 So the definition, the topological space x is first countable in a point x and x. If there exists, if there is a countable system of neighborhoods, so countable u1, u2, u3, so it's countable system of neighborhoods, such that this is neighborhoods. It's too long, so neighborhoods. u1, u2, u3, u4 and so on, such that given an arbitrary neighborhood u of x, neighborhood u of x of this point, there exists some index here, there is, there exists n in n such that u n is contained in u, u n is contained in u. Some of these is contained in u, okay? u is for any neighborhood. We can assume, we can assume, we can also assume that u1 contains u2, smaller, decreasing, okay? Smaller and smaller. u3 and so on. We can also assume that this holds, okay? Why? We can assume that we just take intersections now. So here, it's clear, we take decreasing set of neighborhoods, okay? So we take, if you have this one here, u1, u2, u3, okay? Then we always can do this, u1 and u1 intersection, u2, u1 intersection, u2 intersection, u3 and so on. So this is decreasing then, okay? If it's not decreasing, then doing this, we get a decreasing system of neighborhoods, okay? So what? Is there a problem? What? Okay? So this is sometimes called a accountable, a fundamental system of neighborhoods of x, okay? Fundamental, with this condition probably, because this is a strong condition. That means, of course, un is in u, but then also the others are all, okay? Because they are decreasing and so they are all in u, okay? So then, of course, ui is called in u for i bigger equal to n than all the others which come out. Those are getting smaller and smaller, okay? So this is called a fundamental system of neighborhoods. Fundamental system, not in the book of neighborhoods of this point of x, fundamental. So what we, first countable means we have a fundamental system of neighborhoods accountable. Accountable, fundamental system. Fundamental is this condition, this or this, okay? Axis first countable, if it's first countable in every part, okay? Axis first countable, if x is first countable in every point, x and x. So first countable. There's a local condition, no? In some point, first countable. And first countable then, if it's a local condition. So the important condition probably is this here. Now this is a fundamental system of neighborhood, okay? Decreasing system. And then if you have any neighborhood, at a certain point, this is contained in this neighborhood, okay? And then also everything which comes after this, of course. The main example is, of course, this is modeled on example. If x is metric, a metric space, matrizable, metric, okay? Then x is first countable, of course, x is first countable. Why? If you want this condition, what do we take? We take, so take u n equal to u i equal to the ball, open ball, around x with radius 1 over i, okay? This is decreasing, anyway, okay? And if you have an arbitrary neighborhood in the metric topology of x, then there's some of these balls, no? Contained. For some epsilon and then. So this is the case, okay? A matrizable space, metric space, is first countable in each point. So it's first countable. It seems almost, so this matrizable is not a topological condition, no? It's not topological, it's metric, okay? But this is a topological condition, no? First countable. So metric implies first countable, okay? So this is, for being matrizable, first countable is a necessary condition, okay? Necessary, if it's not first countable, okay? It cannot be matrizable. That's a topological, that's a necessary top, so maybe that's worth to write it. So being first countable is a necessary, it's the same as this, but it's a way to say it. It's a necessary topological condition, topological condition for being matrizable. And to recall the matrization problem is to find necessary and sufficient conditions, okay? Which at the same time necessary and sufficient, okay? But may normally say, well, maybe first countable, that's it, okay? That's the same. And first countable does not imply house softness. House softness you need anyway, okay? So if it's house soft and first countable, no? Then try to prove it's matrizable, okay? But it's not true, okay? Unfortunately, it's not. Well, you can think about an example later, okay? It's not true. It's not sufficient also. We need more conditions. We have to find other conditions, okay? So anyway, this is the important thing, okay? Matrizable space is first countable. In fact, first countable is modeled on matrizable, okay? With this as the fundamental system of neighbors, the balls of radius and whatnot. And then, of course, so remark, the sequence lemma holds if you replace matrizable by first countable. The same proof, okay? So I'll write that. So observation, the sequence lemma, which we did yesterday, no? Holds, if one replaces... So in the sequence lemma, there's a condition in one direction matrizable. Matric, okay? Matric, matrizable. Maybe, I forgot, matric or matrizable is the same, okay? It's just a different point of way. Matric by first countable. Then the sequence lemma holds. We don't need matric. We need first countable. Just in the proof, just replace... So also that I will write in the proof of the sequence lemma. Just replace... So there was... There was this ball, okay? B, D. In the sequence lemma, use these balls for one direction. X, one over n, no? By U n, okay? By U n. So where this is... U n is the fundamental system of neighborhoods, okay? Where U 1 contains U 2 contains... So we need this condition here, no? And so on, it's a fundamental system of neighborhoods. That means every neighborhood contains one of these, and then also all which come after this, okay? It's the same proof, exactly, okay? So just write U n when you see B, D, one over n, okay? And then it's the same proof. So this means that these spaces are not metrizable, but so this means then that they are not first countable, no? So our omega with box topology, that's what we did last time, no? With box topology, it's not first countable. Well, there's... In just one point, we did it, okay? In the point zero, okay? We use zero. It's not first countable in zero, in the point zero, in our omega, okay? Because we use the sequence lemma, no? We use the sequence lemma. So our omega with box topology is not first countable in omega. So it's not metrizable, okay? That's what we did yesterday, but it's not first countable. It's the same proof. And rj, j uncountable, that was the second, is not first countable with the product topology, with product topology, no? Here's the box topology, and now with product topology, that's what we did yesterday, with product topology, is not first countable. Again, in the same point, we use this point, in zero in rj, no? So this may be parenthesis. One point where it's not first countable. Then it's not first countable, okay? It's not metrizable. So it's a nice condition for being first countable. The sequence lemma holds, no? For spaces which are... So if you, a sequence lemma means if you have a point in the boundary, okay, now, a limit point, if you want. And it's Hausdorff, also, okay? And then we find the sequence, okay? Then we find the sequence, which converges. That's the sequence lemma, okay? But it has to be first countable. Or, which is maybe stronger, metrizable, okay? Metrizable is stronger. They're not equivalent. Okay, and other examples. So proposition, or observation, proposition, easy. If x is second countable, then it's first countable. Second countable implies x first countable. So if it's second countable, implies first countable. Second countable is a global condition, no? Second countable means there's a sub-proof. There is a countable basis. That's a global condition, no? First countable is a local in each point, we can say, no? So proof, what means first, second countable? Let b be a countable basis for x. And now we have to prove first countable, okay? So let x be in a point, in this point. Let x be, choose a point, x in x. Then what I take is, I consider b prime. Here I take all b in b. I just take all basis elements, which contain x, okay? All basis elements, which contain x. The others are not interesting for this point. And b is a countable, so this is countable, okay? So b prime is countable, of course. So b prime is countable. So b prime is, we can count it, no? b1, b2, b3, you know, it's countable. So b1, b2, b3, and so on. It is certainly countable. And now we find, so this is already the definition. If you have, ah, we have to prove. And we can order this, no? No, we can make it decreasing, no? Let's say, so you can u1 equal b1, okay? The first of this. u2 equal, what do you take? b1 intersection b2, no? u3, b1 intersection b2 intersection b3, and so on, okay? So we get a decreasing system, u1, u2, u3, u44. And if, of course, I have to prove that this is fundamental. So, but that's the definition of topology generated by the basis, no? If u is any neighborhood of x, then there exists, of course, a basis element. Then there is, then there is. That's the definition of the topology generated by a basis, no? Then there is b in b such that x, which contains x in b, okay? So this means b is in b prime, in the other one, okay? So b is really in b prime, because it contains x. It's the neighborhood of x, okay? So it's one of these. That's it, okay? So given any neighborhood, we find one of these, which is contained in this neighborhood, okay? And then taking intersections, we can make it equal. We get actually a small thing. So this is, second countable implies first count. Second countable implies first count. We need some other condition for matrizable, no? But there's one missing condition, which comes later, which is one of these conditions, T3, T4, okay? T1, T2, T3, T4, no? Which is normal, or regular, normal or regular, one of these two. And then it turns out, together with this condition, first countable is, as we already know, in any case, necessary, okay? Second countable, together with this condition, is sufficient for b matrizable. So we are between first and second countables, so this is an interesting notion, okay? Matrizability is, together with some other condition, normal, which we do later, but it's between first and second countable, somewhere between. It's not clear where, okay? It might be first countable, it's not, it might be second countable, it's not, okay? One is necessary, the other is sufficient. So these are interesting conditions anyway. What else? So this is first countable, okay? Now we know what is first countable. So metric spaces are first countable. Second countable spaces are first countable, okay? Let me think one second, I didn't think about it. So we want the first countable space, we need another condition, of course, no? We want the first, if we want the first countable space, which is not matrizable, okay? So it's not necessary, it's not sufficient, okay? The first countable space, which is not matrizable. The matrizable space is first countable, okay? But we want to see that the other, we want a space which is, then we need another condition, no? Hausdorff, of course. Hausdorff, we want all this, okay? But we want the first countable space, which is not matrizable. Then we need another condition, okay? Which distinguishes, okay? Because other, how to prove, no? And so we have to find another condition. Well, we will see that later, that's not the point, no, okay? So this is first countable, important condition for... Then in the first chapter, which I close now, I just give one more definition, which more or less is easy, but it's used often, and there will be some exercises, okay? I will get exercises, so the next week I'm not in here, okay? So I don't know, you have other lectures, I suppose. Other courses, no? We see, not next week, but the week after next week, okay? I will give today some other exercises. So last definition in this is quotient map, okay? So x and y are topological spaces that we talked a little bit about already. x and y. f from x to y, a function, a function, a map. f from x to y is a quotient map. So what is a quotient map? If a quotient map, that's just a name, it's continuous, of course. We are in topology, so certainly it should be continuous. Quotient, taking quotients means surjective also, okay? Surjective. And the following condition holds, that's the important condition holds. This is okay, but... And the following holds. v is such that v of y is open in y, okay? v is open in y, even though if it's pre-image, it's open in x. f minus 1 of v is pre-image, it's open in x. That's the condition, okay? One direction here is anyway, I don't have to write this, okay? Because it's continuous, no? Anyway, continuous means if v is open in y, then, of course, f minus 1 of v has to be open in x. That's continuity, okay? Interesting is the other direction here, okay? If and only if. So as a remark, given the topology in x, then y, given the topology on x, then y has the finest topology such that f is continuous. That's again such a situation, okay? Then y has the finest topology such that f is continuous. No, we take as many open sets as we can, okay? But the pre-image has to be open, okay? We need this condition as the pre-image is open. So then this is the definition, then. That's the quotient topology also of the name, okay? So this is the quotient topology. Quotient map quotient topology. This is the quotient topology on x, on y, sorry. Because this is the maximum we can take. We say if something is open in y, okay? Then, of course, the pre-image has to be open in x, no? So we take exactly those, okay? We cannot take more. Because then f would not be continuous, no? We take exactly those and this is the topology and this is the quotient topology. And the map is called the quotient map. So the quotient topology quotient map, okay? Just names. But that we saw examples, okay? The finest, the largest topology such that something is continuous, no? That's a typical situation. The closest topology is not interesting, no? Because that's indiscreet, no? If we give you the indiscreet topology, then everything is continuous, no? So here we have to take the finest. That's an interesting topology, okay? And this is a quotient topology. And then this is called the quotient map, okay? That's just the definition. There will be one exercise today which... So I'll just give the definition here. There are some exercises. Okay, so this in some sense finishes the second chapter. That's what I do of the second chapter. And now, because I have to give exercises, I'll start with the third chapter. The third chapter of the book is on connected spaces, okay? So I give some definitions now for connected spaces, okay? And then the next one is compact. But there are various notions of connected and compact, okay? Not just one. So we see two for connected, at least two, okay? There are more. We will see. Connected, arc connected, locally arc connected, locally connected, no? And compact also. Compact limit point compact, sequentially compact. So there are various notions and we will discuss this a little bit, okay? This is... So now I start with connected because I will give exercises. These are two, three exercises, okay? For connected spaces. You have two weeks, so... I don't give too many. So this is a quotient topology. And now I give some definitions and some basic facts about connected. So we have connected what is called connected spaces. Connected spaces. Connectivity. So I give the basic definitions. So definition. So connected is very... Connected means you cannot separate the space in two open subsets, which are non-trivial, okay? There's one open subset and another. And the space is just a union of two open subsets, okay? That's not connected, okay? That would be a separation of the space. So what is a separation? A separation. It's just a way to talk. A separation of the topological space X is a pair u, v. It's a pair, well, u, v of subsets of X. And then I list the conditions. It's just an abbreviation of the conditions. So we don't have to say all of these conditions. Okay? A separation of the topological space is a pair u, v of subsets of X such that. So what are the conditions? That's the way of talking. So the first condition is u and v are open. Well, u, v are open. The second condition is the X is a union of both. So it's a union of two open sets. Well, that's not a big thing. So they should be disjoint also, okay? Otherwise it's trivial, no? So the next one is that u intersection v should be disjoint. So now we have a separation, okay? And there's missing one condition. It's not trivial, the separation, no? Because we have always a trivial open sets, which are the empty and the X. So we don't want those. So u, v are different from empty sets and X. I have to write just one, okay? Because these are complementary sets now. So they are both different from one or both. So this is a separation. These are the conditions separation, okay? Exactly what I said. Separation means you separate in a non-trivial way into two open disjoint subsets, okay? X is connected if it has no separation. That's it, okay? X is connected. That's the definition of connected now. If there is no separation for X. So the first remark, which is the same, might define also the remark, observation, trivial. X is connected if and only if the complement. So these are complementary sets, no? u and v are complementary sets, right? u is a complement of v. They're both open. But that means they are as well right here closed. Because v is a complement of u. And u is open, so the complement is closed also. Also open, okay? So these sets are both open and closed, this u and v. So here you can note that u, v are both open and closed. Both open and closed, okay? That's clear, complementary set. So connected then means the only sets which are open and closed are trivial. That is empty or the whole, okay? So the only subsets of X which are open and closed at the same time, no? Which are open and closed are the empty set and X. These are always open and closed. But connected means there are no others. This is equivalent to the definition, okay? Clearly. Because that's why I write. Note that, so here you note that u, v are both open and closed in this definition. And non-trivial, okay? If it's not connected. If you have a separation. Connected? No. Example, RL is not connected. Proving that something is not connected is easy in general, okay? Why? We just have to find one separation, okay? Proving that something is connected might be difficult. Because we have to prove there's no separation, no? That's another quality, no? Just one separation is easy to find in general, okay? So this is easy. Of course, being connected is useful. Being not connected is not useful, okay? So, clearly. So this is a separation. What is a separation? Somebody has a... So we have to find one separation, okay? So what is a separation? RL is... What? Sorry? Say again, from minus infinity? A open? So A is some point here, no? Choose any point here, yes? And now? Yes. It's a separation. That's what you say, no? That's a separation. It's a separation. Why is it a separation? This is... I mean, the open sets are... Given any point here, you find the basis are half open intervals, right? But given any point here, you... Of course, you find the half open interval, which contains the point, okay? And it's contained here. And here's the same, also for A. Here you cannot close, no? The only point if you do... Don't write it. If you write this, of course, it doesn't work, okay? Because this point, you don't find this neighborhood, okay? So you have to do it clearly. Clearly, you have to do it in this way. Because it's a lower limit topology, not the upper limit, no? So this is a separation. They are open. They are disjoint. They are non-empty. What else? So we have everything here, what you mean. That's a separation. That's not connected. That's easy, no? R, the standard R, that's not so trivial, no? That's analysis, okay? That is connected. But there, we have to prove something. There, we will prove a little bit more general, okay? RL is not connected. The reals with the standard topology, we will see, is connected, no? And this is an important fact from analysis also, okay? It's on the basis, yeah. That we will do. But that's more difficult, as I said, no? It's more difficult to prove something is connected. Proving it's not connected. But it's not so useful, no? It's just an example. Okay. No, some easy properties. Very easy properties. And then I just prove one small theorem. So, it's almost called lemma, lemma in the book. So lemma, okay, lemma, lemma. If X equals C union D, before it was UV, it's a separation. So I don't have to write all these conditions now. It's a separation. C union D is a separation of C, of X, sorry, of X, of course. And Y subset of X is connected. So what does it mean? Y is a subset, but it's a subspace, okay? And if you write this Y subset of X is connected, it means the subspace Y is connected, okay? Of X is connected. That's what it means, the subspace. So here's the subspace topology, no? That's what it means. Y of X is connected. Then what you suppose, what you, what seems reasonable is then Y is contained in C or in D, no? Or Y is contained in, because that's, because otherwise the proof is very easy proof. So we have Y is equal to what? Y intersection C, union Y intersection D, of course. They are both open because it's a subspace topology. They are of course disjoint. The union is everything. So, so yeah, let's write this. Yeah, this is open, disjoint. Both are open disjoint, no? But it's connected. So what? Yeah, something must be trivial, okay? So this implies, so Y is connected. Y is connected, and all this together implies that something has to be trivial, okay? So Y intersection C is empty. And this of course, or Y intersection D is empty. And this of course implies that Y intersection C is empty, it's Y contained in D in the other one, Y contained in D or Y contained in C. That's okay, so this is very easy. Next property, lemma, another lemma, which is small, easy lemma. So we have a collection of subsets, okay? So suppose they are called A alpha in X are connected. This means A alpha are connected, okay? Subsets, subspaces, S subspaces, no? Are connected, alpha in some index set, alpha j, no? Because all these are connected, subspaces, and they have at least one point in common. And the intersection of all these A alpha, alpha and j is not empty. They have one point at least in common, okay? Then the union alpha and j is connected, okay? So in other words, a union of connected subsets is connected if they have at least one point in common, okay? That's another way to say that. A union is the same, but maybe a union of connected subsets is connected if they have at least one point in common, these connected subsets, okay? The intersection is not empty, that's the same, no? If they have at least one point in common, so this is the same, obviously. So what? So they have a point in common. So let's choose such a point. There's one point in the intersection. U of alpha and j. Oh, here's a common point. Suppose now that, what do we have to prove? The union, that's why the union of A alpha and j. We take this union, we have to prove that it's connected, no? Suppose that this is a union, now I might go by contradiction, okay? And say that this is a separation. But it's sort of not a good, maybe not a real contradiction. So what I assume, not the whole. I assume these are open, disjoint, and disjoint, just these two. And then I prove one is empty, okay? No separation, no separation, okay? So open and disjoint. Suppose that y can be written this way, open and disjoint. Suppose P is somewhere, no? P is in C or in D, no? This is a union, okay? So suppose that P is in C, okay? It must be in C or in D. So let's suppose it's in C. But A alpha is connected, A alpha is connected. Each A alpha is connected. This means that A alpha is contained in C or in D by the lemma, okay? The first lemma. This means a connected sub, well, separation. A alpha, I didn't write separation, no? But that doesn't matter. So it's contained here or here, okay? It's contained in C or in D, which might be empty still, okay? But I don't care about that at this point, in C. However, P is in C, no? And this implies, of course, so P is in C. Suppose that P is in C. And this means that A alpha is in C, no? P is also in A alpha. And this implies A alpha is contained in C. But this is for all alpha, no? For each alpha. And this means that the union A alpha alpha and J is equal to C. It's C, we get C, okay? We don't get anything in D. And this means so D is empty. So this means there's no separation. C has no, what? Y has no separation, no? This implies Y has no separation. This is almost a separation, except I prove this is empty or this is empty, no? Y has no separation. So this is the proof. You may, maybe you find it easier to write by contradiction. You write separation, and then at the end you prove one is empty. That is a contradiction because that's not a separation, okay? That's sort of a fake contradiction, okay? Because this kind of contradiction, we prove by contradiction, no? Okay? So let's assume it's not true. Then you prove it's true, and then you write, now we have a contradiction because we assume it's not true, no? But this is stupid, no? That's sort of the very elegant style. Well, okay. So this is this lemma. Now we have one last one. Next, what is next? Now this is interesting proposition. Let A and X be connected. Let A in X, let A be connected. Suppose this is connected, B connected. And there's B between the, and B. So A is contained in B and this is contained in the closure of A. Closure of A in X, of course. So we have something, this is connected and this is contained between A and the closure of A, okay? Then B is connected. In particular, the closure of a connected subset is connected, okay? In particular, that's a special case. We can B take this one, okay? In particular, take B equal to, in particular, the closure of a connected subset. We say subset but it's subspace, okay? But subspace is too heavy, so subset. Of a connected subset is connected. That's the case. That is the case B equal to A closure, of course. So a proof. So this is easy proof. By contradiction now. Maybe fake contradiction with the same kind of, but it's easier to write. Suppose it's a separate, we have a separation. Suppose B, so what have to prove? B is connected, no? Suppose we have a separation. Suppose B equals C union D is a separation. Of B, of course, no? So this is a separation. A is connected. Subspace of X, subspace of B, subspace A, okay? There's this almost preliminary mark. A subspace of a subspace is also a subspace of the large space, okay? A subspace of a subspace of a subspace of a subspace is a subspace of the large, no? And everything in between also, okay? So, okay. Suppose there's a separation of B. So with a lemma, some lemma implies A is connected, no? A is connected. So then, by the lemma, A is contained in C or A is contained in D. Right? If you have a separation and something is connected, then it's in one or in the other, okay? Because it's... Suppose A is in C. Suppose A is contained in C. Okay. Suppose A is in C, yes. Then, we want a contradiction, okay? A is in C. A is in C. Now what is B, no? Then the closure of A is in the closure of C, of course. Then A... Then the closure of A is contained in the closure of C. And recall, this means closure in X, no? If you write this, we mean... Because there are many spaces around here, no? A is in B. B is in A bar. A bar is in X, no? But the closure means in the large space, no? So, closures... Closures... If you write it this way, it means closures in X. In X, who's asking? Closure... If this sign here, no? If you write this sign, we mean always closure in the large... in the X, in the original space, okay? For this inclusion? No, no, no. If A is in C, then A is in the closure of C. Then the closure of A is in the closure of C, right? If A is in C, for example, then A is in the closure of C. Because C is contained in the closure of C. Then the closure of A is contained... Because the closure is... What? Yeah, the smallest closed, okay? No problem. Okay. So then A is in C, and this implies, of course, that... Where are we now? A is in C. So B is in the closure of C. This B is in the closure of C. Because this is contained here and then here, no? B is in the closure of C. So B is equal to... Well, I write it in a different way... B is contained in C. And, of course, it's contained also in B. That's trivial, no? I just added B here, okay? It's contained in the closure of C, it's contained in B. But what is this? That's interesting. Yeah, that's the closure of C, but in B now, okay? This is closure of C in B. This is some lemma, which we proved, okay? A lemma, which has no name. The closure in a subspace, how do you get that? Take your closure in the large space and intersect, okay? That's an important fact, okay? The closure in a subspace, you take the closure in the large space, which you write with this symbol, okay? And then you intersect. Take the closure in the large space, intersect. So this is the closure of C in B, by a lemma. However, what is C? C is this, no? It's closed in B. But because this is separation, these are both open and closed, okay? Here. There's a separation of B, no? We are in B here. But C is closed in B. So this is C since, okay? This is the closure of C. This is the closure, but C is closed. So it's equal to C, since C is closed, okay? It's clear, right? Yeah, C is closed in B. So what we have, this says that B is contained in C. And this, where's B? B is contained here. This means this is empty. And D is empty. So it was not a separation, okay? So this was not a separation. So we have a contradiction, if you want, okay? Contradiction, because we write separation here. So it's contradiction. So that's it. Some other, last easy property. So lemma, this is also lemma. This, in some sense, is the image of connected set is connected. So if f is continuous, f from x to y is continuous. And x is connected. f of x is connected. So in a shorter way, continuous image of, yeah, exactly. A continuous image of a connected set is connected. So let me also subset, no? Then you restrict. It's still continuous. So a connected set is connected. That's what it says, no? In short. So proof, very easy. Well, the only interesting point is, we can assume that f is, that f, we can assume that f is subjective. Assume that f from f of x is y. That is so, which means that f is subjective. Why can we assume that? By taking core restriction to the image, OK? That remains continuous. By taking the core restriction. We will write, I was codomain domain, or restriction, core restriction. And by taking the core restriction to the image, which is still continuous, OK? So we can assume. And now it's trivial. Suppose y equal c union d is a separation. So y is not connected, OK? Suppose y is not connected. So we have a separation. And then, yes, we take pre-images, of course. X is equal to f minus 1 of c union f minus 1 of d. It's a separation. Also a separation. It's also a separation. So this means, that's a proof, OK? But what did we prove? We proved that if y is not connected, then x is not connected, OK? So if y, x is connected, then y is connected, OK? If y is not connected, then x is not connected. If y is not connected, we have a separation, OK? Then we have a separation of x. So x is not connected. So if x is connected, y is connected. Now this is AB equivalent to not A implies not B, OK? It's not by contradiction. Yeah, right. Why we need subjective here? Where do we need subjective here? Well, these are open then, OK, in y now. Otherwise, if it's not subjective, they are open in the image, but not in y maybe, OK? That's all. OK, so the last thing for today, which I want to prove, and this is a nice exercise. It's a theorem. It's a nice exercise for product, and this is important, product, uniform, box topology, OK? And it's a good exercise for these topologies. So I will give the proof. So the theorem is the following. A product of connected spaces is connected. Very easy, OK? A product of any product. So I already used a product. Product, whatever they are called, X alpha, alpha and J. A product of connected spaces. If you say it this way, we mean with a product topology, OK? I said everything goes wrong with a box topology. This goes wrong with a box for the product topology, OK? So I write for the product. If you don't say anything, it's a product topology. OK, that is very short and easy. A product of connected spaces is connected. In the book, you find only finite product, the proof. In this new version of the book, no? In the old version of the book, you find this strangely, OK? But the interesting part of the proof is really the infinite case. So I give the proof, no? It's not in the book, in the new version, OK? It does it for finitely many. This can be proofed in, well, anyway. So there are two cases, proof. If you want to read the proof, I don't know if the old version of the book is in the library, OK? Probably yes, there's some old copy of the book. This is from 1985, no? You have the new one, OK? From 2000, something. I looked and I think he didn't prove. Only for finitely many, finite product, OK? But it's a very nice exercise for a product. And for a finite product, this gets much less interesting because you have product topology and box topology. For a finite product, they are the same, OK? You don't see even which one you have, OK? It's the same. So that's not so interesting. The interesting, the difference comes for infinite product, OK? Between product and box topology. So that's why the infinite case in some sense is much more interesting, OK? Because it's a case of finite product. For a finite product, yeah. It's all proof. Anyway, we prove first the finite case, OK? In the first step. Finitely many. Well, we prove for two. Finitely many, we can prove for two also, OK? By induction, we have for three, for four, for five. So first case, case of a product, well, they're called x, y. X times y of two connected spaces. X and y. So we have a product of two, finite product of two. Well, this is by picture, OK? This is by picture. This is easy. So what we do? This is x cross y. So this is x and this is y. And now I choose a point, which in y, I fix a point, OK? Which is, so let b, well, b doesn't matter, fix b in y. So we have a point b, OK? This is b times, what is this? This is b times x, no? X times b is better, yes. I lost orientation. Coming from the right. So this is x times b, OK? Anyway, x times b is, this is a sign for isomorphism, for homeomorphism, OK? This is homeomorphic to x, homeomorphism. What is the homeomorphism? Forget b, OK? The canonical map, OK? Forget b in this direction, and add b in this direction. That's a homeomorphism between this space and this space. So this point is not important, OK? So this is connected, OK? This is connected, so this is connected, because this is connected, OK? So this is, what you see is connected, no? I take any point in x, x in x, OK? And I take this one, sorry, x, OK? I say, what is this? This is now x times y, OK? What you see is x times y, clear, no? And again, x times y is homeomorphic to y in the same way as this, no? And so this is connected, although this is connected, because y is connected, no? What you see, this is connected, this is connected. This is just x, this is just y, more or less, OK, up to this point, which you can forget. So this means, what you see here, this ends this, this is connected, OK? So x cross b union, union x, small x cross y, this is called tx. So now x is, this b is fixed, no? So x may vary, OK, for any x. So this is, that's what you see here, no? This and union this, no? That's tx, OK? What is this? Connected. Why is it connected? Yeah, they have a point in common, OK? Biolemma, non-empty intersection. But that you can, b is fixed, but you can do it for any x here, OK? So here's another x, OK? So you have tx for each x, no? It's clear that x cross y is a union of tx, x in, that's, you see, no? You have all these lines, these two keep for all the same, but you get everything, OK? But the intersection of these tx, what is that? This line, so this is b times, x times b, where is it? Here, x times b, this is x times b, no? No, it's non-empty, that's all what we need. Yeah, right, non-empty, so this means this is connected. Connect, a union of connected set is connected, if they have a point in common, OK? So this implies x cross y is connected, OK? I mean, do you see the picture, no? Do you see everything in this picture? This is connected, this is connected, so the union of these two is connected, and now you vary this, OK? tx, and then you get everything, but the intersection is non-empty. So you see everything in this picture, the proof, OK? You look at this picture and see the proof. So this implies, and of course, for finitely many, OK, by induction. If x1, xn are connected, then the product, x1 cross xn is connected, by induction, which doesn't work for the infinite case, no? So now comes the interesting case in some sense. This is the final product, no? And this is easy by this picture. Now comes the infinite case, arbitrary case, and that's the interesting. Because here you don't see, this is also, OK, proof, but product, box, all the same topology, no? You don't see the difference, but now you have to be careful, because it doesn't, I will show, by example, then it's not true for the box topology, OK, in the infinite case. It is not true. So this is, the general rule is, for the box topology, nothing is true, almost, OK? Many things are true for the product, but almost nothing for the box. There are too many open sets. There are too many open sets. So you suspect connected, you have problems, no? You're connected. Now if you have too many open sets, no? You find separations, OK, easily. That we will see. There are so many open sets, no? It's not the discrete topology, OK? There are still more, but this is not connected. No, that's trivial. But also in the box topology, you have open sets in the infinite case, no? And then you find easily a separation. That we will do after the proof of the second interesting case. So that's the second case. Now we have an arbitrary product. That's the second part. Well, I have two. Ah, 10 minutes, no? This clock is almost, yeah, 10 minutes, right? So now we have an arbitrary product, product. So x is arbitrary product, x, alpha, alpha, and j. And these are connected. And we have to prove x is connected, OK? And now I try to imitate the proof of the final case, OK? We try to do the same. So let, we fix the point. Try to do the same. Let b be a point in x. Well, I fix the point in x now in y. So b, alpha, alpha, and j. What is it called b? So fix b, a point in x, OK? So for any finite subset of indices, OK? We have finite products here. Now we consider finitely many only. Then we have no problem is connected, OK? We know the finite case, right? But if this would be a finite product, then it's connected. That's what we wrote, OK? So that means for any finite subset, alpha 1, alpha n of the index set, no? Let x, x, alpha 1, x, alpha 1, alpha n. So n is also arbitrary, no? n in n, arbitrary finite subset. Let x, alpha 1 be all, so we define this finite product. So we take all x, alpha, alpha n, j in x. x is a product, right? x is a product. Not as before x times y. I want to define this, OK? This is a symbol, which I define. So what I take? I take all points in x, OK? Such that, such as what? x alpha, the coordinate, OK, is equal to b alpha. I fix this point, OK? It's equal to b alpha for if it's constant, if alpha is not one of these indices, OK? If alpha is different from alpha 1, alpha n. Again, I take all points, OK, in x with coordinates x alpha, OK? For these indices, alpha 1, alpha n, I can take what I want, OK? For the other indices, it's constant. That's why I fix this point, OK? They just take this constant point, OK? All the same. That's the definition of this set, OK? So you say, you guess, what is this homeomorphic to? x alpha 1 times, you forget the constants again, OK? So this, what remains then is x alpha 1 times x alpha n, no? And the rest is constant, OK? It's homeomorphic, homeomorphism. By the way, you say, you don't use here that you have box or product. It doesn't matter, OK? For the box, for the product, this is homeomorphism in any case, OK? It doesn't depend. But this, what is it? It's connected. That we know by the first case. So each x alpha 1, alpha n is connected, OK? So that's what we want to make. x alpha 1, alpha n is connected because it's homeomorphic to this. We still didn't use that we have the product, apologies. This is also OK for the box. So homeomorphism, we have to find the point, however, no? Because otherwise, it's not true for the box. So if you don't use it, then something is wrong with the proof, no? Clearly. So this is connected. And now I take the union again, OK? Over all these. So this is called y. So I take the union over all possibilities. x alpha 1, alpha n. Take union over all these, OK? Alpha 1, alpha n. Where alpha 1, alpha n is a finite subset of alpha 1, alpha n. A finite subset. It's contained in J. Well, alpha 1, alpha n. It's contained in J, OK? And n is arbitrary in n, OK? n in n. So you take all finite subsets of J, OK? And you take the union. Of course, the intersection of all these is not empty, no? There's b. So b is the point which is in the intersection of all these x alpha 1, alpha n. So this means this intersection is not empty, OK? I didn't write this. I didn't write the same here as you, OK? It's the same. So this is non-empty. But as you say, these are connected. Union of connected spaces. This one point is b's common point. So this means y is connected, OK? So conclusion, y is connected. So if we are lucky, this is all x now, OK? Then we are done. But this is not true, of course. If you think about the union of these, then for finite subsets, OK? If you take a point in this union, no? For a finite subset, you have arbitrary coordinates, no? But for the rest, you have always this b, OK? This is not the general point, OK? In general. Because there may be other points in the b, no? So it's not everything. And the claim, however, is the claim is what you suppose? The closure of y, however is x, OK? And that means x is connected. So this implies x is connected. The closure, yes, OK? So we have to prove that the closure is x, all right? That's all we have to prove. The last point, we still have two minutes. So proof of this claim, no? That's the closure. So that means each point in x is in the closure, no? The closure is everything, OK? So we take an arbitrary point. So let x alpha, alpha and j be in x. x is the product, or infinite product, OK? What we have to prove? We have to prove that this is in the bound. This is in the closure, OK? This point is in the closure. So we take a neighborhood. So let x and, well, we take a basis neighborhood, OK? Immediately. And product u alpha and u equal product u alpha, alpha and j. Be a neighborhood of x. Sorry, this is point x now. And be a neighborhood of x. What we have to prove is that this has non-empty intersection with y, no? That's what we have to prove, right? Non-empty intersection with y. We still didn't use that we have the product. But we have to use it. Otherwise the proof is wrong because it's not true for the product. Now we have to use it, OK? So be a neighborhood of x in the product topology, of course. We are in the product topology. This is the point. That's a nice proof in some sense, I mean. So this is the product. What does it mean product topology? That means only finally many, so these are u alpha. Sorry, it's open, of course, in x alpha in any case, no? That we know, OK? Product of open. But that's for the product topology means u alpha is equal to the whole space, no? By the way, so be a neighborhood of x in the product topology. So then u alpha is equal to x alpha, except for finally many indices, no? Except for finally many indices, right? That's the definition of the product topology. What are the finally many indices? How would you suggest to cause them? Alpha 1, alpha n. Alpha 1, alpha n. These are the finally many indices in j, no? Well, you give some name, OK, which we used before. And now we have to prove that the intersection is not empty, OK? That's what you have to prove. So it's finished, the proof almost, except we have to write that what we want to prove is that, so we want to prove that this, yes, u alpha, no? Alpha nj, intersection A, no, how do you call it? Y is non-empty. That's what we want to prove, OK? Non-empty. That's what we want to prove, right? Then it's in the clothe. So we have to find a point, OK? And this point, we have just to find one point. And this is easy to find. Then this y alpha, alpha nj. So I'll just define a point. It's here. If you find a point, then it's not empty, OK? So we have to define what is y alpha. Y alpha is equal to, and we have to be in y and in the product, OK? And so what means in y? That means for finite, except for finally many indices, we have to take b, or a point, or a fixed point, no? Except for finally many indices. So we have to take b alpha if alpha is different from alpha 1, alpha n. Then it's constant, right? We even take our point b alpha. That's good for y, no? But that's good for this, because for these coordinates, if alpha is different from this, we have the whole space here. So we can take b alpha. So we are here. If alpha is equal to this, if alpha is an element of this, then what we take? Yeah, the only point we know then is x alpha, OK? X alpha. That's the only point which we are sure is here, no? So u alpha is open in x alpha. And of course, x alpha is in x alpha. It's a neighbor, no? Because x is in u. It's a neighbor, OK? So if we define it this way, then it's here because of the first, no? But it's also here. So we found a point, so it's not empty. And so this implies that the closure of y is really x, OK? We get everything. And this means x is connected, because y is connected. So this implies x connected. That's a nice proof, OK? Here we, next time I tell you an example that it's not true for the box topology. But we want to see, to understand it now. We are almost done now. It's not true for the box topology, this theorem, OK? In the final case, yes, but not in the infinite case. And then we have just to find the separation, no? And that, in general, you see. So I'll do it next time, OK?