 Welcome to NPTEL NOC introductory course on point set topology part 2. So, module 12, we continue with the study of Paracompat spaces this time partition of unity. So, we shall now discuss one of the most important property of Paracompat spaces namely that they admit a large number of continuous real valued functions. As usual, we shall begin with a definition take any topological space by a partition of unity on x the continuity is assumed. So, I have put it in bracket partition of unity on x we mean a family of functions indexed by i ok. All these functions are defined in the whole of x to a closed interval 0, 1. The first property is for every x you have a neighborhood u x of x such that theta i is 0 on all y belong to u x except for finitely many i inside i theta i will be 0 on this u x in that neighborhood. So, that is like saying that the family is locally finite theta i is the members of these continuous family continuous functions they will all vanish except for finitely many of these theta i's. Since there are only finitely many of them which are value which are non-zero there some make sense even if I write this total sum as summation theta i i theta i this will be a finite sum for each x that sum must be equal to 1. So, this is the second condition the second condition gives the name partition of unity this unit is the constant function 1 here. It is broken up into a family of functions each of them continuous sum total is equal to 1. The sum arbitrary continuous arbitrary family is so what they mean the sum that is why this local finiteness is there actually point finiteness gives you this condition this makes this condition sensible ok. But you will see that local finiteness is important here instead of arbitrary topological spaces if I have say subsets plus Rn then I can talk about partition of unity which are smooth which are c1 c2 and so on. So, you can put those adjectives here as c infinity partition of unity such things are possible because we are studying eternal arbitrary topological spaces there is no notion of differentiability here that is all. The condition 1 we may refer to theta as locally finite condition 1 is the same as saying that family theta i inverse of open 0 closed 1 take this inverse image these things are open subsets of x that family is locally finite is the same thing as condition 1 it follows that at any given point the LHS of the second sum here this sum here is finite sum and hence makes sense. So, condition 2 ensures that sum total is equal to 1 means this open cover this open family must be a cover for x ok every x must belong to something if it is not in any of them then sum total would have been 0. So, that is why this cover given any open cover uj of x we say that the family theta this theta is a subordinate to uj if these open subsets take this family this must be a refinement of this open this family. See this family is indexed by j this is indexed by i ok. So, indexing sets are different it can be different it can be same. So, what is the meaning of refinement for each member here there is a number here. So, that association itself will give you a function on the indexing set we are not writing all that ok when it is crucial you may have to write down those things also the refinement functions and so on. So, here is a theorem take a para compact host of or regular space and uj be an open cover for x then there exists a partition of unity on it which is subordinate to uj. Subordinate to uj I recall that theta i is a partition its support will be contained in one of the uj theta i inverse of open 0 close to 1 ok actually you can you will see that its closure itself is a refinement ok the key to this theorem is the following lemma. So, this is where we are going to use John's lemma also later on. So, first of all the definition is a family of subsets of x is said to be a point finite if each point of x belongs to at most finitely many members of few. So, this is what I have already told last time but we repeat it here because it is necessary ok a locally finite family will be automatically point finite but here is a definition which is now needed for us to proceed namely this new definition take u an open cover for x a refinement v of u is called a shrink of u if v bar that is a refinement of u ok automatically it will be a closure refinement ok. So, this is what we would like to have just not just an arbitrary refinement. So, such a thing is called a shrink if the closures of each v inside v bar is a refinement of u if v is contained inside u for each one that is the ordinary refinement this time we want the closures must closures of this one for each member here must be containing some member in u. So, that is a that is called a shrink ok shrinking lemma is something about just a normal space here we do not bring a paracompactness the paracompact of door spaces are normal. So, some kind of normality is built in there inside paracompactness you put hostage it comes out ok. So, though there are possibilities of proving a result without the shrinking lemma, but we will go through this one so that the shrinking property corresponding to normality is brought out separately ok. So, that is the whole idea of putting this one. So, let take a normal space u be a point finite cover for each u belonging to u there exists open set u f u such that f u bar contained inside u and the family f u u inside u is a cover for x. So, if you call this as v then this v bar is in f u bar they are contained inside u therefore this will be called a shrink you see that is the whole idea. So, f u bar so these are this is a shrink for that one ok the family f u is a cover because in the definition of shrink we are not putting this it must be a cover ok. So, this shrinking lemma tells you that there is a cover which is a shrink of the given family indexing will be the same here for each u you have an f u that f u comes back to you in fact f u bar is inside ok the statement must be clear here, but the proof depends upon using John's lemma. So, how do we use John's lemma you start with a family gamma of pairs v curly v comma g where v is a sub family of u and g is a function from v to u opens of sets inside x this is tau tau is a topology on x ok this is a function now for each member here you will get a open subset here with the property that the closures of g of u are contained inside u for all u inside this sub family v the second thing is all those g u where u is inside v and all those u which are not inside v they cover the whole of x ok see if all u v if it is v is the whole of u then this would be a total shrink since it is not total we have a word here partial shrink ok this partial this part is such that g u bar is contained inside u but we insist upon the rest of them together with the rest of them it covers x. So, such things are called partial shrinks what we are looking at is we do not want anything left here we want the entire function should be defined on the whole of curly u here instead of some subspace v ok. So, that is the statement of the lemma ok. So, how do we prove it? With this family gamma we put a partial order take v comma f and w comma g this is less than equal to that one we are going to define this relation if we will not leave the family is a family of this and the function is a restriction of function v is contained inside w g restricted to v is f why this family is non empty answer is using normality we can see that each singleton u for u belong to u can be taken as a domain for a partial shrink u and rest of the members of u they cover right. So, if you take union of all v not equal to u that is one open set u is another open set these two together they cover the whole of string their compliments will be disjoint closed sets with normality you can take a smart slightly opens of sets containing inside the closure of all these open sets union of all these open sets that will be contained inside this u and so on. So, you get a partial cover ok. So, this is easy easy part ok now if you have a chain inside gamma chain means what a totally ordered subset indexed by some family v alpha g alpha each of them is a partial cover remember they are members of gamma. So, it is a chain I must show that the chain has an upper bound inside gamma so for that I take w as union of all these v alphas that will be some family of u that is fine but now we want a function w. So, w to tau we find f restricted to v alpha to be g alpha because this is a total this is a total order set it is a chain if alpha and beta are given either alpha is less than beta or beta is less than alpha that is the meaning of total order right. Once that is the case g of v alpha will be equal to f of beta is beta is contained in that alpha therefore, this definition makes sense on the whole of f there is no ambiguity here ok. So, we have got a member of member it is not yet a member we have got a family. So, family of v and a function we have to just verify that it is a shrink a partial shrink and then it will be a member automatically this will be the upper bound because see by definition all v alphas are contained inside w and when you restrict this capital F to any v alpha it will be just corresponding g alpha ok. So, we have to show that this is a member of this family gamma for that you have to show that it is a partial shrink ok. So, property 1 is verified easily g u are contained by very definition here because they are all g alpha to see 2 is the important thing ok. How do you see the part 2 namely y all g u u inside w now and those which are not in w cover the whole of x. So, to see we need to use the point finiteness of u I have no other way to prove this one take x belong to x ok belongs to only finitely many of u and u to u n because a point finite right some u and u to u n they take a take finitely many of them all right call them u and u to u n if u i are not in w ok one of the u i is not in w it will be in the in this part therefore x is there no problem it is this part which gives you problem because you have shrunk these things you have made smaller. So, x may be left out so, you have to worry about that. So, the problem is if one of the u i's are inside not entire w you are happy. So, on the other hand suppose if u i's are all inside w for all the u and u to u n then it follows that there is a alpha 1 alpha 2 alpha n and so on to which these u i's will belong to v alpha 1 etcetera right. Therefore, you can take the maximum of this alpha i so there will be one single v alpha to which all of them will belong to ok. So, this is where that this is a chain is used by a property to apply to this alpha it will follow that x must be in g alpha v for some v inside v alpha because it is not in the other part let us talk. So, all that I want to show is that x is covered by these w g of members of u alphas are here. So, now we are it is not here so, we have shown that it is in this part by John's lemma we have a maximal element remember this this itself may not maximal element this is an upper bound for the chain there is a maximal element we do not know what is that maximal element but that is good enough for us ok. So, we shall denote it by u prime f remember u prime is a family of u f is a shrink function now it is enough to show that u prime itself is u. So, that there is nothing left on this part ok this part is not this these are the things. So, there is a total shrink ok that will be a total shrink. Ok this maximal element must have the domain the whole of u for if u is any member of u suppose it is not in u prime then look at this set A which is union of all f v v inside u prime and all v such that v is not equal to v ok all those v not equal to u they are in u minus v prime if if this is smaller then there will be always some such thing right it follows that the complement of this is a closed subset of this is an open subset because this is these are all unions open subset it is a closed subset of u because the whole thing is covering u u is the only one which is missing. Therefore, this set will be a subset of this complement of this one will subset of it. Therefore, there exists an open set which is which we call f u such that A C this closed set contained in the f u contained in f u bar contained inside u normality is used only now normality was used to show that gamma is non-empty that saw so right in the beginning right at the end now the function f extend to u prime union singleton u all other members we have f but one more for u we have defined and we have defined it so now this is a larger member than the member u prime f that is a contradiction why the contradiction because we assumed that u prime is not here okay so I have already made this remark for relevant things you can read from munkra's book that will give you proof of para compact hostile space every open cover has a shrink no assumption of no local hi-nightness on the cover local point hi-nightness on the cover okay but don't confuse it with the statement normality just normal space if you have an open cover it may not admit a shrink okay yeah so here is a comment and then we will stop here the actual proof of this one is now very easy so first let us go through that proof we may assume that u j is locally hi-night okay because para compactness so start with an open cover replace it by locally hi-night refinement so you can assume the locally hi-night and choose a total shrink f u j j upon 2 j okay because it is normal also and now this is a local hi-night so it is point hi-night using normality again obtain continuous functions alpha j from x to 0 1 such that the closures of f u j's are taken to singleton 1 and the complement of u j's are taken to 0 this is normality whenever you have two disjoint closed subsets you have such functions left this u j is locally hi-night if all of that is suitable neighborhood of any given point only finitely many alpha j's are non-zero that is because of local hi-night family therefore the function alpha which is summation alpha j makes sense okay in a small neighborhood of every point exactly some finitely many of them will be same finitely many those things will be left out okay each alpha j is continuous there so in that open set some is continuous therefore this alpha is a continuous function moreover somewhere it must be equal to 1 right because f u j it will belong to some f u j so alpha j of f u j whole thing is going to 1 so one of the index it will be always 1 and rest of the index is it will be either 0 or positive something therefore some total is always bigger than or equal to 1 okay now all that you have to do is take theta j of x equal to alpha j of x divide out by alpha x you have function continuous function which is never 0 therefore 1 divided by that function is also continuous so I am that is why I am using it alpha j over alpha this will be also continuous okay now summation theta j will be summation alpha j divided by alpha x which is equal to 1 what is the zeros of this they are the same thing the zeros of this therefore the support of this will be contained inside the corresponding u j whatever support of alpha j is there same thing will be support of theta j so the proof of partition of unity finally is very simple okay very easy so major thing went in proving the shrinking thing here namely using chance lemma so here is a comment here it is easily seen that para compactness is weekly hereditary namely every closed subspace of para compact space is para compact just for the sake of clarity write it down as an accessory okay however it is not hereditary for the same reason as compactness is not hereditary weekly hereditary yes closed subspace is fine or ordinary arbitrary subspaces may not be hope to see that starting with a non compact non para compact space if there is one we will show that there is such thing you can always add an extra point take the syrupy specification sx to get a compact space which is then para compact as well okay we shall discuss an example of a non para compact space later okay thank you that is enough for today