 Welcome to module 27 of point set topology part 1 of NPTEL course. Today we will give you an attempt to characterize neighborhoods in the case of co-induced topology. Finally, this is going to be a negative result just to caution you what kind of results to expect and where one has to be careful and so on. So, remember open subsets of a coherent topology have the property that intersection with each member of the family if some set has the property that intersection with each member of the family is open in that set then the set itself is open. So, can we have similar characterization for neighborhoods that is the question. So, put in a straightforward way let us look at this proposition here which later Rohan actually will show that this is a wrong proposition is actually has no basis because we think that open subsets have a nice characterization why not neighborhoods. So, take this somewhat special case not very special but easy case namely suppose X has been written as union of some spaces X i's subsets X i's each of X i is a closed subspace of X i plus 1 the next one. So, it is an increasing union and it is accountable because I have put them as integers here and X is given the coherent topology with respect to the family of subspaces X i. It just means that a subset of X is open in this coherent topology is filled only if its intersection with each X i is open in X i. Now suppose B is an arbitrary subset of X contained in another arbitrary subset W such that for each indexing here each index here i if you intersect W with X i that is a neighborhood of B intersection X i in X i. Remember each X i is already a closed subspace to begin with of the next X i plus 1 and so on. Then we have taken X as the union of X i and putting the coherent topology on X. Under this situation we would like to say that W is a neighborhood of B in X. So, this is simpler case you can expect such a thing if it is true for arbitrary union instead of increasing union and so on even in this case simplest case it is false. So, that is the gist of today's talk. So, I repeat a subset W is called a neighborhood of A if there is an open subset U such that A is contained inside U is contained inside W. This is similar to neighborhood of a point if this A is a singleton set then this is a neighborhood of that point. So, that can be generalized to neighborhoods of any set source. In the statement of the above proposition if you replace the word neighborhood by open sets or open neighborhood then the statement follows directly from the definition of coherent topology. Indeed using the definition of coherent topology we are trying to get the same criteria for neighborhoods also that is the whole idea. However, this proposition is hopelessly wrong as you will see by the following simple counter example below. So, why I am giving you this one first of all to caution you with certain kind of results. This proposition occurs in some very standard literature on CW complexes. The classical book is Lundell and Wiengram which had come in 1969 a very standard book very highly respected book which CW complexes etc. I properly studied from this book also and the same kind of mistake actually the same proposition occurs in my own book also 2014 page number 91. So, in these this result is used in proving some other results about CW complexes especially when you are studying product of two CW complexes. Luckily those results can be proved directly without using this proposition that is luck because this proposition is wrong now as I will show you right now. So, one more remark on this in those attempts in proving the above proposition in at least in those two sources which I have quoted just now the mistake occurs when one implicitly and erroneously assumes the following statement is also wrong. So, you can supply easy examples of counter examples for this statement namely take two subsets a and y of a tuple y equal to space x then interior of a intersection y with respect to y that means inside the tuple y equal to space y as a subspace of x is contained in the interior of a as a subspace of x. So, this suffix denotes here x and y you know where we are taking the interior. Interior of a subset is contained in the interior of a larger set that is the correct thing. However, the interior where it is taken that is important. So, if you have taken subspace here and larger space here then there is no containment in general. In other words take a subset of a smaller space an interior point in the smaller space may fail to be an interior point in the larger space very easy examples we have seen such things right. For example, even in the case of real numbers if you take a half closed interval 0 to 1 0 closed in that 0 to 1 0 is also an interior point. However, 0 is not an interior point of 0 closed 1 as a subspace of the entire real numbers. So, this is what you have to know. So, there are easy examples of this. So, such mistakes one should not make which is when you tell somebody it is obvious but you know it is such things can occur while proving something you say now we can verify this one and so on that is not wrong. So, however let us not bother about how this was wrong and so on. Let us just directly prove that why this is wrong by just giving a count example. So, here is the example let r omega denote the set of all sequences of real numbers which are eventually 0 like 1234 and then 000 0, 0.5, 0.5, 0.3, 0.4, 0.7 and so on after 0000 finite sequences but sequences are all infinite but after such a state all of them are 0. So that is my r omega it is a vector space no doubt take accent to be all those sequences real sequences si is 0 i greater than equal to n plus 1 onwards in a very specific subspace of r omega up to nth order up to nth term you can have any real number after that it must be strictly 0 n plus 1 n plus 2 etc. So that is my accent this will be automatically a vector subspace of r omega and x itself is r omega I am denoting this x now to be increasing union of accents x1 consists of only the first entry to be very free all other things are 0 x2 will have first and second entry non-zero and all other things are 0 and so on so it is increasing union of accents under the usual topology r contained is r cross 0 contained inside r cross r cross 0 and so on. So these are all closed subspaces one is closed in the other x1 contained is x2 contained x and so on xn will be closed in xn plus 1 and so on. In the standard euclidean topology for each xn give you a coinduced topology tau 1x from this collection clearly each xn is a closed subspace of this x tau also induced topology from a closed family automatically each member will be closed that is all. Now I come to another subset here sn all those s in xn such that the norm the euclidean norm the infinite norm have taken infinite norm is at maximum because x is sn as infinity to the same thing as maximum of modulus of s1, s2, s3 up to sn after that everything is 0. So that norm is less than 1 by n it just means that each entry is in modulus less than 1 by n strictly less than 1 by n that is my sn and take s to be the union of all these ssens and range from 1 to infinity. So look at this one this is an open subset of xn because the modulus is strictly less than 1 by n. So I have taken the union of these open subsets here how does it look like if you look at x1 which is real number it is open interval from minus 1 to plus 1 when you go to x2 this open interval this part will be till there because I have taken x1 and x1 union x2 but x2 will be x2 will be what is open rectangle of side length minus half to plus half 4 side length will be 1 minus half to plus half minus half to plus half product opposed to open intervals. So that is your s2 but this is hanging out here is from minus 1 to plus half. Now in s3 you will get an open queue from minus 1, 3rd to plus 1, 3rd of size 2, 3rd side length 2, 3rd and so on. So that is my picture of ss1 then s2 and then s3 and so on. The union of all these things is my s. Then this s has the property that if you intersect with xn that is a neighborhood of 0, 0 in xn, 0, 0, 0, 0 that is in xn, 0, 0, 0 this 0 you can take all of them 0 that is 0 here which is 0 of the vector space r omega itself but it is a point of xn also and this s intersection is a neighborhood of xn. For each n we claim that s is not a neighborhood of 0 in x. So go back to this picture. If you take s1 then it is automatically minus 1 plus 1. If you take s2 you do not have to bother this one you can take just this half cube here. If you go to s3, half rectangle here. If you go to s3 you can take this 1, 3rd to 1, 3rd cube here and so on. Those are open subsets. So they are contained in the whole of s1 union, s2 union, s3 or s intersection xn. So s intersection is a neighborhood of 0. But I want to say that s is not a neighborhood of 0. We claim that s is not a neighborhood of 0 in the entire of x. Suppose this is a neighborhood then you have an open subset of x such that 0 is in u and u is contained inside s. This u is now open in x. Then u intersection xn is an open subset of xn such that this u intersection xc is contained s intersection xn because u is contained inside s. Note that the maximal open subset of xn which is contained in s intersection sn is just sn itself. That means in s1 it is the open double minus 1 plus 1. There is nothing more than that. In s2 it will be this box, this rectangle, open rectangle, this portion goes away. That is a maximal open subset. In s3 it will be open cube. Outside this open cube nothing is there. Those are the maximal open subset for each of them. So if there is some open subset they must be contained in these maximal open subset. Therefore the maximal open subset which is contained in s1 is sn. Therefore this u intersection xn must be inside sn for all n. Now this is not possible in s1. Now let take s equal to s1, 0, 0, 0, 0. Anything s1 is not 0. That is all. Someone, some neighbor will be there belonging to you. This implies s belongs to u intersection xn for all n. As soon as some member like this is there it must be inside u intersection xn for all n because this s1 is inside x1. So it seems like x2, x3, x4 and so on. Therefore s will be inside sn for all n. But if s is sn for all n with a very definition of this sn what is it? Modules of all these entries must be less than 1 by n. So this is true for all n. That means mod s1 in particular must be less than 1 by n for all n. Therefore s is 0. But that means u intersection x1, I have taken one point here in x1 and it belongs to you. This u intersection x1 is as soon as this is 0, this s1 is 0. It means it is single term 0 which is not an open subset of x1. x1 remember is open interval from minus 1 to plus 1. So this proves that there is no such open subset which contains 0 as well as contained inside our s. So that is no such s is not an n. So I have already cautioned you about why this is necessary and so on. So that is one thing about coherent apologies etc. You have to be careful. Indeed the problem can be of this kind of problems can occur even when you are dealing with boundary operator or closure operator. So such a simple thing that I have elaborately taught you. So you have to be careful about these operators. Our next topic is quotient spaces. Today we should just revise what is the meaning of a quotient function. Three different ways of looking at a quotient function is our topic just now and then next time we will study the quotient spaces. We will come to topology. Right now it is only set theoretic property that we are going to just recall. Consider the following three concepts in set theory. Start with any given set x then take a subjective function from x to another set y. That is the first thing. The second thing is x itself is written as a mutually disjoint union of some of its non-empty subsets. Indexed by some other set y. So I am writing x as this disjoint union A y y inside y. Each A y is non-empty. A y intersection A y prime. Take any two of them that is empty whenever y is not equal to y prime. So that is the second concept. This has a name which is called partition of a set or a decomposition of a set. These two terms are used. So partition of x or decomposition of x. The third thing is we are given an equivalence relation on x. So these three concepts are mutually the same. They will give rise to one gives rise to another and another gives rise to one in a very bijective way. That is a precisely what we want to say that you will see that these three concepts are equivalent to each other. So let us see how. Start with a function from x to y which is a subjective function on two functions. Then put A y equal to the fibers of q. 2 equals y. y belongs to y. Clearly A y should be disjoint. Being a subjective function each q inverse y will be non-empty also. Any point x will have to be mapped to some y here and then it will be one of the q inverse y. Therefore x will be the union of all these things. So that means that these A y's form a partition. So this is what I have to say. So next given a partition x like A y, y belong to y. Each A y is non-empty. Union A y is x and A y intersection A y prime is empty. So that is an integral. We can define a relation on x as follows. This is the relation. x1 is equivalent to x2 if and only if both x1 and x2 are in the same number here A mi. Verify that this is actually equivalence relation that is not a big thing. So if you see the equivalence relation finally suppose you have an equivalence relation we shall denote an equivalence class of an element of x by bracket x. I can also take q of x equal to bracket x where set of all bracket x forms the set to y. What is y? y is a set of equivalence classes. Now what is q x? q x is the equivalence class of x to that surjective function. So surjective function gives rise to a partition. Partition gives rise to an equivalence relation. Equivalence relation gives rise to a surjective function. Nothing is lost in the cycle. If you start from one thing concept, go through the cycle and come again you will get back the original stuff wherever you stop. This is the precise meaning of this. There is a triangle here, surjective functions, partition, equivalence relation. You can start from wherever you like. Say let us say from here I have a partition, I define the equivalence relation. Then I take the corresponding quotient map. Now I look at the partition, let us call them as B, B y's. But B y will be precisely equal to A y for each y in y where our y was precisely the original indexing set. How? is what I want to show you. So start with y, y has nothing to do with x. This is the indexing set. Here equivalence classes, here equivalence relations are there. Equivalence classes are just if you take x1 bracket x1, it will belong to a particular y. So that is the indexing set. That is A y. So that is what A y is after all here. So now what I am taking x going to is equivalence class. Equivalence class is this A y. So this just y it will be. So q x will be just y. So that is why this y reappears here. So you can do the same thing. Just start. Don't worry about this one. Equivalence relation, get a surjective function, get a partition. That partition will be precisely these equivalence classes. So that is all I wanted to show. All these three different manifests of just a surjective function here is important depending upon what we are interested in. You should immediately shift your attention to or your viewpoint to one of these three things. So that is what I have summed up here. So this is one of the things why a quotient function is called identification function also. Identification world comes from equivalence classes to x1 is identified as x2. If you know this q x1 is equal to q x2. If you know if both x1 and x2 belong to a partition etc. So you have two different, three different terminologies here. One was what are that partition or decomposition. The other one is identification space. The third one is very simple. A surjective function. So all these things you should keep in mind. One simple function theoretic property of the surjective function for equivalence relation is the following. Start with any surjective function. Given any function from x to z, take any function. There will be a function from y to z such that f middle followed by q, this q is the one given by, is equal to f. If this t happens it will happen that q x1 equal to q x2 would imply f x1 equal to x2. Not only that, only if this condition happens there will be such a function f greater. So this is a picture here. q is given x2 y, f is given here. Will there be a function here such that this diagram is commutative? Yes provided. Whenever x1, x2 go to the same point here, q of those two points must be the same here. That is the whole idea. Sorry. Whenever q and q1 x1 and x2 go to the same point here, f of those points must be also the same. So that is the criteria. If there is such a thing it is clear that state two points here they go to the same point and f twiddle of q will be the same point but that is f. So that is necessary is clear. If that happens all that I want to define f twiddle, f twiddle I have defined from y to z. What do I do? Take up a point here. You can think of this as an equivalence class. Where is it? It is coming from q of something. Take f of that to be f twiddle of this. That is all. So if y is q of x, I will define f twiddle y to be f of x. That is all. So this is function theory. This can be used in group theory. This can be used in vector spaces everywhere. This is just function theory. So we are going to use it in most general way namely for quotient maps. It may be recalled that quotient maps are much more in some sense, much more fragile, more general than what you get in quotient groups. Quotient groups are much more rigid. So we will have some beautiful theorems about quotient maps and quotient topologies and so on next time. Thank you.