 Okay, so let's start. We defined a metric on our J. J is an index set, okay? The product, infinite product, countable, maybe not countable. And if you have two points, we define, so we define the metric, it was called, what was it called? Row bar, so this is row bar of x, y. And this is least upper bound, supremum, least upper bound of D bar, x, alpha, y, alpha. X, alpha, y, alpha, alpha and J. Where, obviously, x is the point x, alpha, alpha and J. And y is y, alpha, alpha and J. So the least upper bound. And what is D bar? D bar is a metric on the reals, okay? And D bar of x, alpha, y, alpha, this is a minimum. This is a metric, the standard metric is D, but you may write easier x, alpha, minus y, alpha, okay? We are in the reals, okay? And then you cut at one. So this is a standard metric on the reals, cut at one. That's, sorry, least upper bound, yes. Least upper bound, supremum, okay? Least upper, that's, the book writes L, U, B, sometimes soup. This is a uniform metric, okay? Uniform metric, that I wrote last time, okay? This is a uniform metric and generates uniform topology, okay? Gives uniform topology, uniform topology. But we have the product topology, no? That was the first choice, box topology also. So there's a theorem, and I will write, but not prove. I will tell you why not today, theorem. So this topology is a uniform, it's not the problem. If J is infinite, it's strictly finer than the product topology. And the box topology is strictly finer than the, okay? So if J is infinite, J infinity implies a product topology. It's contained but not equal, uniform topology. This is strictly finer, no? That's what it says, if it's infinite. And the uniform topology is also different from a box topology, okay? So there's all three are different, box. They have three different topology. If J is finite, they are all equal, okay? If J is infinite, they are all different. So that doesn't resolve our problem if the product topology is matrizable, okay? Or the box, but more interesting the product, no? Is there a metric? This is not the metric for the product, and there's a theorem. I'll just write you in the metric, and once somebody says, this is the right metric, then it's easy to prove. And it's not difficult to prove, okay? It's not so interesting, man. The interesting thing is to find this metric, which is the right metric for the product. It's matrizable, okay? So the theorem's the following. Well, sorry, this is RJ. Now we have R omega. We have to go to R omega, okay? So let X, so this is X i now, i in n. And Y, Y i in n. No, the index is a natural number. We go to the countable case. So these are elements in R omega. Now we go to R omega. And then we have D, let D, let, and then we find D. D of X, Y, the distance, and now we do the following. We take the least upper bound, and I write the same as here. D bar, X alpha, Y alpha. This would be the uniform to follow. And what is changed now is I divide by i. Sorry, there's no alpha, yes. i divided by i, i in n, natural numbers now, okay? The index, natural numbers. i in n, this is our index. This is more general case, okay? Arbitrary index. Now this means we have the natural numbers index. And then we can divide by this number, okay? This is a uniform. So let this, then D is a metric. This is easy, it's a metric on R omega. And now we have the product, which induces the product. So now we have a metric which induces the product, okay? This is the product. Well, I mean, this is interesting, of course. It's more interesting if you don't know what happens, okay? If somebody gives you the metric and says this is metric, this is the right metric, then it's not so difficult. I mean, you don't prove it immediately, but once your partner says this is the right metric, okay? Which induces. Then you prove this is the right metric, okay? That's not too interesting and too difficult, okay? The interesting thing is if you don't know anything, then how to find this, no? Then you have to, that's interesting, okay? But I will not prove this because I will go to something which I find more interesting. This is a theorem, okay? So it's interesting theorem, certainly, no? Which is what about if J is not countable, okay? This resolves the countable case, okay? If J is countable, okay? What happens if J is not countable? What happens with the box topology, no? It's not so interesting, but for our omega with the box topology, no? What happens? And what happens is we won't, this is the only thing which remains metrilegal, no? We will prove that for the box topology, it's not metrilegal, okay? For the product topology, if the index at J is not countable, then it's not metrilegal. This is the last, which is metrilegal, okay? We have Rn, okay? The standard Rn. Then we have our omega, the next step, and here the product topology is metrilegal, okay? And then, but the box is not metrilegal, okay? And then we go to uncountable, no? And then nothing else. Of course, there's a uniform, okay? Which is defined by a metric. That's okay, but we have product and box, okay? These two. Okay, so this, I will not prove in this moment, maybe later, but because I want to prove that something is not metrilegal, which is more interesting. Because as I said, here you have a metric, okay? And if somebody says, this is the right metric, then you prove after one or two hours that this is the right topology, okay? The usual lemma. More interesting is to prove it's not metrilegal, because then you have to prove there's no metric, okay? And then you need the condition, as we said. Some kind of condition, some kind of criterion, no? You have to prove, for example, if it's not house stuff, but that's so simple, okay? If it's not house stuff, it's not metrilegal, no? But, and we find an interesting condition, important condition. Okay, so let me start with this. Instead of proving this here, the proof is in the book, of course. So there's a convergence I defined already, no? Convergence, a sequence converges to a point. If given a neighborhood of that point, there exists n, a natural number, such that for all numbers larger than this, the sequence is in this neighborhood. That's the general definition of convergence, okay? I will not rewrite this. And then there's a lemma, which I've proved now. Easy, very easy lemma. And it has a name, because this is, okay, so this is the sequence lemma. So lemma, and this is the sequence lemma. So we call it the sequence lemma. And it's easy, but nice, sequence lemma. So what? So x is topological space. X is the topological space. And A is subset of the x. So there's two parts. If the first part is very easy, trivial exercise. If there's a sequence in A, if there is a sequence in A, which converges to x to a point x in x, okay? A sequence in A in the subset, which converges to a point x in x. Then what happens with x? Then x is in the close-off. Then x is in the close-off. And this is definition, okay? I will make a picture, okay? But this is only definition, okay? Conversely, if you have the sequence. Conversely, suppose that, and now x is in the closure, and we want a sequence, okay? We're talking of limit points, maybe, sometimes, no? What means limit points, the sequence, maybe? So conversely, but this is not true in general, okay? That I said already. Convert, suppose that x is, and now I write matrizable for the moment. Matrizable, okay? So we need a condition now. Suppose that x is, then if x is in the closure of A, there is a sequence in A which converges to x. So then we also have the sequence. There is a sequence in A which converges to this point x. But here we have a condition, here's no condition, no? For the other, we have a condition. So the proof is, of course, directed is very easy proof. If there's a sequence which converges to x and x in A, so we make a picture, no? There's a sequence, we recall the definitions. So what is this, x is A, and here's a point x. And now we have a sequence in A which converges to x, no? So here's a sequence, A and let's call it, okay? A sequence in A which converges to x, no? Then this is the closure. Why isn't the closure? Well, what means closure? In the closure means every neighborhood intersects. So given any neighborhood of x, u, given any neighborhood, but this circle converges to x, no? That means there is a capital N, natural number, such that for all natural numbers larger than this given number, the sequence is in this neighborhood, no? So this sequence is in this neighborhood. So the intersection is not empty, okay? With u of m, A, okay? So this u intersection, A, certainly is not empty. That's all, just definition, convergence and two definitions. What is convergence and what is a point? That's not a definition, that's a lemma, no, proposition. A point is in the closure of a set A if and only if every neighborhood intersects A, no? That we proved, okay? It's not the definition, but it's very close to the definition. So this is the first part, no? The picture, you see it, it's okay. And the second part, if you want, it's equally simple. So now x is matrizable, x metric, x matrizable, matrizable, matrizable. Now x is matrizable and we have the metric D, okay? Metric D, with metric D. That means D induces a given topology, with metric D. And let x be in the closure and we have to find the sequence. Okay, so now what we do is the following. Well, it's clear what to do, we have the metric, we have BD, no? X, one over I, N, one over N, okay? This is the neighborhood of x, okay? So the intersection of this, this is the neighborhood of x. The intersection of this is not empty, no? Because it's a, by, by, what is the closure, no? So it's not empty, so choose a point, okay? It's not empty, so you get a point here, you call it AN, okay? AN, let AN be in the intersection which is not empty, okay? So you read this, let AN be in the intersection, which is not empty since it is a point in the closure. And then, that's our sequence, no? That's our proof in some sense. Then, this sequence AN converges to x, so proof, that is converges, okay? What means convergent? Convergence means, let, let you be a neighborhood of x. So I make it very step by step, okay? This, let you be a neighborhood of x. Any neighborhood of x, okay? Open second, okay? You have to prove, what have you? AN converges, that AN is in this neighborhood for large n, no? Let you be a neighborhood of x. But you have symmetric topology, no, okay? So there's epsilon, so I'm making two steps, okay? Maybe exaggerate it, but there is epsilon bigger than zero because you have the matrix. That's a bar, b, d, around x with radius epsilon is contained in u, okay? Because this is the basis for the metric topology, no? All open bars, okay? You have a metric space, no? Metrizable metric is the same, no? We have the metric, so we call it metric space, no, even, okay? So this is the definition of metric topology, no? And now, of course, I choose n, let n be an n. Very careful, step by step, such that one over n is smaller than epsilon, no? And this means, what that b, d, x, one over n is contained in u, okay? But for n, bigger equal to n. Of course, b, d, x, one over n is contained in b, d, x, one over capital N, and this is in u, always, okay? It's even smaller, you know, this radius, no? Smaller, still smaller. But a, n is here, no? And also in a, okay? So this implies that, what movement do we have to prove? So this implies for n, bigger equal to n, sometimes you have to think, what are we proving, no? A, n is in u, okay? A, n is in u. And this means, of course, then I have no, this means that last line A, n converges to x. But this is also trivial, almost definition, only definitions, no? Only definitions. So this is a sequence lemma, okay? Very easy. So I give you first application. And here, we will extract a topological condition. And here we have still metrizable, okay? That's a hypothesis, no? For the interesting direction, it's metrizable, no? And then it works, no? And now, but this is metric condition, no? And now, we think, well, what is the topological condition such as this works, okay? Metrizable is not so good, that's not the topological, that's metric notion, no? What is the topological condition such that the sequence lemma holds? Instead, metrizable is replaced by this topological condition, okay? No, too weak, much too weak. No, no, you don't know that. That will be first count of it. You know what's second count of this, okay? And here, the topological condition, we replace metrizable, which is not topological, not the metric notion, okay? We replace by a topological condition. And this will be first countable. You can think what will be first countable. That's what I will do later, okay? First, I give some applications. But this, you write first countable here and then it holds. But this is the topological, first countable will be a topological condition, like second countable, okay? And then the sequence lemma holds. So here, in a natural way, you find the notion of first countable, okay? In this context. In the book, it's later, in chapter four, okay? Two, three, four, yes. But I like it to do it here because it's very natural at this point, okay? Good, so some applications first. And then I introduce first countable. You can think what is first countable, you already know, no? What is the good condition? Well, you have to imitate in some sense, what you have is, what is important is this, you know? BD, this is important. Yeah, there's some countability here, okay? BDX1 over N, okay? There's some countability because N, okay? So we will see. But some application first. Some nice applications of the sequence lemma. So the first is the theorem, let F from X to Y be a function, not continuous, just a function for the moment. X metrizable topological space, X metrizable topological space and Y topological space, okay? Y topological space. Not, we don't care here, topological space. Okay, so this is our situation. Then F is continuous, if and only if. If and only if, F is continuous, if and only if. We want to use sequences now, okay? In analysis, that's standard, no? So for every convergent sequence in X, the image is convergent in Y, okay? So the notion will be for every convergent sequence in X, for every convergent sequence XN, which converges to X in X, the image sequence, sometimes, sometimes, yeah, not always, no? In analysis of epsilon delta, no? You have also, what? Sometimes this, one of the definitions, yeah, so what? Sometimes you use epsilon delta or the best definition, sometimes it's pre-images of open sets are open, also analysis, no, they are all in analysis, they are all equivalent in the situation of analysis, okay? For RN, analysis is in RN mainly, no? The image sequence, F of XN converges to F of X. So maybe I should try it. F of XN converges, that's what it means, no? So F of X, if and only if. So one direction, you don't need matrizable, no? One direction is always true, okay? So proof, that is the not interesting condition. This is this condition. This direction, so this one, if F is continuous, no? If F is continuous, then this holds. So, let F be continuous, no? And XN converges to X. How to write? We have to prove that F of XN converges to F of X. So let V be a neighbor of, it's almost automatic without thinking, no? Let V be a neighbor of F of X. And we have to prove that F of XN is in this neighborhood for large N, okay? That's conversion. Let V be a neighbor of F of X. Yeah, the F inverse of V is the easiest, no? Then F inverse of V, so U, which is F inverse of V, is a neighborhood of X, such that F of U is contained in V. This is the local continuity, no? Given the neighborhood of the image, there's a neighborhood of X, such that F of U is contained in V. And you can take the old pre-image, okay? Because pre-image of open sets are open anyway, so. It's a neighborhood of X, such that F of U, what? Sorry. So we have that XN, it's here, right? XN converges to X, so there is capital N in N, such that F of X, no, XN is in U for N, bigger equal to capital N. That's the convergence, no? And this thing for N in N, F of XN is contained in F of U. So, which is subset of V. Sorry, for N, for N, bigger equal to capital N, no? I didn't want to write to this, I wanted to write these two, okay? For N, bigger equal to capital FN, and this of course means that F of XN converges to F of X, okay? So we just apply definitions. Not so interesting, no? And we don't need that X is metrizable, okay? This is a general fact, okay? We do not need, we didn't need that X is metrizable. So this is more general, okay? If the function is continuous, then for any continuous function, convergent sequences go to convergent sequences, okay, for any continuous function. Interesting is the other direction, no? That this is efficient, so this is interesting. And here we need metrizable. Metrizable, maybe also not here, but there are counter examples. It's not true, it's not true otherwise, in general. So what we have to prove, so F is continuous. So here we use this strange criterion, or not strange, no? Continuous equivalent to F of the closure of A is contained in the closure of F of A, okay? F of the closure of A is contained in the closure of F of A, okay? So let, that's equivalent to continuity, okay? And it's convenient here because I have the sequence number, no? So let A, how is it called? A, sorry, B and X. We prove that F of the closure of A, as I said, F of the closure A is contained in the closure of F of A, okay? And this implies, this implies F is continuous. And this is now, again, easy. Once we say, well, if we use this for continuity, then this is also easy now, okay? It's automatic. So F of the closure of A. So let X be in the closure of A, okay? Let X, without thinking, no? Now it's, once you start in the right way, then it's okay. Let X be in the closure of A. And now, important, where is it? X, matrizable, X is matrizable, okay? We'll add an X, so X matrizable, X is matrizable. What does it say? So the sequence lemma, what does it say? The sequence lemma, what does it imply? Exactly what is here, no? There is a sequence. No, no, the sequence lemma is not, sorry, this is not the sequence lemma. This is what we approve. So the sequence lemma applies, there's a sequence, A n in A, which converges to X. So we write this A n converges to X, okay? There's a sequence in A which converges to X. And here we use matrizor, here we use matrizor, matrizor, we use the matrizor here to prove that. It's interesting to see if it's necessary, no, also. If you have to use it, or something. If it's proved it's not, no, it's not true. And then, what is our hypothesis here? We add this direction here, no? So this implies F of A n converges to F of X. Then F of A n, sorry, F of A n converges to F of X. Well, now we have the trivial direction of, by the other direction of the sequence, the trivial direction of the sequence lemma, not by the trivial direction. The less interesting inclusion, inclusion direction of the sequence lemma. What we get, F of X is in, now this is a sequence in F of A, no? Because of course, this is in F of A. If you have a sequence in something which converges to a point, at this point this is the closure of F of A. Okay, that's it, no? That's the proof. F is continuous, no? That's a nice proof, no, this criteria, okay? Now I give some other applications. As I said, we know now, I told you, that our omega with product of all these matrices, no? I gave you the metric. I didn't prove that this is the right metric, but that's not very difficult, okay? Once you have the right metric, and then you prove that that is true. And now we want to prove what is sometimes more, well, it's interesting to find the metric, as I said, okay? What is the right metric, no? If you don't know the metric, then of course, it's interesting how to find, no? What does it mean to you? You have the uniform metric and you modify a little bit, no? And then it works for the product of all of you, no? What is interesting is to prove it's not natural, no? And just be like proof, no? So it's another application, no? So, we prove two things. The first one is proposition, I call proposition, proposition. We have our omega again, countable product of wheels, no? Our omega with the box topology is not metrizable. Our omega with the box topology is not metrizable. So we use the sequence lemma. That's the criterion, okay? If it's metrizable and something isn't the closure of A, the point, then we find the sequence, okay? If it's metrizable. So if you don't find the sequence, it's not metrizable. So what we want to do here is, we want to define a subset, okay? The upper half space, and we have a point in the closure and then we prove there's no sequence which converges to this point in the upper half, okay? And then the sequence lemma says it's not metrizable. That's the whole point, okay? So we use the sequence lemma to distinguish. Metrizable. So proof. So A, so as I said, we define a subset of our omega and the point in the, A, A, and the point in the closure, which will be zero, okay? And then we prove there's no sequence in there. A with the upper half space, so what is the upper half space? So let A be the upper half space in our omega. That means these are all points with positive coordinates, okay? So our Xn, n in n. So of course Xn is in the reals, okay? Anyway, we are in our omega, okay? Xi I wrote before, now it's Xn, but it's the same, no? But Xn is bigger than zero, okay? This is the upper half space, okay? In Rn, no? Upper half space. Avoiding also zero, okay? The upper, open upper half space, no? Upper half space in our omega, that's what it is. And zero is zero, okay? Zero, so it's zero, zero, zero. So maybe this zero is a little bit bigger than this, but we cannot see it, okay? So this is in our omega. That's the vector space anyway, okay? Zero in our omega, no? And this is zero in the reals, okay? Constant. Then clearly zero is in the cloud of A. What is our topology? Box, no? We have to think of the box topology. We end the box topology. So this is in the cloud of A, well, easy, no? If you think a little bit about the zeroes and clouds. If you take any enabler of zero, no? In the box topology, you take a basis enabler, okay? Basis element. That is the product of open sets, okay? In each open set, you find open intervals, because of reals, okay? And then in each open interval, in each coordinate, you find positive points. That's all, no? So maybe I'll write the proof here, okay? Let U be enabler of zero. So then, we'll write this carefully now, almost too careful, okay? So by the definition of the box topology, no? We have a basis element. So this implies zero is contained. And now if you find this product, no? Product UN, N and N, contained in U. Well, UN is open in the reals, no? Okay? And contains zero, no? Because zero is in this product, no? Contains zero in R, okay? It's a product of open sets, no? The box topology, the basis are arbitrary product of open sets, okay? What did I say, box? The box topology, arbitrary product of open sets. The product topology, not arbitrary product. Product of open sets, but you have the whole space, okay? Except for, finally, many indices. That makes a big difference, no? Well, here not, of course. I mean, it's, so then, of course, you can find intervals, no? We are in the standard topology in R, no? This is, UN is open in the reals and contains zero, and also one interval, no? That means that zero, obviously, now I take another product, I take a product N and N, and now I take my, minus epsilon N, epsilon N, okay? Which is a small interval around zero. Minus epsilon N, epsilon N. Contained in product UN, N and N, contained in, maybe I'm exaggerating, no? I start with an arbitrary open set, no? Then by the definition, I find the basis element, which is this one, and then I find also these intervals, no? And now you can write down explicitly the point in the intersection, no? So then, which is the point in the intersection? Epsilon N over two, N and N, is where? Well, it's in A, because it's all positive, no? These are positive, epsilon N, as usual, no? Minus epsilon and epsilon N, okay? So this is in A, no? Clearly, where should it be? And in U, also in U, okay? It's positive coordinates, and it's in U, because it's here, no, in this part. Okay, and this means what we want. So this means that zero is in the closure of, so zero is in the closure of, okay? So this is easy, no? That's not the main part. Now what we want to prove, and that there's no sequence in A which converges to zero. It's in the closure, but there's no sequence. So this is an example, or one of these examples, where you have something in the closure of a subset, but you don't find the sequence, okay? I thought it's not, it's a limit point, it is, no? But you don't get it as a limit, okay? So let's see, and this is a famous argument. I mean, it's not exactly, but almost a famous argument, which is the argument of Kantor, the diagonal argument, okay? Now you see a version of the diagonal argument of Kantor. Kantor, no? Which is, which you proved that the reals are not countable, and you use the diagonal argument, okay? And now you see such a diagonal argument. So what we have to prove is that there is no sequence in A converging to zero, no sequence, okay? Not a single sequence, that's what we want to prove, okay? No sequence in A converging to zero. So we take any sequence, okay? We take any sequence in A, and we prove zero is not a limit, okay? So let, given any sequence, let, how is it called? Let A n be any sequence in A. Let A1, A n be any sequence in A. Take any sequence, so we have A1, A2, A3, and so on, no? That's a sequence in A. A is of course where in R, we are in R omega, no? Anyway, okay, in R omega. So each A n, what is it? It's a real sequence, okay? Each element of R omega is a real sequence. So what? So where A n, so each A n is a real sequence, okay? It's an element of R omega. So this is, well, which index n, I don't know. It takes a second one. n is fixed here, no? And you have x1, n, x2, n, x3, n. And so on, no? This is an element in R omega, okay? The elements of R omega real sequences. So this is, in other words, it's x. And now I n, I in n, no? In R omega. This was this way, so what, okay? You write the first three, I need another index here, okay? X I n, yes, what else? Well, yes, we are, of course, in A, no? We are in A. A, and so the dimensions are positive. We are in the upper half space, positive, okay? Positive upper half space, so. And X I n are all bigger than zero. We are in the positive upper half space of R omega. That's A. Okay, and now we prove that this sequence, which is completely arbitrary, cannot converge to zero, okay? So let U be any neighborhood of zero, okay? Let U, do I have a U here already? Well, I have a here, but I don't care about this, this was before, okay? Let U be any neighborhood of zero in A, in where are we? R omega. R omega with the box topology, no? We have the box topology, with box topology. So again, we cannot describe an arbitrary neighborhood, okay? But we have the basis, no? What does it mean? So we have basis element, which contains zero is contained here, no? The basis element of the product topology, of the box topology, arbitrary product of open space, okay? So then, well, in one or two steps, no? We make one step, no? You find the neighborhood, well, U n, I don't have. So again, is this, well, let's do it in two steps. In product U n, n in n, one second, one second, one second. Yeah, well, contained in U, okay? This is open in, where, in R, and zero is in U n, as before, no, that we did before, no? This is the basis element of the product topology, no? Basis element of the product topology. That is the basis element of the product topology. And U n is open in reals and contains zero, and so we can again find an interval. As before, it's the same as before, no? So this means that, sorry, yes, box, box, box topology, only the box topology, thanks. Okay, then, what is, I'm doing, too much, sorry, I was, that's not necessary, what do we have to prove? We have to find one neighborhood, just one neighborhood, okay? Here I give an arbitrary neighborhood now, okay? That's not necessary, so forget about it. Here I write any neighborhood, okay? That's not necessary, okay? So I just need one neighborhood, okay? Which doesn't fulfill this condition, okay? Just one, and as I said, this is a diagonal, okay? So I give immediately this one neighborhood. Then U is even, no, explicit, okay? Not any neighborhood, just one, and this is, so we have this situation here, okay? With all these numbers, two indices, okay? I and N, N is, yeah, two indices, no? We have a sequence of sequences, no? A sequence of real sequences, well, that's what it is. A sequence of real sequences. Now I take minus x11, I go to the diagonal, okay? Where the indexes, minus, they're positive. Times, minus x22, x22, times, minus x33, x33, times. So in general, it's a product of, whatever it is, no? So it's a product of xNN, NNN, minus xNN, xNN, the diagonal, okay, NNN. Product minus of these, okay? First times, times, times, times. So this is a neighborhood of zero, okay? Because these are positive, zero is all here. It's a neighborhood of zero in our omega. Certainly it's a neighborhood of zero in the box topology, no? Arbitrary product of open sets. So it's a neighborhood of zero in the box topology, such that, and now the conclusion is, such that, such that, well, aN, maybe I was not too careful, this is in this, it's not in U for any NNN. Not a single one is in this neighborhood of these four elements, okay? Not one, and why not? Since, well, we are looking at the Nth coordinate, no? Since the Nth coordinate of this is aNN, no? This is a sequence. Since aNN is not an element of minus aNN, aNN, this is the open interval, no? So it goes wrong in the Nth coordinate, okay? This aN is not in U, it's not here. We're looking at the Nth coordinate, okay? xN, we avoided xNN, okay? This of course means that this sequence aN does not converge to zero. But this is an open sequence, okay? So then the, by the sequence lemma, our omega is not metrizable, by the sequence lemma, our omega is not metrizable. Our omega with the box topology is not metrizable. So here's the application of the sequence lemma. So what is the classical diagonal argument of Kantor? That the reals are not, then let's talk, so. Because, so the classical diagonal argument of Kantor, okay? So we have a sequence of real numbers, okay? Take any sequence of real numbers. Take a sequence, consider a sequence of real numbers. Maybe you know who knows this argument, the diagonal argument of Kantor, not. So it's okay that we do it. Consider a sequence of real numbers. Well, just the idea in zero one, okay? In the interval, any sequence, no? We want to prove that this is not countable. So if you take a sequence, you never get everything, okay? That's what we prove, we don't get everything, okay? You cannot take this, you cannot count this, okay? You cannot, this is not, you cannot count, first, second, okay? You never get everything. That means the real numbers, or this, is not countable, okay? You cannot count this, no? That's what it means, okay? So I take a sequence, okay? I have a sequence, and I will prove this cannot be all, no? The other, there are always other elements, okay? And this is a diagonal argument. So consider a sequence, a n of real numbers, okay? Well, now I write a real number. What is a real number? That's not so trivial, no? But a real number is, so a one is equal to zero, no? Decimal expense, no? It's not unique, okay? So, but we don't care here, we don't. So zero times, okay? A one. So we have x one one, x two one. The n versus second index, no, it doesn't matter, really. X three one, and so on. So where are these numbers? X ij are now, they are between zero and nine, no? There's a decimal, no? Between zero and one. Between zero, one, right? One half is zero, five, okay? One third is zero, three, three, three, three, three, three. Forever, okay? And so on. A two is zero, one, two, two, two, three, two. A three, zero, x, one, three, x, two, three, x, three, three. So, okay? It's not completely unique, no? You have, obviously the problem is nine, nine, nine, nine, nine. Okay, that's one. Zero, nine, nine, nine, nine, nine, nine, nine, nine, nine is one. And one is one, but well, we have zero here, but you can start later, no? Zero comma one, nine, nine, nine, nine, nine, nine, nine. Okay, that's zero, one, one, okay? It's the same. So it's zero, one, one, zero, zero, zero, zero, zero, no? So it's not completely unique. And then I define now diagonal, I now go to the diagonal, okay? That's why it's called diagonal argument. I look, what's the diagonal here? This, this, this, okay? This is diagonal, no? And now I define y or b equal zero point p one, p two, p three. Where of course also this b, i are in, well, I define it, okay? Let b be this element, okay? Such that b i, b n, whatever, b i. I'm confused with i and n, no, there are two indices. It's equal to, well, what is the standard choice, huh? You mind this one, this one, this one, this one? I want to change. I want, yeah, I look at, but I exactly don't want this one, okay? Here I don't want this, here I don't want this, here I don't want this. You say exactly the opposite, okay? But that's the idea, okay? You avoid this, no, okay? So you don't take the diagonal, but you change, okay? So the one choice is take zero, if you say x, i, i is different from zero, okay? If you see something here which is different from zero, then you write zero. If it's zero, then you write one. Something different, okay? And one, if x, i, i is equal to zero, if it's equal to zero, okay? And this implies that, this should, one has to be a little bit careful. I mean, this is just the idea, okay? One has, because it's not unique, no? That's a crucial point. But this implies that, should imply that b is different from a n for all n. Because you look at the diagonal and there it's different, okay? So here it's different, here it's different, here it's different. And okay, so you compare. This is a rough version of the classical diagonal of Kantor, okay? So just because this is a famous point, this is a classical diagonal argument of Kantor. So what is the consequence? This implies r is not countable, no? Since this subset is not countable, okay? No, the last application, the next application, the last application, obviously, no? So we say this r omega is a box topology, no? Well, the box topology is not the most important topology. It's interesting, but not, okay? What about product topology, no? R omega is a product topology, this is my try, no? It's a theorem, I'm also metric, but it didn't prove, no? What rj is product topology, no? For the other rj, no? And so the next proposition is exactly this. If j is uncountable, then if rj is a product topology, it's not metrology. And now here we have the good topology, the product topology, no, but it's not metrology. So the last one, except for the uniform, these are all metrizables, no? But between product and box topologies, the last one which is metrizable is r omega is a product, okay? R omega is a product. Everything else, box product is no longer metroid, proof. The proof is very similar to the other proof. Same idea, except we now, we have to think what one has to define a differently, okay? What is a in this case? So a, let a be, so now we are in rj, no? That's our space, r, rj, no? Let a be a subset of all points, x alpha in rj, x alpha, alpha in j, that's the way I'll write in rj, such that all x alpha, such that this x alpha is equal to zero for finitely many values of alpha, x alpha is equal to zero, can be equal to zero for finitely many, only for finitely many, values of alpha, okay, can be zero for finitely many, and one for everything else, for all others one, and one, maybe always one, no? For no value, and one for all other values. So this is the definition of a. It's a subset of rj, obviously, no? This is now our subset of rj, which will work. And again, we have zero, so now it's zero. So let's write zero, now it's zero, but alpha in j, no? The zero in rj, then zero is again in the closure of a. So we, again, should prove this, zero is in the closure of a, okay? So take any neighborhood of zero, okay? But now I will write the standard neighborhood, the basis element, okay? If we start with an arbitrary neighborhood, then we find the basis element, okay? Which contains, and it's a neighborhood, okay? So I will avoid now, okay? Starting with an arbitrary neighborhood, so. Let, so what is a product of u alpha, alpha in j, be a basis element for the product topology containing zero. And just to prove that the intersection with a is not empty, no? That's what we have to prove. Basis element for the product of will be containing zero. We are in the product topology, so that means u alpha, this implies u alpha is different from zero, is different from the reals, only for fine little many indices, okay? Only for fine little many. Well, alpha one, alpha n, I suppose. Alpha one, alpha n in j. Okay, and now we found the point in the intersection, x alpha. So what does it mean, u alpha, basis element for the product? So containing zero, so zero is in u alpha, no? I should have, so this means zero is in u alpha, no? Of course, and u alpha is open in R, no? U alpha open in R. Well, I didn't write with. The important thing is this then, how so, no? And now I take point x alpha, r, j, alpha in j, sorry. Such that x alpha is equal to, so we define the point in the intersection. What, we have to be careful, in the intersection of, what, of A and, this is our neighborhood, no? And A, okay? A is this strange definition, no? Zero for fine little many and one, okay? So what we do, we take x alpha, one, okay? One, if, if this is the reals, we can take one, okay? No problem. So this means if alpha is different from, from these indices, no? Alpha one, alpha n. Finitely many, then we take one. And for the other one, the only point we know is zero, and zero, if alpha is equal to alpha one, alpha n is equal to one, okay? Right? And that means we are in A, and we are also in this neighborhood. So this means an x alpha, this point x alpha, alpha in j. Certainly it's in A, because one always, except for finitely many, we have zero, no? And it's obviously in this product here. So intersection, product, u alpha, alpha. So this means that this is not empty, this intersection. And this means then finally that zero is in the closure, okay? This is an arbitrary neighborhood, basis neighborhood, basis element. And this implies that zero is in the closure of A again, as before. It's in the closure of A, okay? And now we do the same. We take an arbitrary sequence, and we prove it does not converge to zero, okay? So again, sequence lemma implies not metronome, no, as before. Same structure of proof as before, exactly the same. The details a little bit different, but so let's take an arbitrary sequence. So what was it called? A n, no? In the first proof. So we use the same notation. So let A n be a sequence in A, any sequence. Let A n be any sequence in A, which is the subset of rj. So each A n, what is it? So in some sense, A n, let me write it, I didn't write here, but this is A x in the notation before, okay? The coordinates, alpha n, alpha was the first index now, okay? Alpha, this alpha in j, this is an element of rj, no? Each is this type, no? That's not the sequence now, okay? Because what did I write? J is uncountable, no? Did I write that in the proposition? J is not countable, okay? So this is not a sequence now, no? You cannot give a reasonable order for a second or a third, no, it's not a sequence, but it's an element in rj. But we are in A, what was the definition of A? Definition of A is that we have finitely many zero and one for all others, okay? Finitely many zero, and then one, always one. So we want to see where zero, okay? So we call Jn, let Jn. So I want for each an, I want to have the indices where zero, okay? This is finite set. So these are all x, alpha n. Sorry, all indices here, alpha in j, such that this is zero. X, alpha n is zero, okay, n is fixed now, okay? X, alpha n is zero. And so this is finite, finite, okay? By the definition of A. Finitely many zero. It's not zero, no, no, zero. Finitely, I want, no, it's equal, it's zero for finitely many indices, and it's one for all others, okay? I want finitely many. So this is finite by the definition of A, okay? This is finite, Jn. Well, maybe, because we don't need, this was a proof that it's in bound, we don't need this. So now I take the union, Jn, n in. For each n, I have the coordinates where zero, finite set. Now I take the union of these. So these are finite. This is a countable union of finite sets, this is in J, okay? Yes, this is countable, a countable union of finite sets, okay? It's countable, no? A countable union of finite sets, that's what it is, it's countable. Yes, right, it's a clear, but so we have finite sets. This is some argument counting like this, no? Even a countable union of countable set is countable, no? Countable union of countable set is countable, no? Countable union of countable sets, it's a first countable set. Then the second countable set, the third countable set, okay? And then if a countable union of countable sets, and then you count, how do you count? Yes, you will start on the left, then you go, you count in this way, you know? This, and so on, and you get everything, okay? Countable union of countable set is countable, countable union of finite set is countable, okay? And this is uncountable, right? By hypothesis. So they're not equal, countable, non-countable, no? They're not equal, right? So I take any point in J, let whatever is called beta, okay? Truth, beta in J, beta not in this union. What does it mean? Then for X beta N. So beta is fixed now, no? And X beta N, we avoid it where it's zero, it can be zero, so it's always one now, okay? The zero we excluded now, now it's one, it's always one. Because the zeros we have here, we miss this. So here, she was beta fixed now, beta, and then this is one, always one. We get all this one now. Which topology do we have? Product, no? Product. So I need one neighbor of the product topology of zero, which avoids the sequence, okay? What, here, we have this sequence, any sequence, no? Okay, and we have, I define, no, it's key in this neighborhood. It's one line now, the proof, okay? No, I give you the neighborhood. So I take that, then, who I don't have here, no? U equal pi beta, this is projection on this coordinate, okay? Pi beta minus one of minus one, one, open interval. Minus, I take this open interval, in the reals. This is projection, no, on the coordinate. This is continuous, no? So this is open, so this is a neighborhood of zero, it's zero goes to this, no? It's a neighborhood of zero. This certainly is a neighborhood of zero, because this is continuous, of zero in RJ. This is a neighborhood, no? And it contains zero, such that, such as what? A n is not in this neighborhood, for all n. Such that A n is not in U for all n and n. Since, I say, it contains not a single element of this sequence, okay? Not one, so the sequence cannot converge, okay? There's no sequence which converges, no? Because this is an arbitrary sequence, okay? Since A n is not in U, since A beta n, which is what, one, no? Okay? A beta n, this is one, no, X, okay? I call it X coordinate, no, not A. X beta n is one, it's not an element where, in? So the projection goes to one, okay? Over here with minus one, one, okay? So it's not here. So it's not in minus one, one, no? The end, the beta coordinate is one, okay? And then it takes the projection on the beta coordinate, it's one. So it's not here, okay? So it's not in U. So this means then, again, to conclude, so this implies, this sequence A n does not converge, no, against zero. And now again, the sequence lemma says, implies R j is not metroidized, okay? In this situation, not metroidized. So what is the situation? The situation is that j is infinite, the j is uncountable, and this has a product apology, okay? J uncountable, R j with product apology, okay? So this is the proof of the propagation. Well, it's perfect, because it seems, yeah, also, yeah. It's exactly. So these are three nice applications, no? We have the sequence lemma, and we have, these are three nice applications, no? The first one is the continuity, no? Under some hypothesis, function is continuous. If convergent, if and only if convergent sequence goes to convergent sequence, no? That's the first application. The second one is R omega with a box of what is not metroidized, and the third one, R j. If there is uncountable, the product apology is not metroidized, okay? These are the three applications we saw today for the sequence. And now we have also the situation, this limit point stuff, no? Limit point is a very weak definition. Limit point means every neighbor, limit point of A means every neighbor intersects of X, intersects A in a point different from X. So intersects A minus X, okay? X might be in A or not in A, that's not so clear, okay? If X is not in A, then anyway. But that's very weak, okay? If it's house stuff, no? Then in every neighbor, we have infinitely many different points, but this is still not limit point, exactly, no? And then we would have a sequence, but that's in general, this shows it's not possible, okay? We have a limit point, we have a point in the, we have a limit point here, no? We have a point in the closure, okay? It's not, zero, zero, zero is not in A, no? It's not a point in A, it's a limit point of A, no? It's in the closure, it's a limit point, okay? A is finitely many zeros, okay, it's a point. And zero, zero, zero, everything zero. So it's a limit point of A, zero, it's house stuff. This space is our house stuff, okay? So any neighbor intersects in infinitely many points, but you don't construct a sequence in general, okay? You cannot find a sequence, but this proposition and also by the procedure, okay? You have two examples where we have such a situation that you have a limit point, but you have infinitely many points in the intersection, but you cannot choose a sequence which converges, okay? The spaces are too large because of that, okay? So this notion limit point is a little bit misleading, no? It's not a limit, okay? But it's called limit point in the book, so, okay? That's interesting point, okay? And tomorrow I will give you first count of it. What is behind? This is metric, no? We have metrics in the sequence limit, no? So now we have to replace this metrizable by some topological condition, no? Which allows the same proofs, okay? That's easy. Well, tomorrow I will define first count of it.