 So, this is the cost of topology, okay? There will be two parts as I'm going to give you. So, the first is general topology or set theoretic topology. And the second one is a small part of algebraic topology. And this means for us fundamental group and coverings. This is the main part. The book is monkress. So, this is topology. And this is the second edition. So, that's also the first edition, but you should take the second edition. Because we have other exercises. This you will need copies as everything, okay? And I'll tell you what you need. So, these are general topologies. These are chapters two, three, and four. We will not do everything. But these are general topology. And then this is too much, but you will need these, okay? And then also chapter nine. This is the second part, fundamental group and coverings. Okay, that's chapter nine. Maybe a little bit also from later chapters, but that's not so clear. Okay, these are the four main chapters, okay, from the book. From the second edition. Not the first. The first is a little bit different. And so you need copies of these chapters. You can also take chapter one. Chapter one is set general, which is useful, of course, okay? So I recommend that you take copies of chapters one, two, three, four, and nine. Because there are many exercises. We will not write on the blackboard the exercises. And there are many examples and so on, okay? The book is in the library, okay? It's the beginning. The textbooks, no? You find it immediately. Or you go to somebody from last year and he will have said, okay? So you can copy immediately, okay? So I will start with general topology now. However, I would also, so it will be English. I would like to know about the English also a little bit. So maybe I go over the list. So from Egypt, you are familiar with English, right? Sudan, or this is my language, English. Senegal, that's French more. Vietnam, whatever. Pakistan, that's English. Iran, Uzbekistan, Kenya, Sri Lanka, Bahamas. That's more Spanish or what? The main, huh? English. So I suppose the language will be no problem, alright? Most of you are English, maybe French, but I don't know. You have to learn English. You have to write English. We will give written homeworks, okay? And you have to write. So not today, but maybe tomorrow, okay? And then I will collect, okay? So anyway, let's start. So it starts, I mean topology is a generalization of analysis, real analysis, okay? I don't have a lot of topology anyway, so it starts from this. The definition is quite abstract of topology. So X is always a set. X denotes a set. And topology on X, I use, of course, the notion of the book, okay? I follow exactly the notion of the book, otherwise it will be confused. So topology on X is a collection subsets, a collection T. And now this is a written T, okay? So there will be a difference between printed and written T. A collection T of subsets of X, and these are called open sets, okay? The open sets. There are three conditions. So the first one is that the empty set, that's the empty set, right? The empty set and the whole are in T. They are open, okay? In every topology. Topology is a collection of subsets of these. Then we have the second condition. So this is not very significant. The second condition is that... So you're thinking of analysis, right? Open sets in R2, the topology of analysis. So what you have is that arbitrary unions of open sets are open. Okay, so arbitrary unions of open sets are open. Or more formally, if you have open sets, you alpha in T, where alpha is from some index set. So index set, it's not so clear, I or J. In the book it's J. I very often is an interval, okay? So this is an index set. Then the union is open. So this is a union of all these alpha and J, okay? So this is formally... Then this is in English. Arbitrary unions of open sets. Arbitrary unions means that this is an arbitrary index set. Find it in, it doesn't matter, okay? And then there's a third condition. And this says then that... Find it in the sections of open sets. Find it in the sections of open sets. So now, more formally, this means if U1, Un are in T, so they are open sets, then also the union, the intersection. Find the intersection. U1, intersection, Un is in T. So these are the three conditions. The first one, empty sets, the whole sets are open. Arbitrary unions of open sets are open. Why find it in the second? Well, to motivate that, you all know real analysis, okay? The topology of the real line of R2, okay? And so let me give an example. Informal example. I have to get used to this blackboard. Oops. So, example. Let's take real intervals, okay? Real analysis. So let's take the interval from minus 1n to 1 over n. So this is the real interval. Let's take the intersection n in n, natural number. What is the result? Zero. So real intervals, real, sorry, open intervals, okay? Real open intervals. This is open interval, okay? So they are open. But this is zero, okay? And this is not open in the standard topology of analysis, okay? Not open. So arbitrary intersections in R. And this means R is the standard topology, okay? The analysis topology, okay? We will define this anyway, but this is just standard topology. So that's why we better ask finite intersections. For the standard, for this standard topology of R, finite intersections is no problem, okay? And arbitrary unions is no problem. So, but this is a very abstract definition of topology, okay? Okay, this is the basic definitions of general topology. So examples. So we're going to write x with a topology t. So this is topological space, okay? The set with a topology. But very often we will not write a t. There are many topologies. So examples on the same set. Let's discuss some easy but important examples. So on each set, so x is a set, all of us, okay? Capital X, yeah. So we can take, so x is a general set. And we take t, and this is the indiscreet topology. Let's start E, I, I write I. So this, we need two open sets, that's the empty set and the whole space, okay? So we take the empty set and the whole space. So this is a topology. And if you have questions, you should ask. Also about English, I don't know. If you don't understand some words, you have to interrupt and ask. You have to learn English, okay? There's no way. So you have to learn English and mathematics both, okay? So don't hesitate to ask, okay? Mathematics, but also if you don't understand some words. That's important, okay? Because you have to write. So this is a topology and this is the indiscreet topology on x. On each set we have the indiscreet topology. Which is maybe not very interesting, but then we have the opposite. What is this? This will be td, the discrete topology. So we take everything. So this is, we take all subsets of x, okay? All subsets of x. All the conditions are valid and this is the discreet. This is indiscreet and the discreet. So this is on each set. Then we can compare topologies. So we learn a language in some sense, okay? So we say that, so we have two topologies, let's say. Other examples. So let t and t prime be two topologies on x. And maybe we can compare, maybe we cannot compare. What means compare? So t is finer. Yeah, coarser, finer. Finer, or strictly finer, is finer than t prime. Okay, so that's a definition. Finer than t prime if t has more open sense. So this means t prime is contained in t, you know? If t prime is contained in t. And strictly, so this means subset may be equal, okay? Including equal. And strictly finer, strictly finer if t prime, and now we have to write that to the different part. So this is finer and strictly finer. And the opposite is coarser, okay? So this means t is finer, t prime is coarser than t. So finer, yeah, one second, finer means larger, okay? Yes, what? Finer means larger, okay? More open sense. Coarser means smaller, less open sense. Strange words, finer coarser, okay? Larger, smaller, that's coarser than t or strictly coarser. So these are, it's just a vocabulary, some words. But it's clear, we will see examples. Not all topologies we can compare, okay? In general. Other example. So this is the first example where we give a name because in some sense it's useful and also for exercises, okay? We have some standard examples, okay? This is maybe one, not the most important, but one. This is the final complement topology. The final complement topology. So this, I will define, the final complement topology. So x is a set, okay? A set. And then we have, let's say f, that's okay. We have indiscreet, discrete, final complement topology. Give some name, t, tf. All sets which have final complement. The name just already says what it would be. So these are all u in x, all subsets, u of x, which have final complement, okay? So complement, I write x minus u. There are other words, okay? X minus u. This is a complement of u in x. So I use this notation, x minus u, okay? Because sometimes you use some minus. It's finite. This is finite, except that in general it's not a topology. Because we say that the empty set should be open in any case. So the empty set, x minus empty set should be finite, means x is finite, no? So that's not very interesting finite sets. Well, it may be interesting, but so we better add something, the empty set, okay? Because, sorry. If x is finite, no problem. If x is infinite, it's not a topology. I have to add this. For the whole, it's no problem, okay? If this is x and x minus x is empty, okay? But I have to add the empty set. Then it's a topology. So tf is a topology. Let's see. Well, that's an exercise, easy exercise. Maybe useful because you use some law, which you find useful every now and then more than law, okay? So the first is the empty set and x are in tf now. By definition almost now, okay? By definition. And for x, it's trivial, okay? So this is okay. Then we have arbitrary unions. So you just arbitrary unions, finite interceptions, arbitrary unions. So what do you have to prove? We take arbitrary many sets. So u alpha, okay? Open sets. What does it mean? So u alpha open? Open means now we have this topology, okay? We are working with this topology. There may be other topologies, but if you say open, then that's this topology you're working with. What does it mean? That means that x minus u alpha is finite, almost except this special case. Let's first forget this special case. So u alpha is open. That means that x minus u alpha is finite, okay? Finite complement. But what we want to prove, by the way, we want to prove that the union is, that the union is also open, no? So let's take the union, union u alpha, alpha and j. You should prove that this is open. What means open, finite complement, right? So x minus union u alpha. And now you have demorgan. So this is known as demorgan, one of two demorphs. The other one we see for the intersection. So what is this? Intersection, okay? That's a demorph. This is the first chapter of the book, okay? That's demorgan now, okay? That you should be able to prove, okay? In general one proves this two includes, okay? Take an element here and see it's here, and the other direction, okay? I will not do this now. This is the first chapter of the book. You find all this stuff. But what is this? These are finite, okay? All these are finite. So the intersection remains finite. So this, all this is finite. And this means that this is open. Alpha and j is open. Except that it's not completely true. Because we have unfortunately this special case, the empty set, you know, right? So we have to think a little bit about this. So we have to prove that the union of open set is open, okay? If one of these sets is the empty set, we can forget it. It doesn't matter, okay? The union doesn't, the empty set doesn't contribute to the union. So to make this proof correct, because this is finite, it's not true, okay? To make the proof correct, I have to add something just here. What do I add? We can assume that u alpha is not empty, okay? We can assume, we have to add that. Otherwise it's not true, all right? We can assume that u alpha is not there. And then it's true, okay? But unfortunately we have to add this. Otherwise we write something which is not correct, right? So we have to be careful. Okay, so this is arbitrary unions, and the same for finite intersections. So this is the third point, finite intersections. Well, it's sufficient for two, but we can use n, okay? If we prove it for intersection of two, then we have for three, four, five by induction. But it doesn't matter here. So let's take u1, un, open. So open means just to change in tf. That's the same as, oh, well, here I write open. That means tf, yes. What De Morgan's law says, x minus a union, but you don't write the wrong De Morgan's law. Let me first do this second De Morgan's law. So this means that we have to consider now the intersection, right? The intersection of, well, ui, i from one to n. And we have to prove that the complement is finite, except for this, so x minus intersection, so we have the other De Morgan law, okay? And now we have union i from one to n, okay? If you don't believe this, we have to prove this, okay? That's an exercise. Is this okay? So if this is okay, let's assume this is finite, finite complement, then now a finite union or finite sets remains finite, okay? The finite union of finite sets is finite, finite union of finite sets. And this means that the intersection ui, i from one to n is in tf, it's open, okay? Except if you write this, it's not completely true, okay? Because ui might be empty, okay? This is open, okay? Then this maybe is not finite, but then you have x here, okay? And then we get, well, the union x, which is open also. So this is true, trivial in some sense, okay? So you are, so I have to add here something, okay? As before. So ui, we can assume that ui, because if they are empty, then we have x here and the union is x on this case, okay? This is the main case here. The other case is true. So x minus intersection is the union of x minus thing and now, where's the other one? Let's check if it's correct. Yes. Is this okay? So as I said, if you have to exercise to prove this, then you have these two, no? And here you have these two, no? The union thing, you say, you take a point here, what? You take a point here, it means it's not here, okay? It's not here in this union. Then it's not in u alpha for each alpha. So it's here, it's not in u alpha. So it is in x minus u alpha. So it's in each of these. So it's in intersection. That's the proof, right? Of this inclusion, maybe too fast, okay? But I don't want to write that now, okay? This is the proof of this inclusion. You want me to write that? I don't know. I mean, there are two problems. So you just have another notion for this. You say, this is, you write the union u alpha. Okay, but it's a complement, right? x minus, I mean, it's the same. Sometimes somebody writes this complement, but I write this way anyway. I don't use this, okay? And the book doesn't use this. If you have the complement, it's x minus the same. That's the complement. It's better to stay to the book, okay? To the notion, okay? If it's clear what you mean, it's no problem, right? But it has to be clear. If you write this for complement, of course, I will understand, everybody will understand that you mean the complement. It's just a question of notation. The notation has to be clear and, okay? I don't insist on that it has, okay? But it's better to stay to the notation of the book. This is our basic text, okay? So what is the easier proof? There's an easier proof? Which one? What? Well, it's just a way to write, okay? Another way to write. To complement this. But that's exactly the same. This is exactly the same, I sense. Yeah, but this is not a proof. If you have to prove, I mean this and this is exactly the same in another notation, right? It's just another notation. And you say this is the Morgan's law, okay? So this is clear. It's the Morgan's law. But it's not the proof. If you want to prove it, you have to prove something, okay? You cannot... If you say I want to prove this, so I write this, okay? Then I say that's the same, okay? That's exactly the same. It's not a proof. If you want to prove this, okay? And then I prefer this notation. Why? Because... But it's a matter of taste, okay? As I said, let me say it again. This inclusion here, okay? Take a point here. We have to prove it's also here. That's this inclusion. Take a point here. That means it is not here. Because it's x minus, okay? If you take a point here, it's not here. Okay? Because this is going away. This is clear for me. It's easier to say, okay? So it's not here. What does it mean? x is not in the union. That means x is not in U alpha for each alpha. Okay? So that means it's not here. So it's here in the complement of each U alpha. In each of these. So it's in the intersection, okay? So this is the proof of this, okay? And you see a certain advantage of this notation, maybe, okay? Because it's... Well, what is the matter of taste, okay? For me, this is clear. This is shorter, maybe. But for the proof... And then you do the same in this direction, okay? The same kind of argument. And here you have also two things to prove, okay? Maybe you can do it in a single one. No, implies implies. Or you say if and only if, if and only if. And then... But that doesn't matter. Okay. So this is a final complement topology, which we will see again. Okay. What's this next example? We have another definition. So we will see this final complement. Well, it's very intuitive, okay? Final complement topology. Okay, now we have another definition. Basis for a topology, so definition. And now we have a set again. We have no topology. I mean, in general, in many books, you start with topology and then you have basis for a topology, okay? Here, you start with basis and then you define the topology. So X is a set. Not a topological space. You don't have any topology, okay? Basis for a topology on X. The basis for a topology is a collection, collection B. And this is again written now, like T, topology. It's a collection B of subsets of X. By the way, what means collection? Collection. Huh? So let's forget group. Group is more complicated. Collection just means set in some sense, okay? Collection family. Yeah, group, not in the aljubaric sense. Yeah, right, right. Then group. But group in mathematics is something from algebra, no? So that's, we don't want to use group anyway because this is algebra. Collection is just the word, which means, we may say it's a set of subsets, okay? But that's not so nice. Set of subsets for... For family. Yeah, family would be okay. Collection, family, the most, or set. Well, there's a certain difference between collection, family, and set. Because what may happen, the difference is the following. Collection B of subsets, okay? So it may be many subsets, but it may happen that all these subsets are the same, okay? So we have a collection of 20 subsets, but they are all equal, okay? So it's a collection with 20 subsets, okay? It happens that they are equal, okay? Some of them are equal, no? A set of subsets, okay? We have one, each one one time only, no? So if they are all equal, then the set contains just one element, but the collection has 20 maybe, or maybe infinitely many, okay? So this is a difference. It's more linguistic one, okay? Of family, collection, and set, okay? In a set, you write each element one time. In a collection, it may happen that you write the same element many times, okay? In infinitely many. And so this is a... because we use one collection of family. Family would be okay. And it's used also, no? But in the book, it's collection and collection. Even set, but this is better. Of subset of X such that. Two conditions. In the union, so for each X and X, it's in one of these, okay? Each, for each X and X, there is B in B, and now be careful. This is B in B, no? This is a collection of subsets. So this is one element. So this is a... This is a written B, and this is printed B, okay? So you have to don't mix up these two, okay? So this is a different book, okay? This is a written B, and this is printed B, okay? Sometimes I look similar, I don't know what is written, what is printed, but you have to be careful with this, okay? There's B in B such that... This means such that, no? Such that, such that. That means the union of all... How do I write this? The union B, B in B, okay? The union of all B, B in B, but sometimes this may be written out in a different way. So the union B, I don't know what is better, B in B. So this, and this means the same. Right? So that's the first condition, okay? Which is not the interesting one. The interesting one is the second. The second condition is the... Interesting condition, which says... So if you have two basis elements, B1 and B2, B1, B2 in B, they're called basis elements, okay? And X in the intersection, B1, intersection B2. Then, so I make a picture at the same time. Very stupid, simple picture, of course, but we have B1 and B2 and they have a point in the intersection. By the way, this is... That's another point we have to be careful, no? X, the set X is capital X. An element in X is small X, okay? So we have... This is a small X, okay? So we have small X in capital X. So you have to write carefully, okay? Then there is... There exists a third one, B3 in B, such that X is in B3 and this is contained in the intersection, B1 in the intersection. So here we have the third one and this is B3. This is a picture. Sorry? B3, okay? That's the definition of the basis. No topology. Of course, now you think of the analysis. So motivation for basis, if you... So we take R2, okay? The standard topology from analysis. So what do you think will be the basis for this topology? Yeah, open disk, okay? Take B. And that's an open disk. And then we have this condition, no? If you have two open disks, that we will prove anyway. But we will... Much later when we take about the metric space, okay? So we take a point here, right? And then we found the third open disk, which contains this point, even the center, okay? And which is contained in the intersection. No, this is a picture. So this is this condition, okay? And now we have also... It's clear how to define the topology, okay? Because how to define the topology here? Well, you remember the standard topology from analysis. Something is open. When is it open in analysis? Some set U is open. If given any point, X in U, you find an open ball around X, which is contained in U. Now we have to find this, okay? And this is the definition we take for the topology now, okay? So this is just the motivation for this basis. And now we need the topology, okay? So definition. The topology generated by the topology T generated... Definition of the topology T generated by the basis B. That's what we want to define now, okay? We have a set, we have a basis, and now we want to do topology. So the definition is the following. So U is in this topology generated. So the topology generated by the basis. So U is... And now we take what we said before. For each, for every X in X, X in U, like this element B in B, which depends on X maybe, so I may add an index here, no? BX in B such that X is in BX, subset of... That was exactly... So then we have to prove that this is the topology now. The topology generated by... So proof, so claim, the claim is that T is the topology. So we have, again, our three conditions. So empty set. So what's the definition here, no? U is open if for each, for every X in U, empty set is automatic. There are no points, so no problem. That's trivial, okay? For every X in U, there's no X, so we don't have to verify anything. Entity set in U and the whole, what is the whole? X in X, sorry, X in T is open, X is open, it's easier to say open, okay? This T is more technical. If X is open, why? That's the first condition of the basis. By the first condition of each X is in some, by the first condition of a basis. That says that each X in X is in some basis element, B in B, okay? So this is, what is also very easy is... So this is the main condition, right? Should use that at a certain point. But the second one is also almost trivial, okay? What is this second one? Arbitrary unions, no? Of open sets are open. So union U, alpha and J. So let... So these are open. Now you see it's trivial. So what we have to do is let X, this is our condition for every X in, okay? So let X be in this union. So what we have to do is let X be in set. Well, sometimes it's a black one. It's convenient to use this. Or you write then, okay? But it doesn't... Both are okay. Of course in the book, you will never find this. Because it's language, okay? It's not logic. It's mathematics. So you use language. You don't use... But yes, of course, shorter sometimes, okay? To write this term. What does it mean? X is in the union. That X is in some of these, no? X is in... Alpha is all good, no? Alpha, what? Alpha 0. Yes, right. Because alpha is a... Index, very index. So we have to specify now, okay? X alpha 0 for some... What? Alpha 0 in... J. But this is open, no? So here's a definition of open. So this means there is B, X, and so on. So this implies... Now, also, sometimes, okay? There is. There exists. Right? Implies, there exists. Sometimes it's shorter, okay? Well, it's not good style in mathematics. Okay? It's just convenient as a blackboard. Even if you write exercises, it's better to write, okay? You write then. There exists. There is. Okay? But... What exists? There exists B, X, and B, no? Such that. This double point is okay. Such that. There's a property in that. X is in B, X. Contained in... Put it all right. In... You alpha zero, no? And then this, of course... It's not a stupid to write this, okay? So this means we go... This is a first. Then we have this. Then we have this. Then we have this, okay? Sometimes we write this to make it clear or to emphasize, okay? But it's something that's stupid. And this means that... What do you want to prove? That X, of course, is in B, X. And this is contained now... Also in the union. What is it? U, alpha. Alpha. Okay? And this means that the union is open. And this is the last step. So I started with this. That means that the union is open. I'm sure the union is open. So this is... Okay? We didn't use anything, right? Trivia, okay? We didn't even use the first condition. Here we used the first condition, but here we don't use anything. So where we use the second condition for... for final intersections, okay? I'll prove this. So this is... And I write these final intersections. That's just a title, okay? That's what I'm going to do. So it suffices to prove it for intersections of two elements, by induction, okay? By induction. It suffices to prove it for two elements. For intersections of two elements, okay? It suffices to prove it for intersections of two elements. So what? So let... Give them two elements. B1, B2. B and B. Sorry. What we have to prove is that U1 and U2 be all the same. So what we have to prove is that intersection is open, no? So... Intersection. What does it mean? That X be in the intersection. U1 intersection root. And two U2. To both, yes. So this... Then, now I write then instead of... So it's the same. Then X is in U1. The best... Maybe it's to write nothing, but... And X is in U2, okay? Now U1 and U2 are open, right? U1 and U2. So... Therefore, then B1, B2 in B, such that X is in B1 contained in U1 and X is also in B2 subset by the definition of open. No, this is not the definition. There was a definition, okay? And now X is in... So then X, of course, is in B1, intersection B2, intersection B2. This is printed, okay? The elements. So these are elements of the topology. The topology is written, but these are printed, okay? So you ask, what is the written U? That's not so clear. It's all useless, but... This is a printed U, okay? It seems written. Right this way. But we will not see, okay? Because here we have... This is a topology, okay? Then X in B1. But this implies then there exists... No, it's not so clear. Then, then, then, then. So by the third condition, by the second condition, so here we use the second condition. For a basis, for a basis, there exists, there is B3 in the basis such that X is in B3 which is contained in B1 intersection B2 which is contained in C1 intersection U2. And this means... So this was arbitrary. This means this is open. So... So U1 intersection U2 is open. It's in T. Well, of course I should be coherent in one proof at least, okay? So here's in T, then this is in T. Well, it's open, then this is open, okay? But I will change from one notation to another. It means the same. So this is very formal, right? It's a topology. It's very formal. So we have a basis and then we have a topology, okay? That's a... In other books, you have first a topology and then you say a basis for a given topology is something, okay? So... Let me discuss this also. So T is called the topology generated by B. No? Maybe I wrote that. But that's a way how to talk. So maybe... So T... Did I write that? I don't see it. The topology generated... Yeah, I wrote anyway. So I will not rewrite. Yes. It's an easy lemma. So what you see, this is a very formal... In point set topology, in general topology, okay? The proofs are from logic, okay? There are not many computations, okay? It's a language from, okay, to logic, okay? That you have to learn. In analysis, you have computations, okay? Maybe. Here you have to do small proofs all the time, okay? Because it's close to logic. General topology. So the first lemma. Small little lemma. So let B be a basis for topology T on X. What does it mean? That means T is the topology generated by B, okay? So T is a topology that is what this means. It's a topology generated by B, by the basis B. That B be a basis for the topology, for topology T. Okay? Then we can characterize in a different way. We have a definition of the topology, but we write this lemma. Then T is a collection of all unions of elements of B. Then T in topology is a collection set. T is a set topology, collection. Of all unions of elements of B. Maybe these are the words in the booking section. Of all unions of elements of arbitrary unions for proof. So since B is contained in T. So this means basis elements are open. Why B is in T? What is the definition of T? Definition is something is in T if given the point there is a basis element which contains the point and is contained. Now we have a basis element so given the point here we take B itself. Since B in T. So this is the definition. By definition of the topology T. Since B is in T arbitrary unions of elements of B are in T. Because union we know as a topology arbitrary unions of elements of B in T are open. So this is a collection of all unions. If we take such unions then we are in T anyway. But we have to prove still that we two directions that we get all elements in this way. If we take arbitrary unions we have open sets. Now we prove each open set is an arbitrary union. So since B in T arbitrary unions of elements of B I wrote conversely. Conversely I don't like so much now conversely. For the other direction. What is the other direction? It is very often here is a problem of language in a good way. No computation but you have to concentrate on the language. How to say it in the best way. I don't like this too much. Well, it's clear what it means conversely. Let me write this conversely. Let U be in T O. And we have to prove that U is a union of elements of B. That's the other. For each X in U there is so what it means open for each X in U there is the basis element BX in the basis such that X is in BX and this is in B by definition of T. That's the definition of T. By the definition of T the topology generated by B. And then I'm thinking yes, but I'm thinking about I mean all of this Zen is not nice also a set called U. U is equal to the union BX X in U So this is looks strange but it's familiar one direction is all this BX where is it? BX in B BX in B BX in B such that X is in BX in B anyway contained in that's the definition for each BX in U So this means one direction is clear this one and the other one is equally clear why if we have a point X in U however we have BX for each X so also this circle and this means that this is a union of elements is a union of elements of basis elements basis elements so that's the proof two parts first part arbitrary unions of elements of B are open second part each open set is this little lemma so there's a second lemma and there's a third lemma they are very similar I will prove the second one but then maybe not the third one but because the proofs are very but if you have to write such a proof you have to concentrate that's clear that's the main point so there's a second lemma which is used very often so it's a technical lemma but it's used to compare topology so we have two bases so in general topology will be given introduced by a basis that's why the book first the basis and the topology why because in general it is easy to describe a basis open this in R2 this is a standard topology the basis in general is easy to describe the topology each open set is difficult to describe indirectly but not directly open sets in R2 in the analysis are very complicated extremely complicated the basis is very simple so you start with a simple basis and then you have a topology and you can only define the topology you cannot describe in general all open sets there are too many that's the reason why the book starts with basis and then topology because that's an important passage first you want something as a basis and then you have the topology first you have the topology and then you look for a basis sometimes also in general it's the other way so lemma so let B be prime let B and B prime be basis so it's E let B be primary basis for topologies T T prime on X respectively which means that B is for T B generates T B generates T prime then the following equivalent well I use as it's written in the book then the following equivalent you find everything in the book of course and I recommend to read to take the book once you have copies and to look at the book to compare this then the following equivalent one is finer than the other let's say T prime is finer than T T prime is finer than T I hope you can read is finer than if you cannot read something because I'm writing fast so this is an N not a U it looks very similar sometimes not so clear it's finer you should ask if you cannot understand some English work because that's important the English work is finer than T and the second one is for each B in B and X in B there is B prime in B prime such that X is in B prime and contained in B that's the condition this is the condition which we use often to compare the problem this means T prime is finer than T for each so you start with this topology for each B for this topology at each point you have to find something small in the other topology which contains the point and is contained in this B that means that this topology is finer it's the right direction T prime is finer than T anyway so we have again two directions so let me I'm used to writing this the first is so we know what we approve 2i implies i that's this direction so what? i so we have to prove this T prime is finer than T that's what we have to prove so that any open set here is also here in this topology so we start with an open set here and prove it's here also so let U be in T so let U and we have to we have to prove that it's in T prime what does it mean T prime so let U in T prove it's in T prime what does it mean by the definition of T prime for each point we have to find B prime yeah belong to this so we take a point okay for each and X be in U what we know however so you see it's clear what comes out more or less but we have to write since U is in T since B generates T no that we know there is B sorry yes yes since B generates T there is by the definition of the topology generated by B there is B in B such that X is in B contained in all this called U and now we have this second condition here so we have exactly the same what is written here we have B in B and X in B there is so by 2I there is there exists what is it called B prime in B prime Z X in B prime in B X is in B prime contained in B and of course B still is contained in U okay so I add this contained in U so this means then that U is also in T prime because we found this element and this implies that U is also in and this means that T prime is final T so T prime is final so this is one direction and now we have to use the other direction which means I implies so I write schematic in this way okay so what do we have to prove now I implies 2I so I have to prove 2I so for each B in B and X in B so without thinking let B be in B and X in B or let B be in B and X in B since yeah B is contained in T anyway the basis elements are always open what is our hypothesis T prime is final then T means T is contained in T prime and this is contained in T prime okay so this is contained in T prime by the definition of T prime this is of T prime B prime T prime is a topology by the definition of the topology T prime generated by the diffusion of the topology T prime generated by B prime now I have no well I go on here because this is the other direction by the definition by the definition of course it's a good better law by the definition of the topology T prime generated by B prime so this is in T prime this is open in T prime which means we find the base okay by the definition of the topology there exists remember I write there is there exists as the same there exists B prime in B prime such that X is in B prime contained in B which was this open set record this was an element of the topology and this is the definition and what what do you have to prove I implies 2I yeah exactly T prime is finer than T this is no and now there is still one lemma which I will maybe not prove also the time seems almost finished so I will write it tomorrow of course we need some examples now of bases we see many examples the ideas that the idea is that the basis is easy to describe okay and so we define topologies by defining a basis okay and then we have a topology that's the and even a sub basis we will define tomorrow okay so we have topology, basis and tomorrow we have something still weaker which is sub basis if it's a topology it's okay if it's not a topology we take it as basis if it's not a basis we take it as sub basis and sub basis is almost no condition so everything is almost everything is a sub basis okay and this is a way how to generate topologies and then we need many examples okay what I wanted to ask is so this is very formal okay so what you need is a book okay you need copies the first chapter is good to have also a copy okay but you need the second, third and fourth chapter and then later chapter nine and then we may be something from 11 but that's not so clear so for the moment what is three and four of the book that you need in any case we will not do everything okay it's too long but we will do the main some of the main points of these three chapters two, three and five okay and then later nine chapter nine two, three and four what is this well I cannot you look in the book what is two, three and five chapters two, three and four okay that we will see where we arrive okay because that depends on you also because we may go more or less fast also right that depends more on you then okay if you say something is not clear it's too fast and you have to tell me okay it's too fast for many reasons maybe the English is not clear the way the letters is difficult to read okay too small in general or not I don't write very small but also that might be a problem okay so you cannot read even now then you have to understand the English and then you have to understand the mathematics so that we have to see how fast okay you can go slower sometimes how fast and that depends where we will finish okay but we will not do all the whole chapters two, three and four but the main points yes so have a look in the book now you have to make copies okay for the exercises and examples in any case you need the book chapters two, three and four for the moment okay later chapter nine if you have there's somebody around still from other years last year he will have the copies okay and so it's easier you don't have to take the book and to make each page but you have it in one minute okay right and that's much easier of course but we need it also for the exercises okay there are many exercises and we will use these exercises and as you know we there will be a written final exam okay a written final exam no the exercises are important okay you have to write okay exercises to solve so it's very important exercises if you understand the exercises more or less but there are many exercises so you cannot do anything there are too many sometimes so we will take some easy ones some medium ones not too much some of the medium who of you has done already a course of some topology I mean topology you do in analysis what are the reals are two are L maybe okay real analysis so you have topology some topology you have compact you have connected and we will general once okay from a more formal point of view some who has seen a course of general topology more or less all so that might mean that we may go at a certain point also it depends on you okay so you have to tell me it's too slow or it's too fast we cannot understand the English we cannot understand the mathematics we cannot read okay so we see tomorrow at four again we have other this is the first course you have now when did it start today so today you had real analysis and then you have complex analysis no functions when this week starts three courses all three and then in October they will okay so we see tomorrow