 Welcome to module 41. Today we shall take up three more interesting properties. First countability, second countability and separability. Once again we go back to metric spaces for motivation, especially the Euclidean space itself. So these properties just like path connectivity, connectivity, compactness, lentil of etcetera, they can be classified as smallness properties, but not exactly. We have given a formal definition and this first countability and second countability does not fall into that, though separability does. Now why combining them but they are quite going hand in hand that is why I recommend. But nevertheless there is some big feeling that these conditions put some restriction on the size of the topology. You shall see why. And it is better to leave it like that instead of getting into formal definitions. So what are we going to observe here? First let me take any metric space. The base for a metric space topology is you take the balls of radius positive at all the points. Suppose you are fixing one point and then looking at all the neighbourhoods of that point then instead of taking all the balls of positive radius you can simply restrict yourselves to balls of radius 1 by n, balls of radius 1, 1 by 2, 1 by 3, 1 by 4 and on so on. Two, you know wherever the discussion of neighbourhoods is involved only these balls will be sufficient. So this idea leads to the notion of first countability. With the usual topology on earth we know that the set of rational numbers is countable as well as dense subset. The same thing is true for all Euclidean spaces also only thing you have to do is you take all points such that all the coordinates are rational. That is also a countable set and it will be dense inside Rn. So this gives you another notion which is called separability. Before going to the third one let me state a theorem here for metric spaces and then that will induce the second definition, the third definition. Start with any metric space with a countable dense subset A just like Rn but now I am generalizing it for any metric space all that I want is a countable dense subset similar to q or points with rational coordination and on. With this much of hypothesis I want to conclude the following thing. Then there exists a countable base for the topology x space. D is a metric space so D is a metric and the topology x tau d has a property that it has a countable base. In particular all Euclidean spaces have countable base. So what does it mean? It means that every open subset is arbitrary union but the members coming from a fixed countable set that is the beauty. So that already puts a lot of restricted condition on the topology. It does not say remember it does not say that number of open sets is countable no because you are taking arbitrary union but now all the open subsets are unions of members from a one single family of countable sets. So that is the countable base. So that is quite strong condition. So how to prove this one? It is one line proof. Put B equal to collection of all balls of radius 1 by n not at all the points but only at the points of A where A is a countable dense set. So A is inside little A is inside A and N is inside a natural number. This is clearly a countable collection because A is a countable set and N is a countable set. So there are two different variables here but N cross N is a countable set. This collection B is a base for a topology that satisfies B1 and B2. This does not need any extra you know hypothesis. I know this is this kind of things you have done so you must do that one. Okay that they form a base it means what once you take finite intersection two intersection of two of them and a point there show that there is a smaller one contained inside that at the point and so on that kind of B1 and B2 you have to verify. So this I will leave it as an exercise. All these are opens of sets in tau D right. Therefore B if you take the topology generated by B okay tau B that is contained inside tau D because tau D is a topology after all. So what I want to show is that tau D is contained inside tau B so that this itself is a basis for tau D okay and this is a countable so that will end the proof. So but this is one line proof again. Let you belong to tau D and X belonging to you. What I have to do I must produce a member from this B such that X belongs to that member and that member is contained inside you. So choose R positive first of all such that B R of X is contained inside you are we are used to this one because every member of tau D is union of such member where R is any positive number. But once R is positive there is a N sufficiently large such that 1 by N is less than R by 2 okay. Now A is dense in X I am going to use instead of arbitrary X I am going to shift it to A okay how A is dense in X it follows that take 1 by N radius around X this open ball every open ball will intersect A that is all because A is a dense subset. So B N X B 1 by N X intersection A is non-empty. Now take a point in this intersection then check that X itself belongs to B 1 by N X that is contained inside you. This part is obvious because X is inside this one means distance between X and A is less than 1 by N so it is symmetric so this is here. But why this is contained inside you because I have chosen 1 by N as than R by 2 triangle inequality over the gist of this is that we are going to take this as a definition of having such an A which is countable dense this is going to be separability. And we are going to do that B 1 by N's at each point only countable thing we are going to take this first countability. This gives you a big thing namely there is a countable base for the whole topology it is a global base. So that gives you the third definition which are interested which is called second countability okay so we shall make a definition here. So we will now study these three notions in the reverse order. We say a topological space is second countable okay sometimes you just write double 2 here instead of second okay. If it has a countable base over a topological space is that we separable if there exists a countable dense subset over. Third one we will wait for it let us make some let us make some the third one was the first one which is started with there okay let us make some observations here. Typical examples of second countable spaces are Euclidean spaces all that we have to do is consider the family of all balls with radial with rational radius and centers at rational coordinates. When I say rational coordinates all of them should be rational okay. So you can also see that finite product of second countable space is again second countable because if you take a countable base okay countable base here members from here cross that one they will generate they will give base for the product. So countable cross countable is countable again so finite products are countable every subspace of a second countable space is second countable. If you can generate all the members of the bigger space okay when you take subspace look at say y is a subspace of a look at y intersection members of b b b where b is so that would be a subspace for y okay. So once again this property is very important in the analysis because it allows several countable processes such as taking countable sums and integrations etc because every open subset now will be union of from a countable family right. So you can take u1, u1, union u2, u1, union u2, union u3 so any opens of any union can be written as increasing union. So all these things are possible okay. So we shall first prove two easy consequences of this definition namely second countable type. Every second countable space is separable that is why I took the other way round here okay first I said second countable. The separability automatically comes here so how start with a countable base for x okay. Let S be a subset of x which contains exactly one point from every member b of this curly b. One single point b inside b for each b inside curly b b is the curly b is the countable base for x okay. Since I have chosen one point from each of them this S will be again countable this S is subset of x by the way okay this is a countable site okay. Claim is it is dense S bar is the whole of x. What does that mean of that? Given any point x belong to x and an open set u such that x belongs to u there exist a b belong to b such that x contained inside b contained inside u okay but that means u intersection S is non-empty okay. So that b will have a point inside S right. So b intersection S is itself is non-empty so u intersection S is also non-empty. So what we have shown here is that it intersects every non-empty open subset and that is density. Every second countable space is another thing it is Lindelof. Remember what is Lindelof? Every open cover has a countable sub cover this also is very easy. Why? How? Take an open cover I must produce a countable sub cover right. Yes or no? So how do I do that? So ui is some open cover for x but let us say b is a countable base for x because x is second countable okay. Consider all members of b which are contained in some member of ui okay. So that is a sub family of b some of them may be too large one it must be some member of ui that is all. So only take those things call it b prime. Clearly b prime is countable because a sub family of b okay. We claim that b prime itself is a cover for x. Automatically this is sub cover because we have taken only some families from these members ui each of them is contained in some member of ui. So that will be automatically contained inside that okay. So all that I have to show is we claim that b prime is a cover for x. So let x belong to x be any take x belonging to ui for some i because uis are cover for x. Since b is a base there will be some bn such that x belongs to bn and bn is contained inside ui. This bn is inside b prime by definition because I have taken those which are contained inside ui here. Therefore x is contained inside only those bs which are inside b prime okay. Now for each member b of b prime choose one member ui which contains b okay that form a sub family of ui which is clearly countable because I have chosen one for each bn and they are bigger subsets than original b so they will cover x. Now I come to the first countability. Look at a topola equal space x. So I want to define another concept here which I have not done so far namely by a neighborhood system or a local base there are two names for the same concept okay. Some people are local base some people are neighborhood system. For at a point x belonging to x we mean a collection of open subsets ui of x such that x is inside ui for every i okay. So they are all neighborhoods. Secondly given any neighborhood namely some open subset such that x is inside u there exist a ui such that ui is contained inside. So this part is similar to base okay but everything is happening at a single point x that is why it is a local base that is all okay away from x this x is fixed away from x this collection may not have that property for only for neighborhoods of x belong to you you will have a ui which is contained inside so it is called local base okay. Clearly collection of all neighborhoods of a point forms a neighborhood system you are very generous in taking all of them you then there is no problem but this system may be unnecessarily huge and so we are looking at cases when there are neighborhood systems which may be countable and that is the third definition for today we say a space x is first countable at x belonging to x if it has a countable local base at x if this happens at every point then we will say x is first countable. Now we said second countability implies separability you have seen second countability implies first countability also because if we have global base which is countable global base is always a local base also right therefore second countability implies first countability so second count out of these three second countability is the strongest okay once again a typical example of a first countable space is any metric space no no need to take a Euclidean space here of course Euclidean space are first countable okay that was our motivation all that we have to do is take balls of radius 1 by n at each point point is fixed at take all the balls of radius 1 by n that will be a first that will be a countable base countable local base every second countable space is first countable that I observed already the converse need not be true if that was the case then we won't have two different the inventions here okay easy counter example is obtained by taking a discrete topology on an uncountable set okay an uncountable discrete topology cannot have a countable base think about that whereas a discrete topology each singleton is an open set so I can just take that open set namely singleton as a local base that's all one single set will give you a base local base okay the importance of this local base okay is more or less due to the following fact it allows us retain the notion of sequential continuity intact intact means what sequential continuity was true inside metric spaces now first countable is coming from the metric space that is what we have seen the notion of that one so only that is the key there not the rest of the metric space structure first countability is playing the role of this importance of sequential continuity let us see let x be a first countable topology space then any function f from x to y is continuous at x belonging to x if and only if xn inside x converges to x should imply fx and converges to fx okay proof is exactly same one way is always possible you don't need first countability okay xn converges to x f is continuous fx and converges to x this is always true in any topological spaces the converse suppose for every sequence xn converging to x fx and converges to fx then how to show that f is continuous the proof is exactly same as in the case of metric spaces okay so let us go through that I am trying to put converse here converse assume that f is sequentially continuous at x and not continuous at x then you will get a contradiction okay not continuous at x means what there exist an open set u in y such that fx belongs to you but no open set v in x such that x belongs to v and f of v is contained inside you now I apply this part no open set blah blah blah to each of these basic basic elements here vn local base okay for each n give take an x1 inside v1 such that f of x1 is outside you after that you take v1 intersection v2 that is an open subset containing x so take x2 inside v1 intersection v2 it may be x1 don't don't worry okay take xn intersection xn in the intersection of v1 v2 vn inductively such that f of xn is outside you okay automatically this implies xn converges to x okay but fx in terms of case is very poor it is far away outside you so if fx means cannot convert fx so this contradiction proves the claim so I have a few exercises here and a comment show that every subspace of a first comfortable space is first comfortable second comfortable space is second comfortable this I have already explained same explanation is there for one also as an exercise you write down the details in particular reduce that a discrete subset of a second countable space okay has to be countable this is used in complex analysis I don't know whether you have done this kind of things the set of zeros of an analytic function analytic function is defined on a domain of c right on open and connected subset of c it may be the whole of c it will be automatically countable so I would have been prove that by proving that it is discrete automatically it will be countable show that if f is a family of closed intervals of positive length which cover the whole of x there is a countables of family of x which covers r x x is r here if it is open intervals which cover r countable family will cover because it is Lindelof okay we have proved that r is second countable and second countability implies Lindelof right so open intervals covering the whole of r will have a countables of family covering r but I want it closed intervals closed intervals you don't take single points no single points are allowed okay intervals will be of positive length that's all a b closed interval a less than b take such thing then you will get a countables of family which covers it all so this you have to do is you know only r this is some tricky thing on a uncountables attacks let tau be a co-finite topology okay this is our favorite example is x tau with co-finite topology here first countable is it second countable is it separable check them because you check these things you will understand these these concepts properly okay so that's all for today thank you