it is identically 1 okay when the when the x rate is sn minus 1 distance is 0 so minimum is 0 so twice that one is 0 so as the x approach is sn minus 1 it is it is b it can become 0 actually sn minus 1 is precisely the set of zeros of this function and half ball it is actually 1 so this is the bump function which is not it may not be smooth because it is defined by distance function and so on so that is why you have to be careful but inside acidity is smooth but when you go to the topological space there is no question of smoothness so do not worry about smoothness so put eta ij beta composite psi ij so first takes phi ij it will come into b10 and then composite beta so these are okay where bij are the homomorphism chosen earlier extend eta ij by 0 over the whole of y because these eta ij's are such that on the closure they are identically 0 so beyond that we extend it as 0 that will be a continuous function okay now re-index this family by single letters why bother about double indexing it right in the beginning I could have done that because it is a countable anyway for each i fixed i there are finitely many j's and the what is i i range from 1 to infinity so it is just a countable family re-index it by eta j it easily verified that this family satisfies property 1 2 3 4th and 5th whatever let me just see whatever properties 1 2 3 first one is the values are between 0 and 1 that is what I told you because they are all something composite with beta and beta takes values between 0 and 1 okay and we have verified that each bij this family is locally fine and outside bij all these eta j's are 0 so it will be it will have satisfied this property and support of theta j intersection x6 u alpha because they support each inside u ij's so this is automatic okay the 4th condition we don't know yet okay so that is what we have to what we have done is these these eta j's to take larger values some like some some total okay not individually so that is what we have to be it is easily verified that is 1 2 3 now define eta i could sum up all the it has okay why this makes sense because it is locally finite in a neighborhood it is only finitely many of the swing wheel survive it is a finite sum finite sum of continuous function so it is continuous if it is continuous on every open set it is continuous in the whole space that is why eta is continuous okay but at each point there is at least 1 eta i to which x belongs to at that point the value of this eta i will be 1 whereas we are taking some what are all these non-negative functions so value will be always bigger than or equal to 1 sum total eta x is always bigger than 1 since each x belongs to some half beta j and eta j of x will be 1 at the center in the half the ball take theta j now is eta j divided by eta okay divided by eta makes sense because this is a non-negative I mean always non-zero function okay it is continuous so I can divide by eta and verify that this will require the last condition namely now some total of theta j will be always equal to 1 because numerator and denominator are same for the sum so let us look at a few fallout of this one if you begin with a smooth manifold why the theta j's can be chosen to be smooth the beta can be chosen to be smooth then all these local homeomorphism the child can be chosen to be smooth then you get a smooth okay so this is the smoothness of partition affinity for smooth manifolds naturally we now have many other consequences of existence of comfortable partition affinity such as you do not have to prove you don't slammer separately for manifolds okay so all that I told you you could have done all that you could have done normality second accountability first accountability is there and so on locally compactness is there and put together do some typology and prove compactness now you can do all that many other things also just by using partition affinity t sections in theorem can be proved by this partition affinity okay all this 10 k for example if theta j is a comfortable partition affinity on x then define f from x to 0 infinity by the by the by the formula fx equal to k times theta k okay this will be a proper mapping of x onto 0 to infinity 0 close infinity open a proper map means what inverse image of a compact that is compact okay so this I will leave to you as an exercise easy exercise take any partition affinity then redefine fx as k times theta k and take the sum okay so I have given a few exercises here which are not all that difficulty I will I will just start the the next topic of interest today but we will not go very much deeper into it so so that next time you will have some you know familiarity with this definition namely start with the definition with half space inside rn what is this all those x1 x2 xn such that nth coordinate is greater than equal to 0 so this is subspace of rn I got half space okay upper half space you can say topological space x start with this one is called a manifold with boundary this is a nomenclature I am going to use it's a straight nomenclature now I am going to extend the definition of this local euclideanness okay to cover things like closed intervals you know in my original definition open intervals are etc were manifold but closed interval was not a manifold so I have to include them so this is the this is the mechanism so what we do such that he is called manifold with boundary if it is second countable house door those things are there local euclideanness changes what is it such that each point of x has an open neighborhood ux and homeomorphism fee all this same same as but now from ux to hn instead of rn okay on to an open subset of hn open subsets of hn okay think of half we have closed interval closed at 0 infinity open okay half ray what are open subsets zero closed one open is also an open subset whereas zero closed one is not an open interval inside r inside this half space it's an open set so the half space has more open sets than the the full space okay you have to swallow this one the half space has more open sets than the full space it's a subspace though okay watch out for that so these extra open subsets will be there therefore this is an extension of the old definition namely you are allowed to have an open subset which is completely in the open part of the half space it does not touch the boundary boundaries what for points when xn the last coordinate is 0 okay then that will be an open subset of rn also because if you take xn positive here okay that is hn interior that is an open subset of rn okay so our old definition of a manifold local equidinous is valid here also but this is extended there are it will allow more more creatures okay more manifolds will come here so such manifolds are called manifolds it found okay so let us start from here next time thank you