 Okay, so we're talking about continuous function, functions. So here's the proposition, so x and y are topological spaces, that's not spaces. And we have a function, f from x to y, a function. Then the following equivalent, i, f is continuous. So I abbreviate continuous to i for every closed subset. So continuous means pre-images of open sets are open. That's our definition, okay? Except you save continuous in one point. For every closed set a subset a in x, the condition is one has f of the closure of a is contained in the closure of f of a. So this is reasonable, if you think about this a second here. This is called closure, okay? So that means if a point is close to a in the closure of a, that's what it says in the closure of x. If a point is close to a, then its image point is close to f of a. It remains close to the image, you know? If it's close to a, the image is close to f of a. That's what it says, you know? And that's continuous. That's equivalent to continuous also, okay? Points which are close to something remain close in the image of the situation. Well, so this is easy to remember in this sense. And then we have pre-images of open sets are open and now pre-images of closed sets are closed, okay? Pre-images of closed sets, instead of opening of closed. Okay, so this is proof. So I didn't, I proved the most interesting stuff. So, well, i in plus 2i is not so interesting, but so let's start with it anyway. So this, we prove this, no? i in plus 2i. So f is continuous, no? F is continuous. And let's, we have to prove this, this is conditional, no? So let x be in the closure of a. Let x be in the closure of a. We have to prove that f of x is in the closure of f of a, okay? So let x be in the closure of a. And what means closure, meaning the closure of f of a, each neighborhood intersects, okay? Non-empty intersection, that's in the closure. So let x a, and let, so given a neighborhood, v, and let v be a neighborhood, neighborhood of f of x. It's continuous, so that means, well, pre-image of open sets are open. f minus 1 of v is a neighborhood of x, okay? So what x is in the closure of a, x is in the closure of a. My hypothesis then implies, of course, that the intersection is non-empty. So f minus 1 of v intersection a is not empty, right? And this implies, what does it imply? That the image is not empty, no? So f of f of minus 1 of a intersection a is not empty, of course. And then this is set theory again. Now what is f of a intersection b? f of, it's contained in, not equal maybe, no? So if this is contained in, what did I write? Yeah, v, this v, okay? So this, all of this is, of course, contained in f of f of minus 1. It's contained in v, not equal to v, no? Because not subjective maybe, okay? So it's, but it's contained in v, and f of a is contained in f of a. So this is contained here, it's not empty, so this is not empty. And that, of course, says that, what does it say? What do we have to prove? That f of x, v is the neighborhood of f of x, no? So f of x, this is our neighborhood. So f of x is contained in, is in the closure of f of a. So this is not too interesting. Now we just apply definition, one definition, okay? After the other, continuous closure, closure again. And so, applying definitions. The more interesting is 2i implies 3i. So the next one is 2i implies 3i. That's the most interesting of these inclusions. So what do we have to prove? So we have this condition now. We have this condition, f of the closure is contained in the closure of f of whatever. And we have to prove that pre-images of closed sets are closed. So we start with a closed set, okay? Which is called b. Let b be a closed set, let b be closed. Let b in y be closed. And then to prove that it's pre-images closed, okay? It's called in the book a, so I called it also, f minus 1 of b. So we have to prove that a is closed. This a, the pre-image of b, no? We have to prove a is closed. So let x be in the closure of a. Then, sometimes I write this, okay? Well, it would be better to avoid all those points. At certain points. This means that f of x is in the closure of f of a, of course. Now we have our condition, no? The closure of, f of the closure of a is contained in the closure of f of a. So f of a is contained in b, all right? f of a is contained in b, not equal to b because it's not subjective, maybe, but it's contained in b. So this is contained in b closure, clearly. But b is closed. So this is b. So what we do, we read here, if x is the closure of a, then f of x is in b. So that means x is in f minus 1 of b, of course, no? Clearly. f of x is in b, x is in f minus 1 of b. But this was a. Now we have a strange situation. If x is in the closure of a, then x is really in a, okay? If x is in the closure of a, then x is already in a, no? That means they are equal. I mean, a, the closure of a could be larger, no? But it's equal. So this implies. The closure of a is equal to a. So a is closed. So pre-images of open sets are open. A was pre-images of b. Pre-images of closed sets are closed, okay? That's more interesting now, the proof. Seems not difficult, of course. And then there's missing something. 3i implies i. And that I will not prove. Sometimes we prove, give two proofs sometimes. So what is missing is 3i in plus i. And of course, this one is pre-images of closed sets are closed. The other is pre-images of open sets are open. So just take compliments. By the way, this is easy to prove even this way, okay? Take compliments in both cases, no? Compliment of open is closed. Compliment of closed is open. But I will not write, take compliments. This is more, this is also not so. So I will not. So now we have some collection of examples of continuous functions. So let me start. So other properties. I will not call proposition lemma. I just give a list of other properties. Of continuous functions. Concerning continuous functions, so i. The first is constant functions are continuous. Well, that's not a big surprise, no? Constant functions are continuous? Well, why? If you have a constant function, no? What is the pre-image? What is the pre-image of any set? If it contains the constant, it's all, otherwise it's empty. It's open in any case, no? Empty and X are open sets. So this is, okay. Second one. Let A be a subspace. So let A subset, which is a subspace? Subspace is a subspace. Then the inclusion is continuous. So the inclusion, i, inclusion, i, all is i. So from A, and sometimes you write this way. No, this means inclusion, okay? It's a subset, but it's a map from X is continuous. Continuous. So the proof is, so pre-image is open set. What is the pre-image? I minus, so I write just I minus one. I minus one of U is an open set in X. What is this intersection, no? U intersection A. One second. I got, okay. So we have this, A intersection U, okay? A intersection U, U intersection A, okay? And of course, if this is open, where in X, no? Then this implies, if this is open X, then this is open in the subspace topology, no? That's the definition of the subset. Just take intersection. Open in the subspace topology. So this implies this, no? Open X, then this is open in the subspace topology. And this means i is continuous, of course. And we have a remark, which is in some sense interesting. Trivial, but interesting. In fact, the subspace topology, I mean, if i is continuous, then these have to be open. If i is continuous, and U is open in X, then these have to be open, no? Otherwise, i would not be continuous. All these have to be open. So that means that the subspace topology is the courses topology, the smallest topology, the courses topology, such that the inclusion is continuous. So I will write. This implies i, the inclusion, i, the inclusion is inclusion. The subspace topology, sorry, subspace topology on A, is the courses, courses, it means the smallest. Finest means the largest, biggest, okay, largest. It's the courses topology on A, such that the inclusion is continuous. Okay? Such that the inclusion is continuous. This is in some sense interesting because this distinguishes the subspace topology. That's the right topology. There's no other reasonable way to define topology otherwise. It's selected by this property. That's the subspace topology. For example, if you take the finest topology, the finest topology, then everything is continuous. The finest topology is discrete topology. So that has no sense. What has sense is the courses topology. This distinguishes the subspace topology. If you want the subspace topology, you want the inclusion to be continuous. Clearly, otherwise it's strange topology. You have a distinguished topology, okay? By asking that you take the courses topology, okay? This is distinguished. The finest is not interesting, okay? That's the discrete topology. If A has a discrete topology, everything is continuous. So that's not interesting. But this one is the right one. Okay, so this is the principle to construct topologies. Let's see again in another context. So where are we? 2i, 3i, A, B, 3i. A composition of continuous function is continuous, of course. A composition. F and then G, we write this way, okay? First F then G. That's a composition of continuous function is continuous. So the argument is here, okay? On the right, not on the left. A composition of continuous functions is continuous. So also here, this is not interesting. This should be true. It's clear. Otherwise, there's something wrong. But so first F then G minus 1. So the proof, if you have to write the proof of this, then you write this, okay? Of something which you call, you can give names to the spaces, no? F goes from X to Y. G goes from Y to Z, okay? But that's not so important. Important is if you take now in Z an open set, okay? Pre-images of open sets are open. So this is equal to, so first, so concentrate, which one comes first? First G inverse, no? G inverse of U. I write from this, from right to left. Second function, and then F inverse, no? Well, if G is continuous, this is open. If F is continuous, this is open, okay? So this is open. So this is the proof. If you write the proof, then you can write. So U open, then this is open, okay? So it's continuous. So this is a one-line proof. What is the next? The restriction. So what is 4I? The restriction. F, sorry, I write this. F restricted to A. Well, sometimes you write here a little bit small, but I don't know. From A to Y, of a continuous function, F from X to Y, the restriction of a continuous function, F from X to Y, where of course A is, subset, subspace now, okay? If you have this situation, we have the subspace topology, okay? So I don't say subset, but I write subspace. A, X is a subspace, of course. We have a topology here. Yeah, that's topology. The subspace topology. The restriction of a continuous function is continuous. Okay? Any restriction of a continuous function is subset or subspace. Subset, you say, often, but you mean subspace anyway. It's continuous. The proof is also trivial. So why is that? What is your restriction of F? So I give the proof, because it's a one-line, this means proof. No, just indication of proof. F restriction to A is nothing else than, well, you take the inclusion, and then you take F from A. This is the inclusion to X, and then you have F, of course, from what, to Y. This is F. So that's the point. F restriction is F. This is composition just, no? And then now apply. So the inclusion is continuous. So that's 2i, I suppose. 2i, and then apply. 2i, and 3i is still here. The composition, no? And 3i. So this is also a one or two-line proof, okay? If this, okay, so... I write two other things, but I will not prove everything, no? No, I'll prove this. So 3, 4, 5. Well, maybe first... So F from... let F from X to Y be continuous. F of X be contained in B, contained in Y, or in a subspace, okay? In a subspace of Y. And then there's no notation for this. Restriction is on the... Well, maybe then... F. So it's not F, really, no? Because I change here now to B, okay? Then F, F prime. Well, let me write F. It's the same F, but it's... Normally it's not the same, but in principle it's the same, okay? You can call it F prime. Then F from X, it's easier to write it this way. To B, because it finishes in B, okay? It's continuous. I am tempted to call this core restriction, okay? Core restriction. But there's something always strange with this restriction, core restriction, because the restriction is in the core domain, no? And the core restriction is in the domain. What is the domain of a function? The domain of this function is X. No, no, it's okay then, okay? And here we have the... No, no, one second. The domain codomain. These are just words, no definitions. The domain is Y in general, like codomain. Okay, then I... Yeah, then it's okay, no? That should be also... I got confused between domain and codomain, no? Yeah, you're right. Then the core restriction. That's the core restriction. Then it's reasonable, no? Because you restrict in the domain and now you restrict in the core domain, okay? And so you call it then the core restriction. And that I don't prove. And then we have something similar. So last one, six. You can extend, extension of domain, okay? F from X to Y is continuous. And Y is a subspace of that subspace. Then I call it again F. F, but now I can make it larger, no? That, okay? So this is extension of codomain, then, okay? It's continuous. That's the most interesting, these words, maybe. So this is extension. I call this extension of codomain, okay? This is a restriction of codomain. This is the extension of codomain, no? The image. In the image, you can make it smaller if it's contained there, or you make it larger, okay? Oh, so this I will not prove. And then we have a lemma. And this has a name because it's used very often. The proof is very easy. But it has a name. And we will use it by this name, often. Okay, so this is lemma proposition in the book. But this is the pasting lemma. That's the name, the pasting lemma. So this I will prove because... So this is a pasting lemma, and that's the name we will use. So we have to remember that name. So let A, let X be a union. A, B, where A and B, such that A and B are closed in X, yes. Actually, it will be true also for open. However, it will be used much more often for closed, okay? We will see. The proof works also. Let F from A to Y and G from B to Y be continuous such that on the intersection, they agree such that F of X equal G of X for every X in the intersection, A intersection B. Then obviously we have a function on the whole, okay? So this is called FGH. Then let H from X to Y be defined by H of X is equal to... So F of X, if X is in A and G of X, if X is in B, is well-defined, of course, no? It's well-defined. That's trivial because in the intersection, if X is in A and in B, they are equal, no? So it's well-defined. It doesn't depend on the case here. It's well-defined and continuous, okay? That's a continuous. That's a pasting lemma. We paste together two functions, okay? We paste together two functions. It's like pasta, no? You have the same word, no? Paste together two functions and that's a pasting lemma. Okay, proof. So this will be used very often. I mean, you have other formulations of this, of course, which would be shorter. This is a constructive. You construct the function here. I wrote the construction in some sense. You can formulate if F from X to Y is a function such that F restricted to A is continuous and F restricted to B is continuous then F is continuous. That's shorter, okay? Here you have already the function, no? F from X to Y, okay? And it's clear on the intersection. It's the same, no? But this is a constructive version. You have the two functions and then you paste them together and what you get is also continuous, no? That's a stronger version in some sense. Or you apply this kind of the proof anyway, proof. So we are working with close sets. So we have to prove that and then it's reasonable pre-images of close sets are close, no? That's a point. If you work with open sets, pre-images of open sets are open, okay, for continuity. So we start with a close set, okay? Which is called, whatever it is called, let C be a closed subset of Y. Let C in Y, let C subset of Y be closed. C is a closed subset of Y. And this is a proof that's a pre-image. We have to consider H minus one of C, right? This would be closed. Then it's continuous. That's a proposition, which we proved. Pre-image of closed sets are closed. So what is it? Of course, this is F minus one of C, union, G minus one of C. Because H is F or G, okay? So you take pre-image of F, F minus one of C is a union. And then what we have? F is continuous, no? So this is closed, but where? Yeah, that's the only interesting point here. It's closed in A, first of all. And this is closed in B. However, as you said, A is closed in, so closed in A, this implies closed in X since A is closed in X. And here's the same, of course. Closed in X, no? Again, in X since B is closed in X. And then, of course, it's a union, a finite union, finite union. This is a finite only, no? A finite union of closed sets is closed. So this implies H minus one of C is closed. And this implies that's a proposition, one proposition we proved that C is closed, that H is continuous. So this is the case. And you see, the same proof for open, no? If you have open sets, then what? Then you work with open set here, and this is the same, no? It's stronger. I mean, you maybe have three pieces, no? Or ten pieces. Well, then it works also with closed, no? Here you have two pieces, where you piece together, no? If you have ten, it's okay. If you have infinitely many, maybe here it gets wrong with closed, because infinite union of closed sets might not be closed. So open is always true, no? As many as you want, also infinitely many, no? So there's a slight difference, but we will apply, so you can generalize in various forms, okay? So open is even stronger, then, okay? With closed, you need finitely many, okay? Otherwise, union of closed sets might be not closed. So that's the pacing. Okay, so for the moment, we pass to the next point, which is, we go back to the product opportunity. We come back to the product opportunity, okay? So again, product opportunity again. The difference is that now, we discussed two, X times Y, okay? And now we take arbitrary products. This is an interesting point. So we need some notation for an arbitrary product, okay? So we have the product. This is the product, okay? Pi, capital Pi, okay? Greek. Product of space is X, Alpha. Alpha in some index set, which is always J and not I, because I is the interval very often. So this is a Cartesian product, no? Cartesian product of topological space is X, Alpha. These are all topological spaces. And now we take the Cartesian product, okay? And we want to define a topology here. And this is not immediately clear at first sight, okay? If you have two spaces, X times Y, okay? What did we do? We take, as a basis, open times open, right? Open X times open Y. That's not the topology, but it's the basis. These are rectangles in some sense, no? Open times open, okay? And this is the basis. And now here, it's not so clear immediately now what we take. So it's an interesting point. So I will motivate. So J might be infinite now, no? We don't know anything about J. Find it infinite, whatever it is, okay? J index set, index set. This is an index set, okay? So J is some index, which we do not know. Find it infinite, countable, uncountable. There are many uncountables, no? So we need to, anyway, some notation, okay? We need to write these points, okay? And the notation is the following. So I don't want, I will not define the Cartesian product formally, okay? So we need a notation. So this is a set of all. X alpha. So this is a small X, no? Alpha and J. That's the way how we write. Where X alpha is in X alpha. How we write? Okay, this. If you have only two, you have X1, X2, okay? If you have three, X1, X2, it's three. If you have countably many, you have X1, X2, X3, X4, then the index set, you can take the natural, no? And you have X1, X2, X3, and so on, okay? If you have more than countable, then it's difficult to find an order. You have to write this way, okay? There's no first or something. It's not clear, okay? For the final case, it's easy, no? How to write? You write open, no? Two, three, four, five, first, so on, no? For infinity many, it's also clear. For countably many, okay? For uncountably many, there's not clear what order, no? There's no first one, and so... But anyway, we write this way, okay? This is a typical element in this product, okay? Good. And then we have the projections. So we have to define the projections, so let... So for... Now I take beta in J, another index, alpha is varying index for beta in J, let... And I track the pi, this is a small pi, pi beta from the product x alpha, alpha in J to... I'm fixing just notation, to x beta, be the projection onto the beta's coordinates, okay? Be the projection. So I define, of course, but it's clear, no? The projection onto the... Now, how to write that? Beta's... I don't know, maybe this way, I'm... Sometimes I'm... Coordinate, okay? So that means, of course, that pi beta, pi beta of a typical point here, which is x alpha now, okay? That's a typical point. Pi beta of x alpha, alpha in J, okay? Pi beta of such a point is what? x beta, x beta. The beta... These are the coordinates of a point, okay? The coordinates, they are the x alpha, no? And you take the beta coordinate, okay? So this is the definition of the projection. I'm just fixing notation, okay? And we have to fix some reasonable notation. That's the notation of the book, of course. And now, what we want is... We want, of course, to define a topology here, the reasonable topology, okay? That's what we want, okay? We want to define a topology. For two spaces, we know, okay? We did already. But now, in this situation... I mean, the first, what you maybe say, just take products of open sets, no? The products of open sets, okay? So we want, again, I said we want to... For the subspace topology, we found a canonical way to define this, no? That's a course of topology, such that the inclusion is continuous. And that distinguishes the subspace topology. And here, we want to do the same, okay? We want to define a nice canonical topology. Instead of inclusions, here, we have the projections, okay? It takes the projections, not... Well, we don't have inclusions, anyway, here, in a reasonable way, no? Because what we can only have are projections, not inclusions, okay? Inclusion, we don't know what... We have one coordinate, that's okay. But then, we don't know what to take for the other coordinates, okay? So the other... The only maps which are defined are the projections, okay? For this. So we want to use the projections, now, okay? We want to define the topology. Of course, we want all projections pi beta to be continuous, no? If we want a reasonable topology, these canonical maps defined should be continuous, no? We want all projections motivating and constructing also the topology. We want all projections pi beta, beta and J to be continuous. If this is not the case, then we have some strange topology, maybe, okay? If the projections are not continuous, then... So they should be continuous, no? What does it mean? That means that... So we define... If this is the case, then we have... Then for a fixed beta in J, S beta, S beta. This is a written S, okay? Not a printed S. Then for a fixed beta in J, for each fixed beta in J, for each, for every, whatever. S beta, what is this? We define pi beta minus one, that's the projection, of U beta, where U beta is open in X beta, U beta open in X beta. Then this should consist of open sets, okay? If... If pi beta is continuous, then this consists of open sets, okay? Consists of open sets. That's... Because that's continuity. Premature of open is open, okay? And then, and also... And now I call S, without index. Because for each beta, it should be continuous, no? So I take the union of all this. Union S beta, beta in J. Consists of open sets, and also this has to consist of open sets, no? If each projection is continuous, then this consists of open sets, okay? Otherwise, something is not continuous, no? So why did you write S? So now you think, this is not a topology, no? This is not even a basis for a topology. So it's a sub-basis. So we take... So the problem... That's the definition we can define already, yes? The product topology. So definition, that's the definition. The product topology on this product. Product X, alpha, alpha in J. The product topology is the topology generated by the sub-basis S. You recall, for sub-basis, we have no conditions almost, okay? And this is almost fulfilled, okay? So it's a sub-basis, no problem. Maybe it's a basis, but it's not a basis. So we don't care basis. We take sub-basis. We have to take a sub-basis in general. Is a topology generated by the sub-basis S? S, we wrote the sub-basis, no? So that's the definition of the product topology. So now we have a conceptual way to define the topology. Gen, zen, what does it mean? Then that's the minimum we have to take, okay? Such as the projections are continuous, no? The minimum, okay? So then we can characterize, again, the product topology is the topology, the courses again, no? Yes. Then the product topology is the courses topology such that all projections are continuous. Such that all projections pi beta are continuous. So this is easy, the definition. This makes clear that this is a very conceptual way to define this topology, okay? And this, in fact, is a product topology. Of course, in general, we don't want to work with a sub-basis but with a basis. We prefer to work with a basis. So you recall how to go to pass from a sub-basis. So this is a sub-basis and we go to a basis. How do we pass from sub-basis to basis? Find that intersection. Take finite intersections. So let's write down the basis, okay? So a finite intersection would be, take finite intersections. So I write a finite intersection, okay? Now the indices, so they look all like this, no? The indices might be all different, all equal. If they are all equal by the way, no? Then we can take here the intersection, okay? If the beta is the same, okay? So the interesting case is when they are all different. If two are equal of these indices, then we can make it shorter, the intersection, okay? Then we can forget one and we can amalgamate two, okay? If they are all equal, no? Then we, if they are all like this form, then the intersection is again of this form, okay? You just take the intersection of all these, okay? So the interesting case is they are all different, okay? So suppose, I will suppose this, the beta one up to beta n are all different. In J, of course, are all different. Are all distinct, different, whatever. And then I have such a finite intersection, pi beta one minus one of u beta one. Intersection, intersection pi beta n minus one of u beta n. That's the final intersection. The typical final intersection is this one, okay? If they are all equal the indices, then we just have one single, we take the intersection of all these in the same space, okay? And take the privilege. So this is an interesting case. Everything else we can reduce, make shorter the intersection. Suppose all different. But what is this? Intersection. This is again a product, alpha. How did I write? U alpha, alpha in J, U alpha. This is a product again. Where, however. So I have to, where, however. U alpha is equal to what? Well, if I, if we are in one of these coordinates here, now, beta one, beta n. Then we have to be here. The image has to be here, okay? So we have to be in u beta i. If alpha is equal to i from one to n. If it is one of these, then it must be this. Otherwise, we don't care, okay? If we are not in these coordinates, we can take everything. So it's x alpha. If, if, yeah. X alpha, if alpha is different from beta one. That's what it is. This product, the privilege, okay? And then we know what the basis is. So now I rewrite in a slightly different way, but it's exactly what we did here, no? So the basis are products. U alpha, okay? Alpha in J. They are products. Such that U alpha is equal to x alpha, except for finitely many indices. Except for finitely many indices. Beta one, beta n. And of course, n is varying also. n in n, no? For albuquerie, n in n. n in the natural numbers. Also here, this is. So this is the basis for the product topology. So we, one works, this is basis. Sometimes it's a sub-basis, but basis is better in general. This is the basis. And now you may say, so this is the definition of the product topology, okay? And this is important. We will, many exercises. But now you may say, let me define, you may say, well, this is very, how do you say, this is very strange, this definition here. This, except for, oh well, I have to, sorry, sorry. You have exactly finitely many indices. I have to write U alpha in any case is open in, so that's the first, sorry. U alpha open in X alpha, okay? U alpha, sorry, open in X alpha, that's, we want anyway, okay? Let's see, these are these open sets. And U alpha equal to X alpha, except, and U alpha is equal to X alpha for X, close again, okay? So it's a product of open, okay? Of open. It's a product of open sets U alpha in X alpha, except you have this condition here, finitely many, no? You say, maybe, who cares about, okay? So let's forget this condition, no? So we have a B prime. This looks easier, no? You take product U alpha, alpha and J. And now I just write U alpha open in X alpha. So this condition is easier for other cases, now open in X alpha. And I forgot this condition except for finitely many, no? Okay? This is also a basis. You check easily. This is also a basis for topology on the product, X alpha, alpha and J. This is also a basis. This is called box topology, okay? This is not the product of open. This is called the box topology. So this will be also an important example, the box topology. So it's clear that the first thing is, which is clear is that the box topology is finer than the product, no? No problem about that. So the box topology is finer than the product topology. Because this B is contained here in B prime, no? It's a subset of B prime, okay? This is a smaller basis than this one. So we'll get more open sets in general, okay? If J is finer, they are equal. If the index set is finer, then this condition we can even write, no? Then they are equal. If J is finer, the index set is finer, then the two topologies are equal. Then the two topologies, then they are equal. If J is infinite, in general, they are very different. It depends on the spaces, no? If you take one point spaces, they might still be equal. I don't know. In very strange special situations, okay? They might still be even. But for general spaces, they are very different. In fact, it will turn out, and we will see that the product topology has all nice properties which we want, almost. I don't know. There's nice properties, product topology, that we will see for many instances, okay? The product topology is a very good topology, reasonable. The box topology, everything which can go wrong goes wrong with the box topology, okay? Everything which is reasonable to assume is good for the works and is valid for the product topology and almost everything goes wrong for the box topology. That's what we will see, okay? So the good topology is the product topology, not the box. The box has very bad properties, okay? There are too many open sets in general. Too many open sets, okay? That we will see. Many examples, okay? For the moment, we have the definition. Product topology, this basis, okay? That's the definition, that's not the definition, but that's the basis. That's the basis. We take products of open sets, but these open sets, except for finding the indices, have to be on the whole space, okay? Except for finally. For the box topology, we take arbitrary products of open sets, okay? So these are the definitions of these two topologies. So proposition. First proposition for the product topology. So proposition, important. So let F from A to a product. The usual, it's always product X, alpha, alpha, and J. Be given by the, be given by F of, so we define the component functions, okay? So F of A, a point here, F is in, so we have all these components, no? And we call them F alpha of A. F alpha of A, alpha in J again, okay? So these are the component functions. What I call the components, okay? The component function. In fact, how did you find F alpha? So F alpha, that's F, and then pi alpha, the projection of alpha, okay? So this is F, and then we just project on the alpha, that's the alpha's coordinate, okay? Of F of A, no? So this is, we can define this way. That's the F alpha. The alpha, we look at alpha coordinate, okay? And this is the function F alpha, which we see here. The coordinate functions. The component functions, the coordinate functions, okay? Whatever. This is not a definition. The coordinates, no? These are functions. Components are, this is just, well, this is so far no proposition, just definition. So what is the proposition? Then F, so this, this is product topology, okay? With product topology. If you don't say anything, we always have the product topology, but here I should say. So this comes with the product topology, okay? If you don't say anything, it's a product topology. That's the convention. So now what does it say? So I can write here even, then it's short. F is continuous, if and only if. F alpha is continuous for all alpha. So if you have a function from a space into a product, it's a product topology, then this function is continuous if and only if all coordinates functions are continuous. That's very reasonable, no? That's easy to check, very reasonable situation. F is continuous if and only if F alpha is continuous for all alpha in J. That's a proposition. Maybe it's not too frequent. Then F is continuous if and only if all F alpha continues. The proof is easy. So that's another good reason for the product topology. The first good reason for a product topology. It's if and only if, okay? That's a reasonable topology. And it's not, as I said, it will not be true for the box topology, okay? We will see. I said almost everything goes wrong for the product topology except that the projections for the box topology are continuous also, no? Because it's a larger topology, no? So they are, so proof. The trick. So we take, what means continuous? Pre-images of open sets are open. But we don't have to take all open sets. We can take bases and only for bases elements. But we don't have to take all bases elements. We can take sub-basers, okay? Here we can work with the sub-basers. We don't work with the sub-basers very often, okay? But here, yes. So we have the sub-basers, S, no? The sub-basers. And the typical element is P pi beta minus one of U beta. The sum element in the S, which is the sub-basers for the product topology, no? This is the sub-basers for the topology. That was the union of these S-beta. Now this is in sum S-beta, okay? And then now it's... We have to prove what is continuous. Ah, there are two directions, no? And then we see which direction we proved. So F minus one is a pre-image of a sub-basers element, no? Of pi beta minus one of U beta. The first F composition pi beta minus one is first pi beta minus one, then no? You have to change order here, no, for the pre-image of U beta. But what is this? F beta. So now we have to decide what we want, which direction we want, okay? Which I didn't forget to write. So we have the interesting direction, which is the interesting direction. This is this one. What does it mean? It means all F alpha are continuous. So F beta is continuous. So this is open. Because this is open in X beta, no? Open in X beta. So this is open in A, okay? And then we have that F is continuous. The pre-images of sub-basers elements are open. And this is sufficient, no? We have to, for all open sets, for basis elements, or for sub-basers, no? This is sufficient. F is continuous. So this is the interesting direction. But also only three lines of what? The other direction is, in fact, trivial, no? No, almost easy. So I want to see this, the proposition, how I write here. So this is the other, including this one, no? And now we have what? F is continuous anyway, okay? F is continuous, this hypothesis. However, pi beta, how is it called? Alpha. Pi alpha is continuous. Recall that the product topology is the course topology, such that all these projections are continuous. They are continuous, okay? And this implies that the composition. So F and then pi alpha, what is this? This is F alpha, no? This is F alpha is continuous. So this is trivial. This holds also for the box topology, okay? Because this direction, no? This, not inclusion, how do you say? This implication holds also this one, okay? Holds also for the box topology, no difference. This implication holds obviously also for the box topology. Because the projections are continuous for the box topology, no? They are more continuous even because we have much more open sets. So this implication holds also for the box topology. The other one does not hold for the box topology. So now we give an example. The other inclusion, no? So this one is okay. This one holds for the box topology. This one here does not hold in general for the box topology. So let's see that. Because this is this interesting fact about the product topology. A map into a product is continuous if all components or coordinate functions are continuous. No, this is very retro, very easy to remember also. And now we have an example which is one of the main examples in the book also. So I said they are in the book, they are about four or five main examples. One is the lower limit topology, okay? One is the ordered square. We didn't do too much with this. This is an exercise, okay? But we will see the ordered square. And the third one is this one. So we take the easiest infinite product which you can think of. And that's R omega, for example. Important example. We take R omega. So this omega is a symbol which stands for the natural numbers, okay? This is, so this stands for countable, okay? This is countable. So this is R cross R cross R cross. So this is a countable product, no? Countable. So this omega stands for countable, okay? For the natural numbers if you want. But you don't write n, you write R omega. So what are the elements of R omega? This is a countable product, so now it's easy. Now we have an order, no? The first, the second if you want, no? We have the natural numbers, countable, no? So we have Xn, now we have Xn, n in n. The natural numbers, no? So this is X1, X2, X3, and so on. Where Xn, all Xn are real numbers, okay? Real numbers. So they are all real numbers. One. No, no, one, one, one. Yes, but n are the natural numbers. Zero is not a natural number. Sorry? Zero is a natural number? Well, it was not so natural in the history of mathematics. I mean, you have one, two, three, very long, okay? But zero, that took a long time, no? Before it was established, or not? So it's not very natural. No, no, here I write one, two, three, okay? And for us it's one, two, three, no? That's n equal one, two, three, and so on. That's the natural numbers. Where do the zero come from, by the way? They are not doing history of mathematics. What? These are Arabic numbers, anyway. So it came by Arabic to West, and Arabic, I think, they had from India. They had started somewhere in India, no? And then by Arab, they adopted, no? And then it came to the West by Arabic, yes. That's a story of zero in some sense, no? It was a big thing. I mean, that was very important. The Roman numbers, if you think of the Roman numbers, no? They were not so efficient, no? The Roman, okay? They have these strange numbers, what are they called, what is L, what is 500, or 1,000? No, 1,000 is M, no? And 100 is C, but what is 50? I don't know. Something. Okay, so what are these? These are real sequences, no? If you want, these are all real sequences. These are just real sequences, okay? Sequences of real numbers, no? That's our omega. So this is, this stands in some sense for Rn also, okay? But this is not so nice to write, this Rn, okay? So it's R omega. And now I give the example. So this is, I take a function, f, f. So, by the way, so take R omega, if you don't say anything, this product topology, okay? No, this product topology. But now, suppose R omega has a, well, sorry. I should start. So I take a function f from R, this easiest function which you can think of to R omega. So f of t is equal to t, t, obviously t, no? So if you write it in a way, it's t, it has no index, it's always the same t and for all n in n, you have always t, the same t, no? Constant functions, constant sequences, constant sequences. So that's our function, no? If R omega, so R has a standard topology always, okay? I don't say anything. So here, R always, if you write this symbol, no? Means R with a standard topology, always. Otherwise, you have to write something else. For example, RL with L, no? That's the lower limit, okay? Or RK, that's an example. That's also an important example in the book, which was an exercise, okay? It's an exercise. RK, the K topology, it's called in the book, okay? That's another. But if you don't write anything, then it's the standard topology. So if R omega has a product topology, then f is continuous, no? Because all components are continuous. Then f is continuous, since all components, all component functions are continuous. That's the proposition, okay? My component functions are from R to R, the identity map, which is continuous. Suppose R omega has a product, the box topology, okay? Suppose now that R omega has a box topology. The other one, the bed topology. Suppose that R omega has a box topology. So how does a box topology? The basis is we take arbitrary products of open sets, no? Arbitrary product. So here's a product. Minus one over N, one over N, N in N. So this is in R omega, no? So this is a basis element. It's open, okay? In the box topology. It's open, it's open. This is a basis element of the box. We can take arbitrary products. These are open interval, no? Open interval. Real interval, real open interval. So what is f minus one of this product? So that's the question. What is it? What? No, it's not empty. Not empty, not empty. Not so far from empty, maybe, but it's not empty, no? There's one point, at least. Zero everywhere. One second, what? These are open intervals. So what is intersection of minus one over N, one over N? So N in N. These are open intervals, yes. What is intersection of all these open intervals? Zero. We have open intervals, yeah. But the intersection is zero, no? Also for closed intervals. But these are open intervals. So it's zero. Intersection is zero, no? The pre-match is this one. This is not open in R in the standard, okay? Not open in R. And this means f is not continuous. Now it's not continuous. Because the pre-match of this open set is not open, okay? It's closed, but not open. So that means that the box is strictly fine for omega, so on R omega, the box topology is strictly fine. But this is trivial. So this is an interesting example, okay? This continuity criterion does not hold, okay? For the box topology, no? For the product topology, it's easy, no? Something is continuous. If all component functions are continuous, here it's not true. As a consequence, the box topology is strictly finer than the product topology. That's clear. They are not equal. They cannot be equal. But this is easy to see in a different way, no? In fact, so we have another observation. For example, so observation. Take any product where you don't have the whole space, okay? So I take whatever you want, minus one, one, N and N. Well, I might have, it's easier than this one, okay? I take just this one, okay? Product minus one. So this is, in some senses, minus one, one, omega, no? If you, some notation would be this. It's open in the product, in the box, no? Open in the box. It's open in the box topology. Why? Well, it's a product of open sets, okay? A product of open sets. In R, yeah. It's always the same here, but it doesn't matter, okay? It's open in the box topology. So what I want to say is, it does not contain a single, any basis element which does not contain any basis element of the product of open. Not one basis element. So it's open in the box topology, but does not contain any basis element of the product of open. Not a single one. Why not? If you take a basis element of the product of open, you need the whole space, except for, finally, many indices. The whole space. And that doesn't go well with this. Here, you'll never have the whole space, no? You always have the small interval. So the coordinate goes out, okay? If you have the whole space, you are out of this coordinate, okay? So it does not contain any basis element of the product of open. It's very far from being open. That means that the interior of this in the product of open is empty, no? That is what it means. This implies that the interior, product minus one, one. How do you write? Interior. The interior, okay? The interior of this is empty in the product of open. It's empty in the product of open. Because it does not contain any open set, okay? Not a single open set. Not a single basis element. If it contains an open set, it contains a basis element, right? Because, clear, no? Any open set contains basis element, unless it's empty. We don't care what empty set. But, so this is very, very far from being open in the product of open, no? Sorry again. This is very far of being open in the product of open, okay? It's on the opposite of open, okay? Almost. It has empty interior, okay? So, it's nothing to do with open. The product of open is open, but not in the box. That's how it is, okay? Because for the product of open, the basis elements, you have this, this is not a basis element. You need the whole space, except for the many indices, the reals. And, if you have one real, one time the real, okay? Then, this is too large for this, okay? Then, it doesn't go well with this coordinate. So, there's not a single point in the intersection. So, that means, this is the first feeling that there are many, many open sets in the box topology, okay? It's too large. It's too large. The product topology is a good topology, okay? The bed topology is a bed topology on, okay? But very often the bed top things are more interesting than the good things. So, we use it as an example, no? If you have old pictures, the heaven is always boring, you know? But hell is interesting. So, people want to... Okay. Maybe I should finish. Yes, I should finish.