 Welcome to module 36 of NPTEL NOC an introductory course on point set topology. So, we begin a new chapter today an introduction to dimension theory. So, dimension theory has several approaches longest all these available version we will choose one such of course they are all topological dimension theories. Our aim here is reasonably modest one the depth and width of the subject does not allow us to do much in an elementary course like this. We shall discuss the zero dimensional case thoroughly and then take you to the door steps of higher dimensions. Our final goal will be to prove that Euclidean space R n is exactly of dimension n. If you are interested in more details for a comprehensive study you are welcome to read this book of Hurwitz and Wallmann for this particular dimension theory. So, module 36 the tightly separation of states it has something to do with the separation axioms that we have studied so thoroughly and it has something to do with connectivity right, but this is nothing to do with the you know separability of a metric space or separability of a space as such. So, there are too many different ways the word separation and or its modifications are used. So, you have to be careful here somewhat unexpectedly we are beginning with something to do with the connectivity. We have seen how connectivity crops up from the concept of continuum in the construction of real number. The space of real number is taken to be of mathematical dimension 1 why I am saying this one is whether it is topological dimension or a vector space dimension and so on various dimension all of them you can call them mathematical dimension that dimension is 1. The word dimension physics has slightly different connotation ok. So, in all mathematical dimension the space of real numbers must be having dimension 1. So, we begin our study at one stage before that namely understanding 0 dimensional space ok. For example, if you are studying linear algebra vector space the 0 dimensional space is just a 0 vector space that is nothing more than that. So, that simplicity makes the life very easy with linear algebra, but that is not the case in topology ok. So, we are trying we are going to spend considerable amount of time in studying 0 dimensional space itself. One of the peculiarity of connectedness concept is that the definition is in the negation ok. In fact we define disconnected space and then if the space is not disconnected then we say it is connected. So, we would like to come to the opposite of this property namely disconnectivity ok. So, we are going to discuss certain stronger forms of negation of connectivity namely disconnectivity which may be turned as stronger forms of separation axioms namely fresh air space, T naught, T 1 space, T 2 space, T 3 space and so on, regularity, normality and so on ok. So, all those things will come into picture now allah indirectly. So, let us begin with the definition here. Given any two disjoint subsets A and B in a topological space we say they are separated. If there exists disjoint open sets U and V such that x is the union of U and V both U and V are disjoint right. So, that is why I have written the SQ cup here. A is contained inside U and B is contained inside A B. After that you can assume both of them are open U and V are open or U and V are closed it is the same thing. So, the term here I am going to use is that A and B are separated ok subsets of x of course to be separated first of all they have to be disjoint. People also use the word disconnected you know for this one but that is too somewhat confusing for me. So, I do not want to use that terminology here ok. Such a writing U and V like this is called a disconnection ok. So, what we will call this as separation. So, we will had such a notation already introduced also we are just writing it as x equal to U vertical bar V you remember that. So, that notation also I may use sometime. In any case starting with disjoint closed subsets we have enclosed them in U and V the only thing is totality of U and V is the whole space x is the extra thing otherwise this is just like normality right. So, you are reminded of normality ok. So, what here is just stronger thing namely in normality the union of two open sets you contain you get is not the whole space they are disjoint fine. But here it is the whole space that makes both of them closed also ok. So, this is going to be quite a strong condition. No doubt it implies normality as soon as there are disjoint non-empty closed sets in a space x this property implies that the space is disconnected. So, even that is true whereas there are plenty of normal spaces which are connected right whereas if you have this property ok disconnectivity for every disjoint closed sets ok. Then the space will have to be disconnected. So, this is very strong host darkness, regularity, normality etc. can be termed as local property whereas the one which you have introduced now is a global property ok. So, let us consider the following condition on a topological space to make sense out of these things you should have the topological spaces at least two points. A singleton set you know singleton topology is both connected and it will satisfy these properties also. So, do not get confused with that at least you assume that there are at least two points in the space. Then consider this four different axioms of disconnectedness ok axioms of separation whatever word you want to do there is going to be some confusion ok. So, I can call it as axioms of separation because I have included I have introduced an attention S naught, S 1, S 2 and S 3 just to you know include people who are using this contract terminology I put that as a title also. So, S naught just say connected components of X are all singletons any two distinct singletons in X are separated is S 1 any close subset A is separated from any point outside A any two disjoint close subsets in X are separated ok. So, S naught, S 1, S 2, S 3 you can see that these three things are one stronger than the other, but this S naught seems to be not in this list why should it this is odd man out it looks like it, but this is the one which connects the connectivity with these property. In this section we shall do a comparative study of these conditions the four conditions above ok and indicate their importance. Note that S naught is somewhat different from other three statement let us give it a more descriptive name sometimes I will use that, but for safety I will use it as S naught only ok. So, any space which satisfies S naught will be called totally disconnected ok. I want to caution you that different authors may use same terminology to mean different things. So, that total disconnectedness may be different for example, one of the books which I am very much using and respect I have a lot of respect for this one is Siemens book on modern analysis and topology and modern analysis ok. In this book a space satisfying our condition S i is called totally disconnected S 1 is called totally disconnected not S naught ok. We shall see that property S naught relates the connectivity to the other three axioms of separability here alright. So, let us let us go ahead S naught is seen to be hereditary if something is totally disconnected namely all the singletons are the components then for any subspace singletons will be components. So, it is hereditary it can be seen without much difficulty that S 1 and S 2 are also hereditary ok. S 1 is hereditary is just like Hofstor's name hereditary. S 2 is hereditary is just like regularity is hereditary. When you come to S 3 it is similar to normality just the way normality fails to be hereditary this will also fail to be hereditary exactly same reason ok. But it is weakly hereditary in the sense that closed subspace of an of a S 3 space will be S 3 ok. Starting with a closed subspace and then taking closed subsets inside that they will be closed in the original space also. Then if you take a separation in the original space then you can restrict it to the subspace then you are done. So, that is the proof that it is weakly hereditary ok. Now S 1 implies S 0 what is S 1 any two points any two distinct points there is a separation ok. If there is a separation they cannot be in the same component. So, if any two points cannot be in the same component all the connected components are significant that is all ok. It is quite often people confuse S 0 with S i S 1. Caution S 0 does not imply S 1 we will see as some example soon ok. Unfortunately the word total disconnected is used for this one also. So, that if I some author. So, you have to be careful with that alright. Let us go ahead if X satisfies S 2 then every point in X has a neighborhood system consist of consisting of open as well as closed sets. There are those things are called closed sets and conversely this could have been taken as a definition of S 2. Why? Take a point take a closed set away from that or take a closed set and take a point outside that is the same thing as taking an open set and taking a point inside that. Then what we have you have a closed open and closed set open sets right that open set means what open and closed you take the complement that will be separation ok. So, both ways I mean this argument can be seen both ways. So, S 2 if pinned on if every point of X has a you know base local base consisting of closed set S 0, S 1 and S 2 are all productive also. Once again the proof is similar to proving that house darkness and what is the other one regularity are productive in these two cases. This case is just like proving product of product of connected space is connected it is much simpler than this one product of disconnected space totally disconnected space is totally like that is what we have to prove ok. So, it is similar, but let us just have a look at the first one why why S naught is productive ok. Suppose each X i is a family of you know X i is a family of spaces which satisfy S 0. Now I am taking A is a subset of product space and A is connected ok. Actually I am connected component with more than one point then I should get a contradiction alright. A i A is a subset with connected component as more than one point it is actually just a connected set with more than one point give you a contradiction that is all. Then there exists X y in A and an index i such that X i is not equal to Y i because they are different at least one coordinate must be different. But then if you look at the projection of A on ith coordinate pi i of A this will be a connected subset because it is a image of a image of a connected subset and a continuous function right. So, it is a connected subset of X i with more than one point because X i is not equal to Y i right that is a contradiction because we have assumed that connected components of X i are similar. Therefore, the product satisfies S 0. The converse why the converse is true suppose the product satisfies S 0. But then each coordinate space can be thought of as a subspace of the product space right. So, this argument you have used several times X 1 cross X 2 if you look at X 1 cross one point will be a subspace of X cross Y X 1 cross X 2 but it is homeomorphic to X 1. So, it is subspace because it is hereditary it will be S 0. So, each coordinate space is S 0 which means each X i is S 0 ok. So, I will leave the other things here namely productivity of S 1 and S 2 as a entertaining exercise to you ok. You should do at least that much so that you will be familiar with this concept ok. The earlier you do them the better. So, before reading the next before coming to the next module try to do this exercise so that you will be completely familiar completely you know thorough with the definition actually. In the most general situation none of these conditions imply any other. So, I said this looks like stronger stronger stronger ok. So, let us see why they are stronger what they in what sense they are stronger. Under additional conditions there will be implications one way or the other. This is what we concentrate upon for some time when S 0 implies S 2 implies S 3 etcetera other way around and so on this is what we are going to do. Just like the usual separation axioms, off-doorness, regularity, normality etcetera the first key is the freshiness. If you admit that all the spaces are T 1 spaces then automatically S 3 will imply as S 2 will imply S 1 implies S 0 ok. Exactly same reason as T 4 implies T 3 implies T 2 implies T 1. Exactly same way. First you have to assume that you know it is T 1. Otherwise just normality does not imply regularity. Same way S 3 may not imply S 2. As soon as points are closed S 3 will imply S 2 because in S 2 I have to take only a closed set angle and a point ok. Disjoint closed set and a point outside there will be disjoint closed subset so you can apply S 3 and come back first. Similarly in S 1 I have to take two distinct elements but they can be treated as disjoint closed sets. And I have already seen that S 1 implies S 0 right. Why because any two points are different connected here dis you know there is a separation means they are not in the same connected component. So this part is easier without the T 1 axiom also. This part we have already seen. So under T 1 they are like this. So from now onwards we shall always assume that our space is T 1 space. Then we will try to see whether we can come back ok. And that is where we have to maybe we have to put more and more conditions. Next note that each of S 0, S 1, S 2 and S 3 respectively is strong form of freshness you know half darkness, regularity and normality respectively. Any connected metric space with more than one point by the way why S 0 implies freshness tell me it is I say it is stronger than that. Connected components are singletons. Singletons are connected components connected components are always closed therefore singletons are closed ok. Any metric space with more than one point will serve as an example which is actually T 5 space right but does not satisfy any of the S 0, S 1, S 2 and S 3. Only thing you have to put is connectivity ok. None of the space is S 0, S 1, S 3 are connected spaces. So that is the meaning of why this S 0 is the you know it brings the connection between connectivity and separation ok. So before proceeding further we shall examine some examples now. It turns out that all these examples are separable metric spaces and satisfy S 2. Why I am doing this one because finally we are interested in dimension theory developed by Walman and Horowitz right. So there is separation separability sorry separability means there is a countable den subset. Separable metric space is a blanket assumption ok. So let us examine these examples. They are somewhat out of your way so far. Any countable metric space satisfies S 2 for given any point X and a neighborhood of X choose R positive such as VR of X is inside you. Now let X 1, X 2, X n be an enumeration of points of VR of X other than X itself because VR of X is countable it is a countable whole metric space is countable right. So I am just writing down labeling all the other elements. They may be finite, they may be infinite does not matter the dot, dot, dot I am not saying that it is infinite it is countable that is all. Choose a real number R prime such as 0 less than R prime less than R and such that R prime is not the value d X X i. So d X X i is not the d X X i as i raise over 1, 2, 3 and n and so on will give you as only countable number of real numbers ok between 0 and R there are uncountably many so I choose one R prime which is not for any of them. It follows that the open ball of radius R prime around X is contained inside you because R prime is less than R but boundary of R prime X where in equality occurs that is empty because equality occur they are distinct from all is X i that is why the boundary is empty. Whenever you have an open set with boundary is empty it is a closed set also ok. Therefore X is B R prime X union the complement is also open is disjoint union X is here right and this B R of X prime is contained inside you. So you have got a clop unsubset ok. So what we have shown is every point has a neighborhood system consisting of clop unsubset so that is the S 2 that we have seen earlier. Any subspace of R which is not it does not contain any open interval satisfies S 2 like we have been studying all these examples quite some time Q R minus Q that is irrational counter set these are all examples of S 2 ok namely for each point you can find a neighborhood such as the boundary of the neighborhood is empty that is the that is the nice way of remembering this S 2. The subspace Q power n of all points in R n all of those coordinates are rational ok Q is contained inside you take Q power n and then it is R power n that also satisfies S 2 all that I have to do is given a point use coordinate rectangular boxes ok with one of the coordinate irrational then the entire box will not be inside Q. So boundary of this entire box open box will be empty as far as Q n is concerned. So similarly the subspace I power n of R n of all points whose coordinates are irrational the other way around ok. Other way it is not complement by the way this here all points are rational here all points are irrational so that also argument is same thing ok this also satisfies it. Thank you sir. So from 2 Q and irrational are S 2 therefore products also will be automatically S 2 also one way of. Yeah you can use that here because S 2 is productive right. So you can use that one also no problem. Moreover now comes you see now I am mixing up for this you will have to use arguments you know down to this way not products. The subspace R 2 comma 1 2 up for comma 1 I have used here R 2 1 of all points are R 2 so that upper subscript corresponds to the where we are taking but the lower subscript exactly one of whose coordinate is irrational ok X comma Y you look at either X is rational or Y is rational. If both are irrational or both are rational I am not taking them ok. So this space is also S 2. So to see this one you will have to you cannot use products and so on it is not a product right now. So you have to directly say that for each point you must be able to get a neighborhood such that the boundary of the neighborhood is empty ok and open subsets of the boundary of that is empty that is what you have to do. So do you do that each point of this space can enclose in an arbitrary small rectangle whose vertices are rational ok. If vertices are rational they are not inside R 2 1 remember that only one of them must be rational right that is what the vertices are not inside this one now and then what are the sides the sides have slopes plus minus pi by 4. So they are not they are not coordinate lines now ok. So slides plus minus pi by 4 which just means that they are points X comma X comma X or X comma minus X or minus X comma X of that nature right or X comma plus something and plus something means what that plus comes from one of the coordinates of this end point the vertices but they are all rational. So if X comma Y the difference will be what difference will be again rational so both of them cannot be rational. So slope plus minus pi by 4 means that ok. So the boundary this boundary of this rectangle is not a point is not contained no point of that one is contained is R 2 1 intersection that is empty. So therefore this is S 2 this satisfies S 2 you try to do that inside R 3 then you will have a lot of problems this even this argument will not work this lines with the slope plus minus pi seems to work only for only for R equal to 2 now n equal to 2. But we have other ways of dealing with it here start with 2 indices m and n m less than equal to n that is all ok some numbers again similar notation R n m denote the subspace of points inside R n exactly m of the coordinates are rational the same construction this 2 is replaced by n and 1 is replaced by m ok this space is a similar generalization of the above example. However the proof of this that it is it satisfies S 2 is not straight forward at least the method that we did in 4 it does not work try try it of course you have to wait and then after some stage I will give you the proof ok so that is the involved proof here just take R 3 here ok instead of 4 instead of 2 then you can take n m then the problem will be there n equal to n I think we have already considered this one see that to be fall inside this example q n right here q n you see R n if you take all the coordinates are rational q n so that is already so it is a special case right so the problem is only when it is smaller then you have problem also when n is bigger than 2 so bigger for n equal to 2 we have already done it ok so I think today this is enough next time we will continue with more examples ok so that we are throw with these concepts S naught, S 1, S 2 so thank you that is all today.