 So the last thing we talked about, compact spaces. And the last thing we proved was a closed subset of a compact space is compact. That's easy. And then we proved that a compact subspace subset of a house-dough space is closed. A compact in a house-dough space is closed, otherwise it's not true. Compact, if it's not house-dough, it's not true in general. So we proved some other theorems on facts. So here's a small lemma. So if f from x to y is continuous and x is compact, then f of x is compact. So this results in what the image of a compact set is compact. That's what it says, the image of a compact set, of a compact space or set is compact. The proof is very easy. So we have to prove that what f of x is compact. So let a, a be an open covering. What should be compact? f of x, f of x of f of x. And as usual in y. So I take open sets in the larger space. I don't let a be an open covering of f of x and y. Then the pre-image, then f minus 1 of a. So what is this? Obviously, this is f minus 1 of a, a in a, the pre-image. f minus 1 of a is an open covering of x. But x is compact, so there's a finite sub-covering. So x is compact, implies. Then f of minus 1 of a has a finite sub-covering. And then also a, it takes the same. And then also a. And hence also. So here you take a finite sub-covering, and then you take the corresponding here. Theorem, the inverse, in general, the inverse map of a bijective continuous map is not continuous. However, so if let f from x, y be bijective and continuous. So we have a bijective continuous map. The inverse map, in general, is not continuous. But now we have x compact. If x is compact, and y is hostile, then f is a homomorphism. So compact house of space is the situation is better. So the proof, what do we have to prove? We have to prove that, yeah, we have to prove that the inverse map, f minus 1, from y to x is continuous. So that means that pre-images of open sets are open, OK? So pre-image under f minus 1, f minus 1 minus 1, that's f. So this is equivalent to the fact that f is an open map. That is, what it means, open means maps open sets to open sets, OK? Maps open sets. So this is a definition which maps open sets to open sets. However, in this situation, open map, maps open to open. That's different from continuity, clearly, OK? Continuity is more important, but also this is important. Then there is a closed map. Closed means maps close to close. Continuous means pre-images of open sets are open, which is equivalent to the fact that pre-images of closed sets are closed. Open map means images of open sets are open. Closed map means images of closed sets are closed. That's not equivalent in general, OK? That's not equivalent. However, in this case, this is bijection, OK? And then it's equivalent, because since it's bijective, closed sets are complements of open. So map complements to complements, OK? So this means equivalent f is a closed map. Maps close sets to close sets. Since f is bijective, OK? Since f is bijective, otherwise, no. In general, it's not equivalent. Since f is bijective. So here we can prove that f is closed, OK? This is equivalent. Inverse map is continuous, f is open, f is closed. So then it's easy. So let A be closed in, what is closed? f from x to y. Let A be closed in x. What is the hypothesis? x is compact. What was that? A closed set in a compact set is compact, OK? So this implies x is A is what? Compact. So this is the fact closed set in a compact space, or set is compact. A is compact. This implies f of A is compact. So this is all what we use, OK? f of A is compact. What is this? The image of a compact set is compact. So f of A is compact. Where are we in y? So y is house of. So this implies then that f of A is compact. f of A is closed. And this is what we proved with our set now. A compact set in a house of space is closed, OK? A compact set. In a house of space is closed. And that's it. So that means then we proved f is closed. So f is closed. The map f is closed, OK? And this is equivalent as we said that f minus 1 is continuous. So this is in fact equivalent. So f is only both. OK. So that's also easy. Now we have to prove other stuff, the main theorems about compact spaces. So the next theorem is, the next theorem we prove is a product of, finally, many compact spaces is compact. Product of, finally, many compact spaces is compact. Let's prove this first, OK? So proof. And then you ask for infinity products, no? For arbitrary products, of course. What happens? And it's also true, but I'll comment on this later. First I prove this, OK? Here we don't see the difference between product and box topology, OK? Finite product is the same topology, OK? For infinite products, there is a difference in general between product topology and box topology, OK? And also uniform. So there it's more interesting in some sense for arbitrary products. So proof. A product of, finally, many spaces. It suffices to prove for product of two, of course, finally, many. Suffices to prove the theorem for the product x times y is x times y of two compact spaces, x and y, OK? So we have x times y only. And then we have a lemma, which is very useful lemma. And there will be some exercises also. So this is a tube lemma. Tube lemma. So what is a tube lemma? So consider x cross y. And here we need only that y is compact, OK? Here's also x compact, but that we don't need now for the tube lemma. With y compact, consider this product. So we make a picture and write. So we have x cross y. So this is x, and this is y. And y is compact. And now we consider a slice. So we have a point x0 here, some point x0. And this is called a slice. So this is x0 cross, what is it? x0 cross y, yes. Which is homeomorphic to y, by the way, OK? This is homeomorphic to y. What is homeomorphism? Just forget this point, OK? This is homeomorphism. As a subspace, there's no problem here. So let n be a neighborhood. So we have n. Whoops. So this is n. So I'm going to write n be a neighborhood. If n is a neighborhood, neighborhood, what is neighborhood? Open set containing, OK? Open set containing is a neighborhood of a slice. So this is called slice, slice. x0 cross y, a slice, x0 cross y, x0 in x. If n is a neighborhood of a slice, x0 cross y in x cross y, obviously, OK? Then n contains, then there's a neighborhood, w of x0. So here's a neighborhood, w. There exists a neighborhood, w of x0, such that, now this is not good. So here's w, OK? Such that the product, w cross y, is contained in n. Such that if n is a neighborhood of a slice, then there is a neighborhood, a neighborhood, w of x0 in x, such that that's a tube. That's called tube. Such that's a tube, that's a product, w times y. That's why the name, the tube, w cross y, is contained in n, in this neighborhood. Given the neighborhood of a slice, there's a tube. Such that the tube, such that, is contained in, what is it, n. So in the picture, that's a tube lemma. If you have a slice and a neighborhood of a slice, then we find this tube, which is contained in the neighborhood, which is a product. This is not true if y is not compact. It is not true if y is not compact. Clearly, it's not true if y is not compact. You find easy examples. I mean, you have, for example, well, I will not. You take r cross r. And then you have a slice, which is the y-axis. And then you take a neighborhood of the y-axis, which looks like this. Something like this, which comes closer and closer, if you go to infinity. Then you don't find the tube here, which is contained in this neighborhood, because it comes arbitrary close to the y-axis. So you need some compactness, otherwise it's not true. OK, so this is a tube lemma. That's the tube lemma, and this is also the picture. So proof of the tube lemma. So we cover x0, the slice, x0 cross y, which is homeomorphic to y. This is homeomorphism. This is homeomorphic to y. So it's compact. That's apparent. By basic elements of the standard basis of x was y, u cross v, where u open in x, u cross v. So this is open in x, and this is open in y. This is a product topology, u cross v. By basic elements contained in n, so contained in, we have n. n is given. This is open, contained in this neighborhood n. n is open, so for each point, there's a basis element which contains the point, and it's contained in this open set, which is n. So we find many of these, for each point, we can choose one of these, for each. But this is compact. So what do we get? We get an open covering of the slice in x cross y. So we get this content in n, so we cover the slice. So we have an open covering x0 cross y in x cross y. But x0 cross y, homeomorphic to y is compact. y is compact. So finally, many suffice to cover. So this implies that we need only finally many. Finally many, u1 cross v1. So finally many of these suffice. But it's compact, this implies, finally many of these names. And let me give them names, u1 cross v1, un cross vn. Suffice of these neighborhoods, suffice to cover x0 cross y. So we have only finally many here. So it looks like this. Finally many. And we may assume, obviously, that x0, which is our point here, is contained in this first here. We can assume that x0 is in all of these. So ui, i from 1 to n. Why can we assume this? Otherwise, if x0 is not in one of these, then we can forget this. Because it doesn't help to cover x0 cross y. So just forget it. OK? So we can assume that x0 is in ui. Otherwise, it doesn't help to cover x0 cross y. And then we have our neighborhood. So let w be equal to, so we have finally many neighborhoods, ui, and it takes intersection of all of these. Then w equal u1, intersection, intersection, un is a neighborhood clearly of x0 such that, and this is good. This is our w, such that w times y is contained in. OK? That's the proof. Of course, we have to see this also, OK? But is it OK? So this is the proof of the tube level. And we will give some exercises tomorrow which use this two-plan or generalize the two-plan, OK? Now we prove the theorem. So proof of the theorem, so what do we have? x cross y, we have, no, but both are compact. So x and y are compact now. x, y are compact, both. We have to prove that the problem is compact. So let a be an open covering x cross y. And we have to find the sub-covering. For each x and x, the slice x cross y, so it's now x before it was x0 fixed now. It's also variable, of course. The slice x cross y, which is homeomorphic to y, is compact. A is an open covering of everything. So it covers also this slice, obviously, OK? The slice is compact, and hence, and therefore, covered by finitely many elements in A, OK? And therefore, and hence, covered by finitely many elements A1, AN, whatever, A1 AM in A. A is an open covering of x cross y, but also of the slice, small x cross y, so finitely many suffices to cover. And now we take n, let n, so nx, now I give index x, is equal to A1 union, then nx, A1 union, which depends also on x, of course. Everything depends on x. AN is a neighborhood of the slice, x cross y. That's obviously a neighborhood, OK? There are two plamma now, OK? The two plamma implies that there exists a neighborhood. There is a neighborhood. Wx depends on x, also, everything, OK? Wx of x such that Wx times y is contained in nx. That's exactly the two plamma, OK? We have a slice, we have a neighborhood of a slice, and we find a tube, which is contained in the neighborhood. There's a neighborhood, Wx of x cross y, OK? So that's just a two plamma. But now we have also an open covering of x. That's the main point. Now, and this implies, now we take Wx, x in x, is an open covering of x. But now also x is compact, OK? x is compact. So this implies that, finally, many suffices to cover x, OK? Finally, many. Wx1, Wx. So what was the name for? This was n, here, now. So I call it m, now, OK? This depends, anyway, on x is n, OK? So M, finally, many, Wx1 suffices to cover, so what? So x cross y is x is Wx1 union Wxn m cross y, which is Wx1 cross y, union Wxm cross y, OK? That's the case. By construction, each of these is contained in the union of, finally, many elements of A. When you see the proof, then by construction, each of these is contained in a union of, finally, many elements of A, of our open covering. If you look at the proof, you see exactly this, OK? And this implies, of course, that x cross y is covered by, finally, many, OK? For each of these, we need, finally, many only. There are, finally, many of these. And this implies that x cross y is covered by, finally, many. Well, get smaller and smaller by, finally, many elements of A. Ah, yeah, it is still. One second, I made a mistake, no? Yeah, by notation. nx is neighbor. There's a neighbor Wx of Wx times y is in nx. And now we have Wx cross y1. This, each of these is contained in nx1. So this is, in fact, it's equal, because we anywhere have everything. So in this nx1 union, union nx, nx1, union nxn. And now, each nx is the final union of these. OK, that's the proof, OK? So this is true in general. So this is a famous theorem, Tichonov, Tichonov theorem. An arbitrary product of compact spaces is compact, OK? Obviously, if you write this in the product topology, OK? In the product topology. In general, not in the box topology, OK? In general, not in the box topology, in general. You will see an example, not now, OK? It's not true for the box topology. So this, I will not, we do not prove. It's too long. There are different proofs. I mean, what we want, what would be useful here, what we, that we will do anyway. We will use sequences for compactness, OK? But in general, sequences are not sufficient for compactness, OK, to characterize. Every, that we will see later. Every, what means compact? Any, any sequence has a convergent sub-sequence, OK? That is not the same. This is sequentially compactness, OK? So it's different. It's, it's not as strong as compactness, OK? And so we cannot use sequences. And then to prove Tychonov, one can generalize sequences, OK? And this is called filter. And then something is compact if and only if every filter has a convergent sub-filter, OK? Then one can use it. And then it's a very conceptual proof, OK? Also for Tychonov. We come back to sequences later, not now. And also through some example. That's not true in the box topology, OK? We see some example, not now. Instead, we have to prove some more standard things about compact spaces. Next theorem. There's a proof in the book, by the way, of Tychonov in a later chapter, OK? But this is far too long and one has to concentrate. It's not so pleasant for my taste. So it's not in this chapter, which is the third chapter, 2, 3. Yeah, in chapter 3, no, yes. It's not in this chapter, the proof. Because it's longer, and if I prefer proof, I prefer proof using filter, OK? Which is more conceptual in some sense. One can try to generalize this proof, but that's not so easy. It's not so OK. OK, instead of doing this, we prove another theorem, which is the following. Let x be an order set which has the least upper bound property. Let x be an order set with the least upper bound property. So we don't need the other condition here, OK? It's not a linear continuum, but with the least upper bound. Then in the order topology, each closed interval is compact. In the order topology, which is the only topology which we see here, in the order topology, each closed interval now, closed interval, is compact. So this means real, closed intervals are compact, OK? That's, of course, important. They're connected, they're compact, OK? Both are at the basis of real analysis, not these two facts. One is used in the intermediate value theorem, right? And the other is used in the maximum minimum value theorem, which we see immediately after, OK? So this is in the basis of real analysis. Closed intervals, closed real intervals, are compact and also connected, OK? What? Well, interval. No, the other one are rays. Not bound. Intervals are bounded. Interval is AB. Maybe open, half open, closed, but bounded. Interval is bounded. Otherwise, it's called ray. A infinity is not an interval. That's a ray, OK? And minus infinity, A. That's, we don't call an interval. So let A, it's a proof. So it's a proof that each closed interval is compact. So a closed interval, AB, let A be smaller than B. And we have an open covering, as is usual, AB, an open covering of the closed interval, AB. In the closed interval, OK? I want to remain in this interval, OK? By open sets, well, the subspace topology is generated by intervals again, by open intervals and half open intervals, OK, on one side or by the whole, of course, OK? So an open covering of the closed interval AB. And we have to find, again, a finite sub-colonial. So this is maybe a remark, which I have written here. So I added a closed interval, open covering, in the subspace topology, which, however, is the same as the order topology. That's not always the case, no? In the subspace topology, which, however, is the same as the order topology on AB, OK? Which is the same as the order topology in this interval, OK? That's not always the case. We have one example where this is not true, where the subspace and the order are different, OK? That's the ordered square. The ordered square, the subspace topology, has r cross r ordered. And the subspace and the order topology are different, OK? They're completely different, very different, OK? The interesting example is the ordered square, which is the order topology. The subspace topology is a completely, there are too many open sets, OK? But here it's the same for this interval, as you see. The point is that in this ordered square, the ordered square is not an interval, OK? I mean, in itself it's an interval, but in R2 it's not an interval. They're missing all these stuff, OK? If you have only the ordered square, it's an interval in the order topology, OK? But in R2 it's not an interval, OK? Here it's different because this is an interval, OK? So there are no missing points. That's a big difference between, well, whatever. OK, so what we have to prove. So let's see. So we define C. These are all points Y in AB such that AY is covered by finitely many elements of A. So we take all points Y here such that AY is covered by finitely many elements of open covering, which is always called A. Obviously with A and C. So this is not empty, no? That's A. The point A is covered by one, OK? So A, A. That's an interval, OK, one point. That's OK. So let C be the least upper bound. Let C, small C, be the least upper bound of capital C, OK? Take the least upper bound. Least upper bound property, no? Any, every non-empty bounded is clear here because we are anywhere. Everything is in the interval. Subset has the least upper bound. The smallest upper bound, no? Yeah. So call it C. It's similar to the proof for connectedness, OK? So the first thing what we want to prove is that C is in C, OK? The least upper bound. So two steps. First step, I, we prove the claim is that C is in C. So first claim, C is in C. That means that A, C is covered by finitely many elements. That means A, C is covered by finitely many elements of, that's the first claim. So we prove this first. So choose one element of our covering which contains C. So let, this is called, whatever it is called, I don't find it here. Choose A in A, such that with C in A. So I make again this, this picture, which we did for the, so this is, how is it called? Our space, our set is X, no? X, OK? This is X. That looks like the reals, no? It's not the reals, but it's the other set, OK? And so we have our interval AB. And now we have C in A. We have some C. This is the least upper bound. And we can assume, obviously, that C, so I can, we can assume that this is bigger than A. Otherwise, it's trivial. Because if C is, if C is equal to A, it's covered by finitely many elements of that script. We can assume, clearly we can assume that C is bigger than A. So C is bigger than A. Otherwise, it's trivial, OK? A, A is covered by finitely many, by one, right? Since A, A, that would be the case C equal to A, OK? It's covered by, this is A, no? It's covered by one element, OK? Of A. So this is OK. So we have this situation, again, OK? And now C is in A. A is open, OK? C in A, A open. C is in A. A is open. So what? We have the order topology, right? That means we find something here. On this side, I don't know. It might be B, right? But I don't care about this side. Here, we have larger. Here, we may be equal. Anyway, it's open. So there exists a D, which is D. There is D, smaller than C, such that what we see, DC is contained in the interval DC is contained in this open set, which was called A. OK, right? So this part is contained in A. So now, I say that A D is covered by finitely many elements of A. So why is that? I mean, what was C? That's the least upper bound of all elements, such that the interval is covered by finitely many. So they are actually close to C. Elements here, OK? Such that the interval from here to this element is covered by finitely many, OK? It's the least upper bound. So then A D, clearly, is covered by finitely many, OK? Because they have points arbitrarily close to C, such that the interval is covered by finitely many, OK? The elements arbitrarily close to C, such that the interval from A to this element is covered by finitely many. Then A D is covered by finitely many, OK? So that's by the definition of C. It's covered by finitely many, OK? And then, of course, A C is covered by finitely many elements of A. Why? Well, just as A, one element, OK? We have this part is in A, OK? Just as A, this A. So then we are up to this point, no? And this means C is in C, no? Clearly. And this is C is in C. So this is the first step. The first claim, C is in C, OK? You look at the picture, OK, and you see it works. What did we use? We used only the least upper bound property. Yes, this one, no? A is open, OK? This A is open contains C. We assume that C is bigger than A, OK? Otherwise, it's crazy. So this means open. What means open? Well, you may say that D might be A. That's OK. This D might be equal to A, for example, OK? Maybe you say maybe there are not too many points here, right? But D, in any case, might be A, for example, OK? That's also OK. So it's not problem. We don't use linear continuum. We use only the least upper bound property, OK? We just use the definition, but it's open, OK? A is open, contains a point. So there's some interval, OK, which contains this and is contained, OK? Yeah, some basis element. Of course, this D might be equal to A, OK? No problem. If there are no points here, no? Then D is equal to A, maybe, OK? But that's not a problem. OK, second step, second claim, first claim, second claim. C is in C, and the second claim is C is equal to B. Then AB is covered. This proves the theorem. So this implies AB is covered by, finally, many elements A. So this is the theorem, OK? C equal to B. OK, so suppose C is smaller than B. Suppose, by contradiction, that C is smaller than B. And now we have, again, our picture. Always the same picture, four times, OK? So this is X. This is A. This is B. And now with C smaller than B, OK? So there's some C here, and we are sure that this is we suppose, by contradiction, that is smaller than B, OK? C is smaller than B. I don't care about this side, OK? Here, maybe, equal, I don't care. I'm interested on this side now, OK? But we have the same notation as before. C in A, A open, OK? Let, so it's the same. Let, as before, I write the same. Let C be in A in A. So what do we have? First of all, I lost my, what do you want to prove? I get lost. This is our picture now, of course, no? Now we do the same as we did before on the other side, OK? So A is open, again, right? A is open. What does it mean? It means we find some D on the other side now, OK? So there is D bigger than C now, such that the interval from C to D, this interval. I don't care on the other side, OK? It's contained in A, which is open. And we're looking for a contradiction now, but it's clear a contradiction now. Because now, AC is covered by finitely many, OK? AC is covered by finitely many elements, finitely many, I abbreviate now, I'm always finitely many elements of A. Why? This is the first step, no? So first claim. So AC is OK. But then what about AD? Also by finitely many, no? This implies that also AD is covered by finitely many elements of A. Why? Way up to C, it's OK finitely many, no? Just at A, maybe that's not sufficient, no? Just at A, then we come up to D, but maybe not D itself, OK? And maybe A contains D, then that's it already, OK? And maybe one other element of A which contains D, just at A, and maybe possibly one more element of A which contains D, which contains D. This is if D is not in A, OK? If D is in A, if D is not in A, then it's covered by finitely many elements of AD, OK? But D is bigger than C, right? And that's the contradiction. What was C? C was the least upper bound of all elements such that the interval is covered by finitely many, OK? So this is a contradiction. So definition of C, OK? Definition. So that's the proof. Very similar to the proof connected, OK? These pictures here, no? You look at one side and then you extend a little bit, OK? Corollary. So again, we didn't use the linear continuum. In this proof, we use only these upper bound, OK? Property, so corollary. Each closed interval, I, each closed interval, each closed real interval, AB in the real, is compact. Closed real intervals are compact. That's important, of course, OK, for analysis. And that is what? Well, no, for game, yeah. So this is a better approach, OK, with compactness. We don't use, we prove finitely well, OK, immediately. But we don't use it. So this is a nice example. And in general topology, there are always strange examples, OK? So the ordered square is a strange example, but it's compact, OK? The ordered square, I ordered square. This is a closed interval, OK? This is 0 cross 0, 1 cross 1, OK? It's compact. Why? This has a least upper bound property, no? For connect, it's a linear continuum. Both are linear continuum, this and this, OK? So it's compact. So this nice example from analysis, this strange example from general topology, OK? The ordered square is compact. This is an interval. The space is the ordered square, OK? It's not an interval in R2 with the ordered topology, OK? That's a strange situation, OK? We are considering only the ordered square, I cross I, OK? We are not looking at R cross R. In R cross I, it's not an interval, OK? Because they are missing all these points above and below the ordered square, OK? So that doesn't work here, OK? So here the subspace topology and the ordered topology are different topology, OK? They are not the same topology. The ordered square is a subspace of R cross R. I cross I is a subspace of R cross R with the ordered topology. That's one. And the other one is the ordered topology directly on I cross I. That's the ordered square, OK? These are very different topologies, OK? So we give some implication of this. We give exercise tomorrow that the ordered square, we prove already, you should know that the ordered square has a linear continuum, OK? It has a least upper bound property. You look at first coordinates, then at second coordinates, OK? That we did. It's connected also, no? It's connected. And now it's first countable. That's no problem, OK? It's first countable. That's very easy. RL is first countable, very easy. Both are not matrizable, OK? RL was an exercise, which I will do at the end maybe today, OK? There was an exercise. And here's the same exercise now. This is not matrizable, OK? And it's not so trivial to prove it's not matrizable. So you can use compactness, OK? It's compact, and then we prove using compactness that it's not matrizable. Cannot be matrizable. Because a compact metric space, the reason is easy. The exercise is a compact metric space has a countable basis. But this ordered square does not have a countable basis, OK? That's the idea. That will be the exercise. A compact metric space has an order, OK? First, I go on with it. So Heineborel, you take it as a definition of compactness. But that's not our theorem. It's not our definition. So this is Heineborel. So what does it say? A subset R of RL is compact. A subset R, A, a subset what? A subset A in Rn is compact, if and only if. If and only what? If and only if it's closed and bounded, OK? A is closed and bounded. Bounded? We have to be careful now. Earth? Yeah, which metric? With the Euclidean metric, OK? Is Euclidean or the square metric? The square metric, right? With the Euclidean. Of course, each metric we may cut at one, OK? So each metric is bounded, then, right? No, no. What? It's a vector space, yeah, that's linear. No, no, no. What do you mean equivalent? What is the metric? It's a discrete metric, for example. What metric? Yeah, on Rn. On each set, you have the discrete metric, right? That's generally the discrete topology, you know? There are many metrics, there are many topologies. It's not so clear. So in the end bounded, in the Euclidean or all metrics on Rn, generally the same topology is not true. It's not true. The discrete metric, generally the discrete, the norm, yeah. In the Euclidean, one second, what do I have to say? A is closed and bounded. In the Euclidean are the square metric. Yeah, you have scalar product norm and metric, OK? And it's not always clear. Metric does not come from a norm. A norm does not come in general from a scalar product. So there's this proof of Heine-Bohel. No, it's a theorem. We never use this, OK? There are two directions. One is not very interesting, which one? If it is compact, then it is closed. But that's not interesting, OK? So this should be the easy direction. So A now is compact, OK? We know already that A is compact. And then we have to prove it's closed and bounded. So we have an open covering. A is contained in the union ball. I forgot, what was the Euclidean metric? D distance, maybe, OK? So this is Euclidean metric. 0, n, n in n. That's trivial, right? This is A compact. So this is an open covering, right? Open covering of A. This is an open covering. A is compact means it's contained in finitely many of these. Then takes the maximum, OK? So A is contained in BD 0, n for some n in n. And this, of course, means that B, A is bounded, right? It's in some ball. A is bounded. What else? Closed. Closed? A is compact. A compact, yeah, right. A is closed as a compact subset of a house of space. Of the house of space. So that's what we need, bounded and closed. So this is easy, OK? What is interesting is the other direction, of course. Because here we have to prove it's compact. So A, now A is closed and bounded. This implies closed. Bounded means A is contained in bounded, in some ball here, OK? But now I take a square, a big square, OK? So I take bounded means also it's contained in, so I take minus nn to the n in Rn, OK? This is bounded, OK? It's bounded. This is a square, hypersquare, OK? This is a hypersquare. So if you want a square metric here, you have to do this. And now we use everything which we proved, OK? This is the non-trivial direction. So minus nn in R is compact. That's what we proved. Real intervals, closed intervals, are compact, OK? And this implies that minus nn to the n is compact. What is this? Final product of compact space is compact. The final product of compact space is, what do you want to prove? So A is contained in this, and this is compact. A closed, a closed in a compact is always compact, OK? This implies A is compact. So here we don't need house stuff, OK? This is trivial. The closed subset of a compact space is compact. I mean, the proof is trivial, no? If you have a closed subset of a compact space, if you have an open covering of a closed subset in the whole space, you just add the complement, which is also open. And then you have a final cover of the whole space. And then you have final sub-cover. So the proof is very, so A is compact. And that's it, right? That's what we've proved. The last thing which I prove here is maximum minimum value theorem. And this is also trivial, almost, no? Here we don't use anything. So minimum maximum value theorem. It's analogous to the intermediate value, OK? Which is also, we use only definition. So now we proved Heineborel, no? We never used Heineborel, OK? So to prove that anything compact, OK? We never used it. This is a proof. So maximum and minimum value theorem. And now you see this is a trivial theorem. Minimum value theorem. Very important, but almost trivial in the sense that we use only a definition of compactness. Nothing else. So what does it say? Left f from x, y be continuous. Let it be continuous. Y is ordered in the order topology. Where y, we have to compare elements, maximum, minimum, no order. Where y is ordered with the order topology. If x is compact, then we have to prove first to apply the theorem. We need that x is compact, OK? And that might be the work. If x is compact, then there are points, then there are points cd in x, cd in x, such that f of x, for any point, it's f of c between f of c and f of d, for all x in x. So this means f assumes its maximum and minimum, no, that's in words. So this means that f assumes its maximum. Well, maybe it's stupid. f assumes its maximum and minimum. Maybe it's not useful to write this, but I mean, that seems to be trivial. It's maximum and minimum. Otherwise, it would not be a maximum minimum, but it would be a upper bar. Yeah, right. So that's better. OK, so proof. So x is compact, f is continuous. That means the image, f of x, is compact. The image of a compact set is compact. And this implies, so he calls a, maybe, f of x is compact. This implies that minus infinity a, a in a, is an open covering of a in y. It's an open covering of a in y by open sets. These are open sets, no? In y, rays, not interval, ray, open rays, OK? These are open, and it's an open cover. Why is this? No, no, sorry. Suppose that, I forgot to write something, OK? Suppose that f has no maximum, by contradiction, OK? That I forgot. Suppose that f has no, has not a maximum. Then this is an open cover of a, y. a is f of x, no? This is some f of x, OK? Each a is f of something, but it's not the maximum. So there is some a, another a, which is larger than this, OK, because it's not the maximum. This is some f of x, but this is not the maximum. So there's some f of y, which is larger than this a. So this a is contained in minus infinity, the next one. So we get everything. There's always, these are the values, OK? There's always some value which is larger. So this means this is an open covering of a in y, OK? a is compact. So finally, many suffice. But finally, many, we have the maximum, OK? This means a is contained in minus infinity, a0, some a0, a0, a0 in a. That's a very strange contradiction, no? This is a0, no? This is an a, but a0 is not here. It cannot be equal to one of these, OK? Because we are missing a0. So contradiction. So this is a trivial proof. One minute, so I give the proof that the exercise, OK? That's one of the exercises I want that you know how to do that. It's very easy. So exercise, x is metric, matrizable, and has a countable, then subset, d. Then x has a countable basis. x is first countable, no? That's the same. No, no, second countable. That was an exercise, right? And I know I make it one minute, so the proof is the following. So the countable basis, b, it's bd. So where's the metric? d, oh that's, the metric is d, one second. So countable, then subset, d I don't like, because this d and this d norm, bd, why? No, we need, we cannot ignore, because we have trying inequality later, so we have to write the metric. Then it's not nice to have d of d, so we better change that. y, 1 over n, y in y, n in n. This is countable. This is countable, and this is countable, OK? So it's countable, countable, b is countable. And the claim is that b is the basis for x, OK? So that's what we have to prove. It's certainly countable. The important thing is b is the basis for x. That's what we have to prove, OK? So I make my picture, the usual picture. What do we do? We have some open set, OK, u open. And we have a point, x I don't have, no? x in u, right? And so we have to find the basis element, which contains the point, and it's contained in this open set. Then it's a basis. Then we use a lemma, and says it's a basis. We don't have to prove the conditions here of the basis. We just apply the lemma, OK? If we have this situation that we now find something here which contains this and is contained here, then it's the basis, and it generates a topology. So the lemma says everything we want, OK? We don't have to prove it. Anyway, we have a metric space, OK? So what I get here is let some ball, OK? There is n in n such that bd x 1 over n epsilon, some epsilon, now make it smaller. 1 over n is contained in u. So here the radius is 1 over n. n is fixed now, OK? m, m, m. Here I used n as a variable. So there's m in m such that this, OK? Better change than n. That's a metric topology. Given a point in an open set, we find an open ball of some radius epsilon, which contains the point, the center we can assume, and contained here, and then we can make 1 over m, this epsilon, OK? No, yes. So there's a point here. But I need another step, not immediately. So I go through half the radius now, OK? So I consider, so let I have 1 over m, OK? This is m, 1 over m. I go to 1 over 2m, OK? So let y be in bd x, and I go through half the radius. 1 over 2m. So now we find it's dense, as you say, OK? It's dense, so in each open set we find something, OK? Right? Since it's dense. So there's some y here. This is x, and here's now y. And this is in y, OK? Then the first point is that the distance between these two. So x clearly is in bd y, 1 over 2m. The distance between x and y is smaller than 1 over 2m, OK? By choice, no? That's true. And the second one is the ball d y, 1 over 2m. It's contained in bd x, 1 over m, and so it's in u, OK? And this is the element of our basis b. So this is an element of b, right? So why is this? So now I make my picture, OK? Triangle inequality. Now I have to write the metric anyway, d. So I take the ball. So this is 1 over 2m. So now I have this one, OK? This is the ball around y, OK? And then take an arbitrary point here in this ball, in this one, z. And then what you see, the distance between z and y is smaller than 1 over 2m. The distance between y and x is smaller than 1 over 2m. So the distance between z and x is smaller than 1 over 2m plus 1 over 2m. So it's smaller than 1 over m. And that means this z is in this ball, OK? Should I write? So if you write, then you write in this way. So let z be in ball d y. Of course, you have to write, OK? At this point, you cannot say it's, well, it's almost trivial. But if it's an exercise, you have to write, all right? Cannot say it's clear. Then I say it's not, I don't believe. So then the distance of z and x is smaller equal. So you have to write this, OK, more or less. The distance of z and y plus the distance of y and x, which is smaller than 1 over 2m plus 1 over 2m, which is equal to 1 over m. And this then implies that z is in bd x 1 over m. And that's it. So it's very convenient to start with 1 over m, then go half the radius, 1 over 2m. And then you find the point y, and then it works, OK? So it's short. Of course, the corollary is, that was the exercise. Now, why is this interesting? So corollary, sorry, corollary, that you have RL is not matrizable. Interesting, but first countable, OK? It is first countable. So that cannot be used, but first countable. So this is one of these examples, which is first countable, but not matrizable. RL is not matrizable. Proof, Q is dense in RL. Well, it's dense in R. So it's dense in RL, OK? Q is dense in RL. And RL does not have a countable basis. That you have to know also. Does not have, is not first countable, is not what, is not second countable. That's sufficient, OK? We have something dense that is not second countable. If it would be matrizable, it would be second countable, you know? That's exactly the exercise. So that's a proof that RL is not second countable. And we will do the same next week for the ordered square. The ordered square is also first countable. It's not matrizable. But the proof is different, OK? It's different. And then we have still the third example, which is this proof is still different. These are interesting because you have to work in each example, OK? It's never completely trivial. The last example is even more strange than the ordered square in some sense. No, no, not really. Well, we will see. Anyway, it's finished. Sorry. 10 minutes.