 So the last thing we did is, I forgot, product topology for arbitrary products and also box topology. Both are interesting for us. Product topology is important, but box topology is interesting to compare. So we start, still another, this is the metric topology. So we talk about metrics a little bit. So what is a metric that you know from another, so I'll repeat. So a metric on a set X is a function. So metric, a metric is a function d distance from X cross X to the reals. So the distance is a real, real value such that, so if you think a little bit, then you have the three conditions. One is the distance should be positive. So the first is the distance between any two points should be positive and zero only if they're equal. So the distance is equal, dx, y is equal to zero if and only if x, x equal to y. That's reasonable distance, no? The second condition is symmetry. So the dx, y is equal to dy, x for all x, for all points. So the distance doesn't depend in which direction you go, no? So this is symmetry. It's symmetric, the distance. And the third one is the triangle inequality. So this has a dx, y, well, whatever, x, y plus, well, let me write it this way. dx, z is smaller or equal to dx, y plus dy. These are the three conditions of the metric, no? So this is a triangle inequality. Triangle, so we define the open ball, give them epsilon. For epsilon bigger than zero, this means real number, of course, as usual. We have set the ball, this is the open ball with respect to the metric d. That's a metric with center x and radius epsilon. So this is open ball with center, that's what it is, center x, x and radius epsilon. So what is it? Well, these are all points, y and x, such that the distance between x and y is smaller than epsilon. That's the open ball, so this is the notion of open ball. Bd, x, epsilon, okay? If the metric is clear, sometimes we will not write the d, but if there are different metrics, then we have to write, of course. Which one do we need? So this is the open ball, and then we have the collection of all open balls, B equal, let B, so yes, B, all open balls. So Bd, x, epsilon, and now we take all possibilities. So x for all points and all epsilon bigger than zero. So this is the collection of all open balls with all centers, all radius. Radius, whatever, what? Yeah, probably, it's not the language, of course. So as you suspect, it's a basis, okay? It's a basis, we have to prove something here. It's a basis for topology on x, and this is a metric topology. Of course, metric topology associated to the given metric d, okay? If we have two metrics, we have maybe two different, maybe the same topology. Associated to the given metric d, so this is a metric topology. If we have a metric, we have a topology associated, okay? We have to, well, there's a little lemma which is very easy. Anyway, let's write it. If y is in the open ball around x with radius a, with radius epsilon, then there is delta bigger than zero such that the open ball around y, this radius delta is contained in this one, it's contained in bd, x, epsilon. So the proof is by picture, proof, so what do we have? We have, well, the open ball might look different, or we give examples, of course. But so we have the center x and we have radius epsilon, okay? That's b, the open ball, the center x over there. Now we have a point here, y. And so we want to find the open ball around y, which is contained in this larger one, no? This is point y, okay? And it's clear what we have to take for the radius, okay? The largest possibility is of course, so that delta, let delta from the picture, what will be delta? Will be epsilon minus the distance between x and y. And that should work if this picture is correct, okay? In fact, it works, it's always a triangle inequality, no? So we have to prove this inclusion, right? Let me take a point here and prove it's also here. The standard sets here, okay? Take a point, let z, so let z be a point where bd y delta, no? Bd y delta. And then what we are interested in is that this would be also in bd x epsilon, x epsilon, right? So we have to see what is the distance between z and x, no? Between x and z. It's symmetric, anyway. So and then we have the triangle inequality, no? We can, we have y, so it's dxy plus dyz. That's exactly the triangle inequality. This is smaller equal to, we need some smaller at this point, no? So dxy, where's x? So this is smaller than epsilon, no? What? Sorry, I'm- This is smaller than delta, no? This is smaller than delta, no? And delta is what? Epsilon minus d, so this is equal to epsilon, right? So this means that z, so this means that z is in bd x epsilon, okay? So you should write the steps, no? So we can, I mean, in exercises improve. Okay, so this is a little lemma and this implies then also that it's a basis. What do we have to prove for a basis? So b is a basis. So if you want corollary, b is a basis to prove. There are two conditions, no? The first one is trivial. Each point is in some basis element, or it takes a center, no? The center is always in the ball, no? So we have to do the second one and make picture. So we have two basis elements. This is bd x1 epsilon 1 and we have a second one, only picture, bd x2 epsilon 2. And then we have a pointed intersection, okay? Which is y, and then apply two times a lemma, no? Y is in this and in this one, so you apply the lemma two times and then you get what you want. So the lemma implies, of course, that bd for some, there exists delta 1, delta 2, no? Such that bd x1 delta 1 is contained in bd, in what? Sorry, x1 is contained in, not x1, y, okay? Bd y, delta 1 is contained in bd x1 epsilon 1, yes? And bd y delta 2 is contained in bd x2, epsilon 2. But it's a constant, it's a notation. I mean it's trivial, but that's all right, okay? And then you take delta as minimum, of course, of delta 1 and delta 2. And this implies that b y, bd y, delta is contained in the intersection, in both, which is clear here. Now, you take one ball for this one, and you take the other one for this one, no? And you have to take the minimum, it's called bd x1 epsilon 1, intersection bd x2 epsilon 2. So this is the proof, and this is the picture. Examples, not examples, first the definition. Which is an interesting definition, so definition. So a topological, now we have a topological space. A topological space x is matrizable. Well, if it's matrizable, it's clear. If there is a metric, if there is a metric d on x, such that, so I don't prove it, such that, okay? Such that, a topological space x here for one, that's the metric topology. So here, sorry. Here for some time, we might write xt, the given topology, okay? So x is the topological space, x is the topological t, okay? x equal xt. It's matrizable, it's a metric d on x, such that the metric topology associated to d is equal to t, so the given topology. Such that the metric topology associated to d, to the metric d, of course. If there's a metric d on x, so the metric topology is equal to the given topology, t on x. So this is matrizable. And it's one of the interesting problems of the most interesting problem topology, so problem is problem, the problem. So given the topological space x, no, not given x, find necessary and sufficient condition, sufficient topological conditions, okay? Topological conditions. Find necessary and sufficient topological conditions. Which suggests a topological, any topological space, such that, well, I have probably this English today, how do you know what here? Such that topological space x, x is matrizable, okay? It's not, I'm not content. Such that topological space, any topological space is matrizable. This, so we give example. This is one of the interesting problems in the history of topology, okay? Of general topology. Topological conditions, okay? Which are sufficient, necessary, and also sufficient. So let's give examples. So examples. So consider x, a discrete space. So a discrete space means a discrete topology, no, x, a discrete space. Discrete space means with a discrete topology, no? That's a topological condition, discrete, no? That's purely topological condition, discrete, okay? Being discrete means that points are open, all points are open, no? All sets are open, all points are open. So is this matrizable or not? So d of x, y is equal? Yes, one, zero. One if x is different from y, no? And zero, of course, if x is equal to y. So this is a discrete metric, no? This is called the discrete metric. This is a discrete metric, which induces a discrete topology, okay? Which induces a discrete topology. Why, because obviously the ball, so this is d, okay? The ball around x in this metric with radius one, what is that? It's just x. So this is open, okay, in the topology, in the metric. So this means points are open, so this is clear, no? So being discrete is sufficient condition, no, for spaces, okay? If it's discrete, it's matrizable, no problem, okay? But it's of course not necessary, I mean, that would be strange, no? There are many spaces which are matrizable, which are not discrete, of course, no? Are two, okay, the standard analogous. So then, coming from the other direction. Here we have, we found sufficient condition, no, being discrete. And now x is indiscreet topology, let's consider next indiscreet. So this is the first example if you want, this is the second example. Yeah, right. So what we know, so observation, so what you say is the following observation. If x is matrizable, then a dilemma. X matrizable implies x house stuff, x metric, okay? Matrizable metric then you can use in a similar way, no? If it's matrizable, there's a metric D, so it's a metric space, okay? The metric is not unique about the topology, okay? So a house stuff proof, that's again our lemma, no? No, no, so by picture again. So we have two different points, house stuff, no? X and Y, and then we have to find disjoint neighborhoods. And what we take is the distance and half, no? Then this is sort of this and this. So they should be disjoint, okay? So we take, so the point is that the point is that the ball x is matrizable. So we have a metric D, matrizable, okay? It's almost D, around x with radius. So what we say here, what should be the radius? The distance between x and y, y delta, epsilon. There is over two, no? Half the distance, okay? So epsilon, intersection B, D, y, epsilon, this is empty. That's two, okay? So if you have to prove this, this is an exercise also, no? If you have to prove to have to write something, no? How to write that? Well, here maybe, what is the easier by contradiction, no? That's the easiest, in this case, contradiction is easy. In general, maybe it's not so much. Suppose, so what you write is this, more or less. Suppose that is a point in BDx, epsilon. Unfortunately, one has to write something in an organized way always, no? That's an important point for general topology, okay? Not computations, but so intersection BD, y, epsilon. And so this means this is not empty, okay? So you find the point. Suppose that is a point in the intersection which is not empty. Once you write this, then now the proof is okay, okay? You have to start in a good way always, no, to write the proof, no? To start with the right thing, that's not so easy, no? Because one has not a good order sometimes with the arguments, no? One starts a mixture of hypothesis and what should be proved. So one has to concentrate on that. That's important, okay, for mathematical style, no? So once you start in the right way here, then it goes almost automatic, no? You want a contradiction anyway. So the distance between x and y is smaller equal distance between x and z you compare by the third point, no? So you have this situation where you have a point here, no? This is z, and then something goes wrong here, no? With this triangle. So the distance plus distance z, y, which is smaller, so distance. So this is x, y over 2 epsilon, no? It's 2 epsilon, that was epsilon, 2 epsilon, epsilon plus epsilon, no? But this is equal to, what is epsilon plus, 2 epsilon is dxy, right? That's a nice contradiction, no? dxy is smaller than dxy, okay, so contradiction. So this, x is matrizable, so it's house off, okay? Sorry, sorry again, well, I don't understand what you want to say. This is a contradiction, no? Contradiction, whatever is the sign for contradiction, okay? We are saying the distance. No, yes, but it's too complicated, I don't like that. Here you have dxy is smaller than dxy. No, no, no, no, that's a contradiction, okay? For the order, that's the real order. You don't have that, x is smaller than x, okay? So here we have the contradiction, but don't make it too complicated, okay? Here you have dxy is smaller than dxy, and that's- Zero is not less than zero. Zero is not less than zero, zero is equal to zero, no, no, it's not a question of zero, right? Zero is not smaller than zero, no real number is smaller. Then another different real number, okay? It's the same real number, sorry, okay? That's part of the axioms for order, okay? We never have x is smaller than x, okay? For any order, that one has to impose, okay? That's what we wrote for order. X is never smaller than x, no? If we want this, then we, this means smaller, this means smaller or equal, okay? And this means equal, okay? And this is really smaller, okay? And you never have x smaller than x, for any order, okay? Obviously, not for the real order, no? If you have different, you see the order, okay? For the reals, then it's clear. But for any order, you have this axiom that it's never smaller, x is never smaller than x, okay? That's not a question of zero. You say zero is smaller than zero? No, we also can't have zero, because x and y are different. Yeah, it's not zero anyway. He wants to say it's zero now, and then they are equal, and that's the contradiction. That's what he wanted, no? He wants to conclude it must be zero, and then they are equal, and then it's a contradiction. But the contradiction is here already, no? So it's better to finish once you, yes. So zero is a strange number, and now if you think zero is, could be smaller, zero is very small, but, well, anyway. So the indiscreet topology, so what? What is the answer? Yeah, that's almost the answer. That's almost the answer, almost the answer. But there are always these nasty special cases, so you have to be, the real correct answer is, if it doesn't include. Yeah, right, if it does, that's what one has to say, no? If it has at least two points, then it's not house stuff, okay? So if x is discreet, x indiscreet space, this is, so how to write, it's house stuff if it's one point, of course, no? But it's also not. Sorry? Then it's also discreet, yes, yes, of course, yes. Indiscreet space, it's not how, well, it's right in a good, in a reasonable way. It's not house stuff, except if x has only one point. That's right, it has only one point. That's not very interesting space with one point, no? That's also discreet, there's only one topology, anyway. So now we found what condition, necessary condition, okay? Being house stuff is necessary for being material. Being discreet is sufficient, okay? Discreet space is house stuff, but they're very far, no? There's two conditions, okay? On one end discreet, on one end, on the other end, what? Okay, other examples. Now, we have the standard examples. We should talk about the standard examples also in this context. We come back to this problem. That's one of the interesting problems. So the standard space, of course, is Rn. So let's consider Rn, Rn, this analysis, okay? So we have the metric, so it's a norm. We have, if we have a point, x equal x1, xn in Rn, I define two metrics, oh, maybe three. If you turn the point in Rn, then we have the norm, no? The norm, that's the square root of x1 squared plus xn squared. In fact, we have the standard scalar product, okay? And this is x, okay? This is a standard scalar product in Rn, no? This is a standard scalar, which I didn't write, but if you, and then to any scalar product, you have a norm, right? And this is a norm, a Pythagorean norm, right? It's Pythagoras, no? The length of, in R2, this is, yes. Okay, and so you define the distance from Rn times Rn to the reals. This is dx, xy. So x is x1, xn, y is y, of course. So what you take is, that's the definition. You take x minus y and the norm. So in other words, that's the x1 minus y1 square plus xn minus yn square. And then you have to take the square root, this is it. So this is the Euclidean metric, okay? This is the Euclidean metric. This we call the Euclidean metric. That's all this, no? That's interesting in mathematics, no? You have a scalar product, then you have a norm, and then you have a metric. Of course, you have to prove for, I will not prove, this is another, no? So this is a metric, okay? There are also conditions for norm, which I didn't say, no? But this is a metric, okay? And the problem is a triangular inequality, okay? Which is not immediate, okay? But this is analysis, so I will not go, we are doing a topology, concentrate on this. So this is a metric. So this, the Euclidean metric and this is a metric. And I will not prove this. Well, it's all in a trivial, no? For the first two, it's trivial, and then the triangle inequality. So what is the ball here? That, the Euclidean metric, what is it called? But what is the name? No, the special name. So the ball is, yes, that's a standard. This is a Euclidean metric, no, okay? And so we can make, this is really in X, we are in R2, of course, here, no? And epsilon. So this is the round ball in R2, and then in Rn it's, this is the Euclidean metric. On R, so we have the topology, okay? So the standard topology in Rn, which is set, we defined the standard topology on Rn, defined maybe for R2, okay? Is the topology generated by the Euclidean metric or equivalently the product topology, okay? Where R has the standard topology. These topologies are the same. The standard topology is a topology generated by the Euclidean metric, which is equal, or equivalently, which is equal to the product topology, okay, which is equal, however, so you can define it in this or in this way, which is equal to the product topology. On R cross R, N times, no? On R times R, where R has the standard topology, no? The standard topology is generated by open intervals if you want, open, but open intervals are balls. Euclidean metric on R, okay? You take the center of the interval and then you have the radius, okay? So open intervals in R are open balls with respect to the Euclidean metric, no? They're clear, yes? So the interval from A to B in R, real interval, this is the ball, so we have the Euclidean metric on R around, what is A, what is the center? A? A plus B over two. One is also here to concentrate, what is the radius? B minus A over two. So this is an open ball, okay? Open intervals, so it's the same topology also here, the standard topology is generated by, so we have to prove here something, by the proof, by picture. So I make proof, okay? So let's do it for R2, okay? It might picture. So we have to prove that we have two topologies, the product and the Euclidean metric topology, okay? So let's start with the product topology. What is the basis for the product topology? Open times open, and we can take a basis for R, so we take open interval times open interval, that's the easiest way. So this is open interval times open interval, okay? This is a basis element for a product topology, okay, on the wheels, right? That's okay, or we can take open times open, but we can restrict to basis, and the nice basis of our open interval. So it's open interval times open interval, okay? And then we apply the lemma, no? So we want to, we have two basis, no? This is the basis for product topology. Well, on R2 here, but it's the same for Rn, basis for product topology, okay? These are open rectangles, no? And then we have to take a point, any point, and then we have the second basis. That's the basis B, okay? The open balls. And then we have to find an open ball which contains this, and it's contained in this, okay? And now it's clear, you go in some sense, you consider the distances to these distances, no? And you take the minimum of these distances to the, and then that this ball is okay, okay? So this is the ball. And so this picture shows that the Vick's topology is finer. Sorry. This implies that the metric topology is finer than the, this implies the metric topology. Yes, the metric topology is finer. Equivalent is not the right word for topology. Equivalent is not a good word. Equal, no? Yeah, equal, equivalent. Equivalent? So the metric topology is finer than the product, topology. On R2, no? This picture, this is the picture. And then we have the picture for the other thing. So then we start with the basis element for the metric topology, and we take any point. So here's the center. You take any point. And now we have to find an open rectangle, right? Well, yeah, so we can take the radius if you want, and then what is the efficient way to know I'm tired? And in any case, it's clear that by the lemma, we have also this, no? And now we can take rectangles here, and this is not the center, no? Not the center. I wanted to find the, and now of course you find here open rectangles, but I wanted to find the optimal. Well, maybe it's not, that's okay, okay? So this implies, what does this imply? The product is finer, is finer than the metric, okay? So they are equal. I'm just gonna exercise this lemma, okay? You have two bases, no? And the one is finer than the other one. If given a basis lemma for the other one, a point, you should find the basis lemma of this topology, which contains the point, as you can take. This we have applied two times, okay? One in this direction, one in this direction. Okay, but of course there are many different topologies, not topologies, metrics on RN. So another one is this one. This is called the row from RN times RN to R again. And now we define row of X, Y, X, Y as before. So this is maximum, maximum of X1 minus Y1, XN minus YN. It's a metric. I will not prove that. It's a metric. This is a, this is even still easier than the Euclidean, okay? Because there you have this square root now for the triangle inequality. It's also pleasant, the square root. Here you have just, it's a metric. And this is called, in the book, it's called the square metric, the square metric. Why square metric? But I prefer to call it the maximum metric, okay? Or maximum metric, what does it say? Just names. I think maximum metric. It's clear about, because it's a maximum, okay? That's a maximum metric on RN. So what? So what is the ball now? Yeah, that's square. That's why it's called square metric, right? So what you take is, let's take R2. We make picture in R2, of course. So we have, we want the ball, rho. This is rho, sorry, it's not delta. Delta thinks D, no? Here I wrote, okay. I was thinking of D, delta, no? But it's rho, the square metric. In the book it's called, my name is just the name. So rho. I don't know why rho, but anyway, it doesn't matter. So around X, with radius epsilon, no? So let's see what it is. So it's, so here's epsilon, epsilon, epsilon. That's very small, but okay? You go epsilon in each direction, okay? And then you have the square, okay? So it's a square. Epsilon in this direction, epsilon in this direction, epsilon in this, and epsilon in this, no? And in R, N, you have the same. You go in epsilon in all directions. So it's not a square, it's a cube, or it's a hypercube, no? But it's called square metric, because this is a square. And now you have to, so this is, by the way, a special case of rectangle, no? So it's almost like the product basis for the product of polarity, no? Except it's a little bit more special, no? But it induces the same, the standard topology, no? You make the same kind of picture, okay? I mean, you can compare with a product, or with a Euclidean metric, no? But it's the same, no? Except here, this seems already a square, okay? And, well, anyway, so the maximum, which induces, which induces the standard topology on R, right? So this is S before. And then you can vary, no? You can take other stuff, no? You can take, well, maybe you can take the sum of the distances, or whatever, no? And then you get other kind of pictures, no? But they, in general, have to induce the same topology always, okay? If you can play around and define many metrics, which, but in general, they define all the standard topology, okay? And now comes an interesting point, because, so a question. Now we, yesterday, R omega, no? This is R times R. This is a product of reals, and how many? Countable, okay? Real sequences, we introduced yesterday, R omega, no? Is matrizable? That's the name. So Rn is matrizable, no? We have the Euclidean metric, okay? We have the square metric. We have many metrics, which induce, okay? Now R omega is matrizable. And that's an interesting question, okay? This is the next case, no? Rn finite product, and now we have infinite product. So we make it a little bit, so we want to generalize these metrics, no? But of course, to generalize the Euclidean metric is not so, well, you can generalize, but you need convergence, no? Because you have the sum, no? Okay, and the sum, if infinite is not so good, so you need factors, so you have convergence, and what seems to be easier is to generalize the square metric. Of course, you have infinitely many distances here, no? So it may still be infinite, no? But any metric, so this is an interesting remark, so we will generalize this one, the square metric, okay? This is a maximum metric, and now we will have a sup, supremum metric, okay? On R omega. However, it should not be infinite, okay? So what we have to do, we have to cut the metric and to bound it, okay? That's no problem. So that's easy, this is before. So lemma, if D from x, x to R, is a metric on x, then also, and now it's called D bar, yes, then also D bar from x, x to R, and now we have to find what D bar is. So D bar of x, y is equal, so we take the distance D x, y, we take the minimum, sorry, minimum, okay? So we take the distance which we have, and zero, minimum, then it would be zero, no? The minimum, zero is not so good, one, okay? So what is this? We cut the metric D at one. If it's bigger than one, then it's always one. If it's smaller than one, then it's the original distance, so okay, so this is, it's minimum, so this is the metric D at one, at the point one, okay? You say we don't want anything larger than one, okay? Here we cut, then we take one, always, okay? Then also D bar, minimum is a metric on x, and this is called the bounded metric associated to D, because now it's one is the maximum, no? This is bounded. So this is always smaller equal to one, okay? I cannot write this here, but this is smaller equal to one, of course, also. It's a metric on x, the bounded metric associated to D. It's very easy to prove that this is metric, okay? However, what we want, so it goes on, it's a metric on x, the bounded metric, which induces the same topology as D. So the topology is the same. This bounded metric induces the same topology as D. This bounded metric induces D bar induces the same topology on x as D. So why is that? Informally, the small balls are equal, no? So we can write this since the ball D bar x epsilon is equal to be D x epsilon if, yeah, less or less equal, less or equal is also okay, no? If epsilon is smaller equal to one, okay? Also equal one, because we take all parts which are smaller, so it's the same metric, no? And note that the topology is the same, and note that if epsilon is bigger than one, no? Then the ball BD x epsilon is what? It's x. This one, if epsilon is bigger than one, then all points have a distance at most once over here, okay? D bar, sorry, yes, D bar, six. D bar, of course, okay, D bar, yes. It looks more like D, okay? And the topology is determined by these balls, no? You then need, I mean, what is open metric topology? Open set, if given any point, you find the ball, and you can make it as small as you want, okay? And so that means that if you know these ones, that's efficient, this is not interesting anyway, and the large one for DD accepts another important for the topology, okay? You can always take small, okay? Small ones, as small as you want, because you just restrict, okay? And it generates the same topology. Okay, so that's good, no? And that means we can generalize the, now it's easy, we can generalize the maximum metric, no? Because now we take the supremum, and it's a maximum, but we have to make this, no? This, we have to cut. Okay, so let's do that, so definition. So we take any index set, we were talking about our omega, but we can take any index set, okay? Let J be any index set, and consider, and consider, so all spaces are J, are J, are omega, are J, no? Are J, so what is this? This is a product, if you want, also reals, but how many, okay? The index is this one, no? So in other words, these are all points, are, alpha is always called, no? Alpha and J, where are alpha, are reals, okay? These are not sequences, no? If J is uncountable, you cannot, it's not a sequence, no? It's not in a good order, no? But anyway, it's this. Consider this set, and now define a metric, and now we can have a metric here. So this is, I should use the name of the book, so this metric is rho bar, because it's rho is a square metric, no? So we define rho, from R J, times R J to the reals, so no. So the points we write in this way, okay? So we have two points, X equal X alpha, alpha and J. So this is here, and this is Y equal Y alpha, alpha and J. X alpha and Y alpha are reals, no? That's real number. So X alpha, Y alpha are just real numbers. That's here, no? Here's R, like reals. So the distance is defined as we said, no? So the distance between X and Y, these are X and Y, no? Now we cannot take the maximum, we have not finitely many, but we take the soup. And in the book, maybe it's not written, yeah, it's not written this way. So how do you call this? Supremum least upper bound, maybe, least upper bound. So let's go least upper bound, so least upper bound. Sometimes soup, supremum, okay? Dependent, what language? So least upper bound, or supremum, soup. Of what, well, of the same as before, X alpha minus Y alpha, alpha and J. No, no, no, no. This, sorry, here, this might be infinite, no? Of course, we are in the reals, the distance. So I have to cut at one, okay? So we have D bar of X alpha, Y alpha, sorry. I have to take, where D bar is the bounded metric on R, okay? Where, so instead of the maximum, I take the least upper bound, here it takes the distances, okay, so what is this? This is the minimum of X alpha minus Y alpha. This would be the collusion metric, no? X alpha minus Y alpha, and one, cut at one. We have to cut at one. So this is D bar, here, okay? We cut at one, the standard metric on the reals. So this is smaller equal to one, no? So this is the name, two names. So this is called row bar is the uniform. That's the name of this metric. Row bar is called the uniform metric. Uniform, well in this case on R, J. And sometimes, so uniform metric. And sometimes I call this, I'm used to call it also the least upper bound metric is not so nice, it's so long, it's a super metric, it's a supremum metric, okay? Instead of the maximum metric, we have the supremum metric, okay? Are you used to this, what, supremum? Yes, or supremum metric? But in the book it's uniform, okay? So supremum is, you recall, what is it, no? It's called the uniform metric. So this is used as a topology on R, J, okay? And the question is what is the topology? So it's a uniform topology, but what is the uniform? By definition, this induces the uniform topology, the uniform metric induces a uniform topology. This induces row bar induces the uniform topology on R, J. So here's the definition of uniform topology. So now we have three topologies, it seems, maybe three, it's also clear. Now what are the three, there's a product topology? We have, in any case, we have the product topology, okay? And we said this is maybe the nice topology. We have the box topology, this maybe is a bad topology, or the bad guy, okay? And now we have a certain uniform topology, okay? So, proposition, so we want to compare, of course, these three topologies. Maybe this might resolve our problem, right? We want to see if R omega, if it's a product topology, that's the first, okay? It's the most important, it's matrizable, no? Maybe this is a metric, but it turns out it's not a metric, okay? So the proposition is the uniform topology on R, J is finer than the product topology, well. The uniform topology, the uniform topology, uniform topology on R, J is finer than the product topology. It's prickly finer if J is infinite, okay? If J is finite, all topologies are the same. The product, well, then it's the maximum, no? That we said already, on Rn, on Rn, well, it's not exactly where we have to cut here, no? That's the only difference. But it's almost like the maximum metric, right? The square metric, except we cut at one here, but that doesn't matter for the topology. So on Rn, the three topologies are the same, well, I will write. It's prickly finer if J is infinite. And then there's a remark, which is not part of the proposition, on Rn, the three topologies, on Rn, the product topology, they are all the same. The product topology, the box topology, well, one second. The product, that's too long, maybe. The box, the product, box and uniform topologies coincide, other thing, okay? Here we have just one. And that you should also, you should be clear. We know already that, I mean, there's no difference between the product and the box here, okay? Open times, open times, open. Finally many, so we have no condition. And the uniform topology is the uniform metric. That's a soup metric, but here's the maximum metric, okay? Except we cut at one also, but for the topology, it doesn't matter. So that's the same, the three topologies. On Rn we don't, if J is infinite, then it's prickly finer, then the uniform is prickly finer than the product topology, okay? And then there's not so much interest in the box topology, okay? It's also not the box topology. The box topology is prickly finer than the uniform. So it's in the between the two, okay? In the middle, yes. Every what we mean by norm, we don't talk of norm. Just, this is in Rn. And norm is something, we didn't define what is norm, if you have a scalar product, you have a norm. But then there's the axioms for norm. There's the axioms for a scalar product. There's the axioms for norm. And then there are axioms for, so what is the question? If every metric comes from a norm, yes. What do you mean by norm? Just the axioms of norm, we didn't define norm. No, they are not equivalent in general. In area? Yes, I mean, well, maybe we are more familiar with scalar products, okay? The norm, if you have a scalar product, you have a norm. And the norm determines the scalar product. That's linear algebra. That's not our, I don't want to talk. The norm determines the scalar product, okay? But they are, what is the question exactly? If all norms induce the same topology? Equivalent. What means equivalent? Every norm I think we're like equivalent. What means equivalent? When we take the true norm. Yes. You can compare. Still, yes. So they induce the same topology also then. But that's not our subject here, okay? These are interesting questions, of course, no? I mean, if you have a scalar product, you have a norm, you have a metric. If you have a metric, do you have a norm? Do you have a scalar product? No, in general, okay? There are other conditions. But this is linear algebra. This is not our, okay? Maybe analysis it even more. So we, I didn't even define norm, okay? I defined on our end. But I didn't say what are the axioms for norm in general, okay? What is the norm? So there are three axioms for norm, no? And what's that linear algebra? Don't, we will not want, yeah, okay. So let's prove this, okay? Proof. Well, it's easy. Proof. Finer, first we have to prove finer, no? So what means finer, what is the uniform topology? It's finer. So we should start with a product topology, okay? So let, and we can start with a basis element, okay? Not just a general open set, because given a point with any very fine basis element, no? So let product u alpha, alpha and j be a basis element for the product topology. So what does that mean? That means we are in the product topology, not in the box, okay? So that means that, so this implies u alpha is equal to r, except for finitely many indices, no? Let me call, give name, alpha one, alpha n in j. However, u alpha is open in r, all of us, okay? u alpha open in r. That's product of open sets in the ring. It's open in the ring. Finitely many indices alpha one, u alpha is however open in the rings, e to u alpha. Because if we don't have this condition, then we have the box topology, okay? Then that would be a basis element for the box topology, no? If we forget this condition, just u alpha open r, then we have the box topology. Arbitrary product of open sets is a box, okay? This is a product topology. And given a point, okay? And let, so we take a point. Given a basis element, a point, we should find the basis element of the other topology. What is the other? The uniform, okay? So let, the point is called x, x alpha, alpha in j, be an element in this product. Okay, so we have to take care of these indices. Alpha one, alpha n. Where we are not the whole space of maybe, okay? For alpha one, alpha n in j, so these are these indices. Alpha one, alpha n in j, choose, so anyway, it's clear that x alpha is in u alpha, no? Clearly, no? And this is open in the reals. That's our situation, no? Hello, so for this choose epsilon i, how is it called? Epsilon i bigger than zero, epsilon i bigger than zero, such that the ball, d bar, this is this French metric on R, around x alpha i, epsilon i. So this is, here we are in the reals, no? In one coordinate. And this is this metric, cut at one. It's contained in u alpha i. So this is, what is this? This is an open real interval, no? Open real interval, that's what it is. It's just a real interval, oh. And of course, we have here x alpha i here, no? So we just choose an open interval around this, which is contained in u alpha i, okay? But we write it in this strange way, okay? Interpreting, interpreting it as a ball with respect to this metric, the interval, no? No problem, open real interval. And epsilon the minimum, take the minimum, let epsilon be the minimum of epsilon one, epsilon n. So these are, these n indices, one up to n. Then the claim is, then the ball around, and now we have the other topology, no? The uniform topology, uniform metric. Of course, x is in the ball around x, x is our point, and you take epsilon, this is okay, and this is contained in our product u alpha, alpha and z. This, we started with these things, okay? Well, if this is true, then we have to think one second about this, of course, it's contained here, no? This, of course, then implies that the uniform is finer than the product. So this implies the uniform topology is finer. So I will not write, first, let's see. So take a point here, y, okay? In this book, we have to prove it's here, no? And this is to check coordinate by coordinate, okay? So let y be a point here in this ball, okay? So the distance between x and y is smaller than epsilon. This is a soup metric, no? That means that, no, no, this is not important. Let's see, then we should find it, no, no. And what, this is d bar, yeah, this is the bar, this is cut, I don't know. You can think of epsilon as small, if you want, okay? There are large values, and well, should I? That's not important. So take a point here, okay? Y, so y would be y alpha, no? Alpha and j, b in this ball, x, epsilon. What does it mean? This means, of course, that's a distance. I didn't want to write, no, I'm not started to write. That means, of course, that's a distance is smaller than epsilon, so the distance, which distance, rho bar between x and y is smaller than epsilon. No, clearly, that's the definition. That means that, now I go to these coordinates, okay? What is this distance? This is, what was it? The least upper, the soup, least upper bound of d bar, x, alpha, that's the definition, no? Alpha and j. So you take the distance in the coordinates, okay? And cut at one. That's just the difference, but cut at one, okay? And so that means that, what does it mean? That d bar, so the soup is smaller than epsilon. So d bar of, now let's take our coordinates, x, alpha, i, y, alpha, i, these are alpha, one, alpha, n, no? It's smaller than epsilon. The soup is smaller than epsilon, okay? So this must be smaller than epsilon. And epsilon is smaller, what, than epsilon, i, no? Smaller equal. And this, of course, implies that this ball, bd bar, what do you want? So that means that, sorry, this point, y, alpha, i, is where? Is here, okay? What do I want to prove? Component wise, it's here and it's u, alpha, i, okay? It's in u, alpha, at a certain point, it's better not to write, but it has to write, of course, okay, it's u, alpha, i from one to n. So first this and then this, no? Okay, first this and then this, for i equal one to n. For these coordinates, it's okay, okay? We are, we have the right coordinates, we are here, okay? And for the other coordinates, we have the real, we have everything, so we don't care. It's, anything is good, okay? So, for alpha different from alpha one, alpha n. If we have another coordinate, okay? Not one of these. Then u, alpha is what? R, and then, of course, this implies that x, alpha, is in u, alpha also, okay? Because it's everything. So u, alpha, so x, alpha, no, no, y. We are interested in y, okay? Where y, alpha is in u, alpha, which is no condition here, because it's everything, okay? I didn't want to write what I wrote, okay? So this is a proof that, so this, all this implies this inclusion here, no? Okay? So this implies, no, you can conclude that b, roba, x, epsilon, it's contained in product u, alpha. So what, it's very easy, no? You take a point here, we look at the coordinates, and we find them all here in the right u, alpha, and then it's here, okay? So this is fine, okay? We have two minutes, because we have to prove strictly fine, no? Well, that's like school. So no, we have to prove strictly fine. And strictly fine, I mean, they're not equal. So what we have to do, we have to find one open set, which is open in, in which one, which is strictly fine, in the uniform, but not open in the product, yeah? Just one is sufficient. And the one which I write is b, roba, zero, one. Well, one is always the largest, which is reasonable. It's open, it's obviously open in the uniform topology, no? It's a basic element, no? What, that's the best. But not open in the product, but not open, why not? So take any point here, whatever you want, okay? Take any point here, in this ball, okay? And then take any basis element of the product topology, which contains a point. Any basis element of the product topology, no? Basis element of the product, we want to see that we don't find any basis element of the product topology. Cannot find, which contains a point, maybe zero, maybe it doesn't matter which one. And it's contained here, okay? We don't find that, why not? Because if you take a basis element of the product topology, for, since it's infinite, it's strictly finer if J is infinite, okay? So there's one coordinate. In fact, infinitely many, where we have the reals. But now you have the coordinate here, okay? Maybe zero is the point, which we have, okay, zero. But in the reals, what is the distance which we can reach? In the reals, we reach distance with whatever we want, okay, but we have cut at one. So we reach one, we get distance one. In this coordinate, we get distance one. But this metric here is least upper bound, okay? It should be smaller than one, but we get one, okay? We get things arbitrary large, then we cut at one. We get one at distance. And then the soup is equal to one, but not smaller than one. Here, it has to be smaller than one, okay? This means smaller than one. Probably, but not open the product topology. In fact, again, what is, there's not a single basis element in the product topology which is contained in this. That's again, so it's very far from, okay? So that's maybe worth to write. In fact, not a single basis element in the product topology of the product topology is contained in this one. So that means that the interior is empty again, okay? It's very far from being open, this one. So this implies also that the interior of B row bar zero one is empty in the product topology, okay? In the product topology. So it's very, very far from, these topologies are very different, okay? The product and the uniform. So that was the first trial, no? Which didn't work. We don't know our omega still, okay? We found a new interesting topology, okay? Uniform topology. It's also not a box topology, that's what we're gonna do. I should give exercises, okay? And then collect. Did you write something? I should, I don't know if I should do with this or I write some exercises, okay? And again, I will give something which you should write very carefully. And something where you should think or only about because it's difficult to write, okay? Like this also, no? I mean, we can talk about it, I know it's late, but. Let me also access it. Page 111, 813. Page 118, six. And here, no. So these have to write and seven only the result. So I don't insist here, okay? Only, you should of course think about that, no? Seven only the result. But otherwise, you concentrate on the other first, okay? But this is interesting exercise. So only the result and then the page 126. Three, so these are the four which you should write very carefully, okay? These four here, okay? And here you think also about, so there's five only result. These is an exercise for product and box and this is the same exercise for uniform, okay? So they're very similar to the exercises, but somewhat difficult to write. And the last one is four B. Of course also A, but it's not so pleasant. I mean, four B on the result. These are more, you should think about these exercises, okay? So these are somewhat special, these exercises. You can do also four A, but maybe you find that difficult. I mean, it's not so, okay? You will see this exercise, okay? And you should think, what happens? The sequences, okay? So the important one which you should write very carefully are these and here you should think and try if you can get out the result, okay? In some way.