 OK, so let's start. So we have S omega. So this means well-ordered, uncountable, well-ordered, uncountable. And each section of S omega is countable. Every section is countable. So then we said this is the minimal uncountable well-ordered OK, the minimal uncountable well-ordered set. It's also unique, but that's, by the way, general properties of a well-ordered set. We had some general properties of well-ordered set. So each element has an immediate successor, OK? Or it's the last element, OK, the maximum. Or it's the maximum. I'm thinking more of this, OK? Or it's the maximum. If it has the maximum, OK, then, of course, this is the last element, but that's the other one. It has an immediate, why is this? You take everything which is larger, and there's the smallest element, and this is the immediate thing. But not immediate predecessors, OK? As I told you, the well-ordered set starts as a natural number, OK? It's like the natural number. There's a smallest element, OK? Then there's a second element, immediate successor, OK? Then there's a third element, a fourth, and so on, right? Then you go on in this way. So you construct the natural numbers, OK? It starts like the natural numbers. And then you have, as we said, this first omega, OK? If this is not ever, so this is like the natural numbers, right? You count, 1, 2, 3, 4, 5, and so on, OK? And if this is not all, then there's everything which is larger, there's the smallest element. This is omega, OK? This is called omega, just omega. And then you start again and again and again, OK? So you see. And there's no other way. Each well-ordered set has to start in this way, OK? And it goes on in this way. Each element has an immediate successor. The only question, as I said, somebody of you also said, is this is a minimal, uncountable well-ordered set. But what is the cardinality? That's the problem, no? Because there's a continuum, my hypothesis, no? So the minimal uncountable. Countable is the natural numbers. That's easy, OK? The real numbers are uncountable, OK? Right? The real numbers are uncountable. However, that's sort of an embarrassing question for my point of view. The real numbers, no? You take a subset. The real numbers seem something not too difficult, well-known. I don't know how well-known, no? You take any subset, no, of the real numbers. You take a subset. And then you ask, well, this subset might be fine, of course. That's easy. Then it might be countable, no? The rationals are countable, right? The rationals. Then it might be like the reals itself, OK? The whole. This is a continuum, no? But then you ask, is there something in between, OK? So it's larger than countable. It's not countable. But it's smaller than the reals itself, than the continuum, OK? Is there a subset of the reals of this type? Can there be a subset? And then the answer, which is somewhat disturbing, is on the basis of the standard set theory, you cannot answer that question, OK? You can assume that there's nothing. So either it's countable or it's a continuum. That's the continuum hypothesis, a version of the continuum hypothesis. But as well, you can assume that there's something in between, OK? So there's something which is not countable, but it's not as large as the continuum itself. So it's in between, OK? And you can assume both. You cannot decide that question. And that's somewhat disturbing, no? Because you think the real numbers are, maybe it's the wrong question. I don't know. The real number is something familiar, no? We work, OK? But if you ask this kind of question, then you have a problem, OK? So we don't, this is uncountable. But we don't ask, what is it? It's like the reals, OK? The cardinality, OK? The number of answers, the cardinality. It's like the reals, or it's smaller than the reals, no? And that, sorry? R is not the order. R is not what? Well ordered. Well, the standard order on R is not well ordered. But as we said last time, that there is well ordering. So you have, you can, the well ordering theorem says that there is an order on the reals, such that the reals are well ordered also. You can do that. That's not the standard order, of course, OK? It's very, but it exists. This, in turn, depends on, as you say, on the axiom of choice. And then you have, again, the problem, OK? The same problem, no? The axiom of choice. You can say, that's good. That's not so good, OK? And here's a continuum hypothesis. I'm not an expert in logic, OK? So I don't know that. But it's somewhat disturbing that this seems now a rather natural question on subsets of the real numbers now, on cardinality, how many elements, OK? But there you get immediately problems. OK, anyway, that's general problems. Each element has an immediate successor. That's the least upper bound property. That's also the same, no? Why, what is the least upper bound? So if there is an upper bound, you take all upper bounds. And this has the smallest element. So that's the least upper bound. So this is OK, because it's well ordered, no? Well ordered means every non-empty subset has the smallest element, OK? So if it's bounded, some subset is bounded, then you have just all bounds, no? There's non-empty, there's a bound. And then you take the smallest, and that's the least upper bound by definition, OK? So it has the least upper bound property, every well ordered subset. And then we have a lemma also, next small lemma. Now I go to S omega. If A in S omega is countable, then A has an upper bound in S omega. And then also least upper bound, right? But that's OK. So there's an upper bound, and then also least upper bound, of course. So this doesn't look like the real numbers, no? For example, it's a standard order. You can go to infinity in the real numbers. There's no upper bound, OK? The natural numbers have no upper bound. But this is not well ordered, of course. The real numbers with the standard order is not well ordered. But here you go to infinity. Here you cannot go too far, because any countable set you have find an upper bound. The proof is very easy. So I take the union. So A is a countable subset. So I take the section of A, A in A. I take the union of all sections of these elements, OK? That's a section, a theory called section of the elements which are smaller than A, OK? So this is a countable union of countable sets, no? Countable union of this is countable, OK? And each section is countable. Countable union of countable sets. And this means it's countable, OK? So this implies countable. So this is a subset of S omega. S omega is not countable, so they cannot be equal. So it's different here. They're not equal, OK? And now if you think a second, then any element in S omega, any element in S omega which is not here, is an upper bound of A. Every element of S omega minus this union is an upper bound of A. Well, if it would not be an upper bound of A, then there's something larger in A. But then it's somewhere here, OK? Again, if this would not be an upper bound, OK, this element, then there's something in A which is larger than this element, OK? But that means this element is in some S A. So it's here. But we take an element which is here minus this, OK? So any element of this is an upper bound of A. And that's it, that's all. So there is an upper bound. Section means everything which is smaller than this element, smaller. Minus infinity, yeah, but we have well-ordered, so we can, so this is minus infinity A. But this is maybe not the best way to write this in this situation, because we have a smallest element, OK? So it's really alpha A. This is the smallest element, OK? Minus infinity we write for the reals, but here we can take the smallest element. It's well-ordered, so it has a smallest element. That's this one, OK? There is a smallest element. OK, and then we have proposition. Now, this is our example of a house of space. As omega is a house of space, well, it's order. Every order topology is house of space, which is not compact, limit point compact. So this is the first example of a space which is not compact, but it is limit point compact, proof. Well, it is house of, that's an exercise. Every order topology is house of, OK? If you think of house of, that was an exercise, probably, no? If you think house of, what means house of? House of means that, that's an exercise. Now, house of means any two points, OK? You could compare, right? This is larger than this, let's say. So this is an order set. And then we have to define disjoint neighborhoods. And so there are two cases. You have to distinguish two cases. One case is there's an element in between, OK? There's an element between these two. Then what you can do is you take this, OK? So let me call A, B. I'm doing an exercise, OK? So it says C between. Then what are the disjoint neighborhoods? They are B minus infinity A. Now I write minus infinity, no? It's not, it's ordered. Well, we don't care about that. And, no, sorry. C, yes. And C infinity. These are disjoint neighborhoods, OK? And the second case is there's nothing in between here. We have just A and B. And this is, there's no other element. And then, of course, you take this and this. So you take minus infinity B and A infinity. The intersection of these is the interval from A to B, open interval. But this is empty in this case, OK? These are the two cases. It's not empty, it's empty, OK? So this is here empty, this interval here. So it's household. Any order topology is household. That's the proof, OK? Order topology is household. There's two cases given two elements. There's something in between and there's nothing in between. And then it's convenient to write it in this way, OK? That's the exercise. Three line. Well, that was not the point. The point is it's not compact. That's still not the point. It doesn't look too compact, right? Because a huge set of space, order topology, it's very large. So not compact. So this is household. That was the exercise, OK? They're not compact, not compact. So this is not so interesting, not compact. Not compact means you have to find one open covering, just one open covering. What we take as an open covering, we take S a, all sections, a in a, a in S omega. Or S x, let me call it S x, x in S omega. So again, you say this is minus infinity x. But in this case, it's alpha x, no? Because we have the smallest element, right? Sorry? No, no, one second. Omega is not in S omega, OK? There's no omega in S omega. So this is, there's no omega in S omega. This seems a section of omega. But where in S omega bar, no? Then we have S omega bar. And that's S omega, and you add omega. This is with omega, OK? It's like the section of omega somewhere, where? Maybe here. This is the largest, OK? So it's a section. Now it's a section. But omega is not here. So it's an open covering. Why is it an open covering? That means each element is somewhere here, OK? So take any element. Then you have to find something which is larger. So it's in the section, OK? Since there's not the largest element, since S omega has no largest element. Why doesn't it have the largest element S omega? Otherwise, it would be countable. Because any section, the closed section, is countable also, OK? Otherwise, it would be countable. Otherwise, it would be a closed section, not open. These are open sections, OK? But closed means one more element. Doesn't matter, OK? Otherwise, it would be a closed section. These are open sections, OK? And hence countable. It's not countable. And therefore countable. And thus, therefore countable. That's not possible. So this is an open covering. So if it would be compact, OK? If S omega would be compact. No, no. How to write it in the best way? So if there is a finite sub, there is no finite sub-covering, OK? Why not? Since otherwise, if there would be a finite sub-covering, there would be many of these X, OK? But then with the largest, then it would be a section, an open section even, OK? And that's countable. But for many, since otherwise, it would be an even open section. You don't need countable even. There's also this strange contradiction now. Otherwise, it would be an open section S A, OK? Equal S omega. But A is not here, no? But A, of course, is a section of S omega, OK? So we have a very strange contradiction. It would be an open section. And countable, of course. And also countable. That's another contradiction. So this is, OK, that's not compact, OK? That's easy. Non-compact is easy. You find in the examples, you find an open covering which has no finite sub-covering, OK? So the intro, it looks not very compact, no? It's too large. But it's limit-point compact. So S omega is limit-point compact. That's the last. What? The construction of S omega is uncountable, well-ordered. And every section is countable. That's the construction of S omega. These are the properties of S omega, which, OK? But it's easy now to construct. I mean, if you have a well-ordered zero. Because if there's an uncountable section, OK, you take Z. If this nu has an uncountable section, you go smaller and smaller, OK? So you take all points such that the section is uncountable, right? If this is empty, that's it. Then all sections are countable, right? If this is not empty, it has the smallest element. So there's the smallest element such that the section is uncountable, OK? And then you take this set, this section. So every section after this is countable, OK? That's the construction of S omega. These are the properties. It's uncountable, well-ordered, every section is countable. But it's easy to construct. Once we have an uncountable, well-ordered set, you immediately have something like this, OK? Either it's already true that every section is countable, then that's it, OK? If not every section is countable, then you take all elements such that the section is uncountable. This is not empty, then, and it has the smallest element. So you take that section. This is uncountable. But every section which is smaller is countable now, because this is the smallest where it's uncountable. That's the construction of S omega. Last time, OK? That's what I wrote last time. These are the properties. The important properties of S omega is the minimal well-ordered uncountable set, OK? And it's even unique up to bijections which preserve order. It looks unique if you look at the picture. Every well-ordered set starts in the same way, OK? So there should be some uniqueness. It's not trivial to prove uniqueness, but it's unique. So that means it's a very canonical object. It's not casual in any sense, OK? Casual. It's very canonical. It's unique. If you say we want the minimal, in some sense, uncountable well-ordered set, then that's unique, OK? It exists and it's unique. So it's very canonical. It's very something which is present, OK? You have to talk of well-ordered sets. That's all, OK? You have to know what is well-ordered, OK? And then you have this object. And it's not casual in any sense, OK? It's very canonical. In the same sense, the natural numbers, what are they? They are the minimal well-ordered set which is countable, but each section is finite. Yeah, exactly. That's the same, OK? The natural numbers are the unique well-ordered set which is not finite, countable, which is countable, OK? But every section of the natural numbers is finite. So here you go from finite to countable. And here with S omega, you go from countable to uncountable, OK? So it's a similar step, except that the natural numbers are easy in some sense. And this step is difficult, so you don't have to ask what is the culinality. Then you get into this logical problem. Is that clear? The natural numbers is the minimal infinite well-ordered set such that each section is finite, or the minimal countable, OK? And here we have one step further. Now we have the minimal uncountable such that each section is countable, OK? The natural numbers are a very natural object. This seems a little bit strange, but it's still, it's no choices, OK? So now we come back to topology. S is limit point compact, OK? And that's a nice proof. So what is limit point compact? Limit point compact means that any infinite subset, no, let A be an infinite subset of S omega. And we have to find the limit point, right? We can assume that A is countable. If we find a limit point for countable, then we find a limit point also for, OK? We can assume, so this is a, we can assume that A is countable. If we find in this situation a limit point, then we find in the general situation, each uncountable set contains countable subsets, OK? Choose a countable subset of A. Choose any countable subset of A, countable subset of A, OK? It's infinite, so you take, it's not empty, you take one element, OK? Then you take the complement, you take a second element, a third element, no. OK, then you ask, do we need here the axiom of choice in such arguments? You never asked this in general, no? But now we talked about axiom of choice. So it's not so clear, OK? Sometimes you can avoid in some, this one trick, OK? Sometimes you cannot avoid, OK? So don't care about this, OK? So what? Here's, where's the lemma? The lemma, what does it say? A is countable now, quiz countable, OK? So A has an upper bound, has an upper bound in, let's call it B in S omega. It's not omega, omega is not there, no? Omega is not in S omega. But the lemma says it has an upper bound, right? That's what is in the lemma. So this means that A is contained where in alpha. This is always the smallest element, alpha omega, smallest element of S omega, I call alpha. I don't know what is the standard. And then you go up to what B? This is an upper bound, OK? What did we say? That S omega has the least upper bound property, no? It's well ordered. So S omega has the least upper bound. Since it is well ordered, S omega has the least upper bound property. And what does it mean? That this is compact. Closed intervals are compact. Alpha B is compact. Now this is a theorem, closed intervals, if you have the least upper bound property. That was the theorem on compact spaces, no? So it's compact. But then it's limit point compact. So this means A B alpha B is limit point compact. A compact space is limit point compact. So this means A is a subset. Now A is an infinite subset, OK? This is countable. This is infinite. It's countable here, but it doesn't matter, OK? Countable infinite. So A is limit point. In alpha B, this is limit point compact. And hence, in S omega, we are in S omega, OK? It has a limit point here, but this is in S omega. And hence, in S omega. That's where we are anyway. So it has a limit point in S omega. This upper bound of A, yes, probably, one second. Yeah, that's a good question. So the upper bound not, but you have to the least upper bound, OK? We have the least upper bound property anyway, OK? So that one second, one second, that would be. So let A be an infinite subset. We can assume that A is countable. So there's an upper bound. That's a lemma, OK? But we are in a well-ordered set. So there's a least upper bound. If we have an upper bound, we have a least upper bound, OK? So you say, if this is the least upper bound of A, then it should be a limit point of A, right? So let's see the situation. So this is a proof, OK? This is a proof. But this might look easy at first thought. I don't still see where. Where should I? Well, this. So what you say is, we have this A now with alpha. Then we have A, some infinite subset now. So we have an upper bound, B. That was our B, an upper bound. But then we have a least upper bound, OK? So we have a least upper bound of A. So that means we're coming somewhat close to this, OK? With A, which is infinite. So there's a least upper bound. And then you say, so we have to construct a limit point of A, no? That's what we want. What? No, no, no, no. By definition of limit point. Forget about matrizal. So what it means? We take any neighborhood of B now, OK? Limit point means any neighborhood of B intersects A, OK? Take any neighborhood. What is the neighborhood of B? The neighborhood of B, we are in the order topology. So it's some open interval, OK? On the other side, I don't care too much. But that's something like an open interval, right? And now you say that there must be points here. Because if there would be no points here in the intersection, then this was not the least upper bound, OK? Then this would be the least upper bound. So this least upper bound, you say then B is the limit point of A. And that's done then, if this is true, OK? Right? Yes. So let's compare the two proofs. I don't see a mistake. This is OK, no? That seems OK. I don't see a... Well, it's not exactly the same. This is easier, no? Because what you hear, you have an upper bound. Then you hear the closed interval, right? And you apply the theorem, OK? You apply this least upper bound property, of course. But in an indirect way, no? Here you apply directly, OK? Here you first indirectly say if it has the least upper bound property, then closed intervals are compact. And that's a theorem, OK? That's not a triviality, OK? And then you say, well, if it's compact, it's limit point compact. That's another... That's not difficult, but still you use something, OK? And then you finally have a limit point, OK? So you say you take immediately the least upper bound of A, and this is a limit point, then, yeah. That seems to be easier. Now, I would not now, tomorrow, have a look at the book, how he presents the proof, OK? Yeah, of course. Yeah, there should be a proof. That S omega is limit point compact, yes. So tomorrow, I will have a look at the book now, and we do that tomorrow, yeah. But I don't see... So that was a good remark, OK? Yeah, you take the least upper bound, using the least upper bound property, which we have anyway, and this is a limit point. So we have a limit point of A, OK? And this is too complicated, sir. If this is true, this is too complicated, OK? Because you use first the theorem, and then a compact in plus limit point, and this is very indirect, OK? So tomorrow, I want to see the book, OK? First, what he says. Good, nice, OK. So that's it. S omega bar. So this finishes, I don't know, the corollary, of course, corollary, S omega is not matrizable. Why not? Of compactness and limit point compactness. If it would be matrizable, compactness, limit point compactness, sequential compactness are all equivalent. But it is not limit point compact, it's compact. So it's not matrizable. That's a nice proof that it's not matrizable, right? S omega bar, so this is alpha omega. Now we have omega, right? Now omega is there by definition, OK? It takes the largest element, OK? That's like in the next few numbers, you take this. It's outside. What? Yes, it's outside, that's omega. Because otherwise, it would be countable, no? If it's in S omega, yeah, it's outside. But this is a closed interval. So this is compact. This is compact. So closed interval. This is a closed interval. So this means S omega bar is compact. Then we have to use this theorem, OK? Which we use here already, OK? S omega bar is compact, yes. But it's not matrizable also. I mean, this is sort of difficult. Well, of course, this is not matrizable because this is not matrizable, right? So it's interesting now to compare. This is a subset of S omega is not matrizable, no? So S omega bar is not matrizable. Because the subset of a matrizable space is matrizable, right? It's the same metric. The subspace of a matrizable space is matrizable. You take the same metric, OK? That's restrict to the subspace. So it's compact. But here it's much easier to prove that it's not matrizable. This is easier. We don't need S omega. Why is it not matrizable? Well, S omega bar is not first countable. S omega bar is not first countable. In which point? Ah, that's another thing which I proved. S omega is first countable, OK? Sorry. Yeah, that's what we'll do next, so. It's not first countable in one point, in omega, OK? In omega. OK, why is that? So what means first countable? First countable means we need a fundamental system of neighborhoods that's countable. And then we can take intervals, of course, OK? We can take intervals which finish in omega, OK? This kind of fundamental system of neighborhoods, otherwise, it's not first countable. Otherwise, there is a countable system of neighborhoods, a countable fundamental system of neighborhoods. And I can make it smaller, these neighborhoods, no? So I can assume that I finish with the largest element. Anyway, I have to finish with this, OK? And then you have some xn, OK? So open intervals, not half open intervals, OK? This is the largest element now, OK? The order topology, we have to take these, OK? So we get this, but this is countable, OK? So this xn, n in n, has a upper bound again, OK? So xn is countable. Of course, it's countable. What does it mean? It has an upper bound again, OK? Countable subsets have an upper bound. It has an upper bound in, sorry, xn, where I want this xn in s omega, OK? Not equal to this one. That's the only one, OK? So this is in s omega, not s omega bar, the only element. It has an upper bound in s omega. How to call this? x in s omega. That's always the same lemma, OK? It has an upper bound x in s omega. What do I want to prove? I want, otherwise, there is a countable. So now I take, I get lost. That's omega bar is not first countable, in omega, in omega. Otherwise, there is a countable system of, otherwise, there would be, of course, a countable fundamental system of neighborhood, xn omega, and then that's omega. So I have a neighborhood x omega. This is a neighborhood of omega. Upper bound x larger than xn for all n, OK? It might be one of these, OK? But even in this case, no, no, I take it larger than xn, OK? If it's one of these, then take the next one, OK? So you find an upper bound, which is, because it's a neighborhood of, which does not, this is a neighborhood, no? And which does not contain any of these, OK? Which does not contain any. Any neighborhood of the sequence, which is xn omega. The x is larger than xn, OK? I want x larger than all these xn. And then this neighborhood does not contain any of these, OK? So it's not, that's a contradiction. That's not the fundamental system of neighborhood, OK? I found a neighborhood which does not contain any of these, OK? So it's not first countable. It has an upper bound x, larger xn, so also without parentheses here, in s omega. Then, and this implies that x omega is a neighborhood of omega, which does not contain any neighborhood. So this is a contradiction, because here I wrote countable fundamental system of neighborhoods, OK? So here we have a contradiction, no. It is not countable. It is not fundamental, OK? This xn is in s omega, right? The only point in s omega bar, which is not in s omega, is this omega. So this xn is in s omega, OK? And the lemma says if you have a countable subset of s omega, then this countable subset of s omega has a upper bound in s omega. That's the lemma for s omega. For s omega bar, it's trivial. We have the maximum element of omega, OK? Look at the lemma. The lemma says if you have a countable subset of s omega, then this has an upper bound in s omega, in s omega. S omega bar is the maximum, OK? That's trivial. But we are in s omega, no? Yeah, that's a good point, anyway, OK? We are in s omega here, and the upper bound, let's try it in s omega. The upper bound is in s omega, not in s omega bar. In s omega bar, we take omega, no? That's the upper bound. That's the maximum of it, OK? So what? It's not first countable in this point, OK? It's not first countable, so it's not metrizable. So this implies then s omega bar is not metrizable. But this is easier, OK? We don't use theorem here, OK? We show it's not first countable, OK? It's not metrizable. Also, since s omega is not metrizable. But this is much more difficult, OK? Also, since s omega is not metrizable. But this is more difficult because we are using the theorem, OK, that for metrizable spaces compactness, limit point compactness, and sequential compactness, I equivalent, right? This is not too difficult. But this uses the back number lemma, no? And one has to prove something, OK? That's a nice theorem. And now you say maybe s omega itself is not first countable, OK? So that's the last thing, no? That you have to know. That's like RL, OK? So proposition, honest, I forgot to do this. That's a nice exercise, also. S omega is first countable, but not second countable. No, no, no, no. No, no, you need something more. You need the dense, or you need compactness, or you need whatever you need, OK? The three proofs for s omega, for RL, and for the ordered square are all different, OK? You use all this other stuff, no? These are three interesting proofs, no? These are the main examples, also. So it's first countable, but not second countable. Proof, similar, very similar as for RL, OK? Similar as for, but I will do this because RL is the same. It's first countable, but not second countable, OK? That you have to know, also. So I give this as the last thing. So first countable, that's easier, maybe. First countable, let's prove this. So first countable is a local condition, no? In any point, OK? So let x be in s omega. And we have to find the, let a be in s, I don't know, a be in s omega, OK? And we have to find the countable fundamental system of neighborhoods, for a. So what do you take? That should be, in RL what you take? In RL you take the point, and then the point plus 1 over n. You go smaller and smaller, no? You remember that, OK? In RL, you take the point, and you take plus 1 over n, OK? Then it comes smaller and smaller, and this is a fundamental system of neighborhoods. So here you don't have plus 1 over n, no? We are not in the real, we don't have a sum here, OK? So what we take? We take, we have a, we have a prime. This is immediate successor, OK? Immediate, I call it immediate successor of a, which exists, OK? Each element has an immediate successor, but not an immediate predecessor, in general, OK? Successor, yes. And then I just take everything, OK? Here, it should contain a, a neighborhood of a. So I take x here, x smaller than a. So I suppose there's one problem. If this might be the smallest element, no? But then that is open anyway, OK? Suppose we can assume that it's not the smallest element. Otherwise, we don't have anything here, right? If this is the smallest element, then we can assume that a is not the smallest element of S omega. If it's the smallest element, what is the fundamental system of neighborhood? That's, no, no, no, no, it's much easier. If you have the smallest element here, right? Alpha of S omega, no? It's open. It's open. It's discreet in this point, OK? So the next point, you have these next points, OK? So what is enabled? It's this one that contains only alpha. It's discreet. It looks discreet here, the space. It's very discreet, OK? The first point where it's not discreet, where is it? At this point, small only. It's not discreet, OK? Because here you have neighborhood, and then you have infinitely many, OK? But this is the first limit point of S omega, this one. So we can assume the immediate section. This is a fundamental system, a fundamental countable system. Why is it countable? Well, it's in the section, no? This means x is in the section of A. This is countable. The section of A is countable. So it's a fundamental countable system of neighborhoods of, what is our point, A? Well, you take any neighborhood of A, then you find an interval. Because it's the order topology, right? You find an open interval. And then what you do is you just take the right border, you take A prime, which is the smallest possible, OK? And on the left, you have something anyway. You get everything here anyway. So this is a fundamental system of neighborhoods. That's all. First countable, not second countable. Then the problem is, if A is the smallest element, then you don't have anything here, OK? It's all empty. You have A prime, and the only element which is smaller. But I wrote an open interval here, OK? I wrote the open. I have to write open intervals here, OK? Then there's nothing. It's empty, OK? There's no law. So this case I have to do separately, OK? But this is a very trivial case. If it's the smallest element, you need one neighborhood only. And that's the element itself, OK? That's the fundamental system of neighborhoods. Countable, finite, OK? Yes. So not second countable. Let B be any basis of S omega, no? And the proof is not countable. That's what you have to prove. It's not second countable. Yes, OK? No countable basis. So what? For each x in S omega, x is in some interval, x prime. What is this? x prime is obviously immediate successor, OK? The next one. Immediate successor. That's the next one. And here it takes the first element, alpha. Smallest element of S omega, alpha. Yeah, it cannot take alpha, OK? Because that's too large on this side, OK? You have to take everything between, OK? But here it takes the smallest element of S omega. So this is the neighborhood of x. This is open, no? Smallest element. So this is the neighborhood of x. So B is the basis. What does it mean? There exists. So this means there exists. There is a Bx in B. Well, this is a typical situation. We have an open set, we have a point, we have the basis. So such that x is in Bx and contained in alpha, x prime. That's basis, OK? Since it's a basis, we don't know this basis, OK? We don't know anything about this basis. But it is the basis. So if you have an open set, a point, you find the basis element which contains the point and is contained in the open set, as usual. So then we have a map S omega to B. x goes to Bx. For each x, you choose such a Bx, no? Again, I'm getting nervous, no? Then you ask X of choice here again. In general, you don't care, no? But now we talked about X of choice. So this means for each x, choose Bx, OK? We find, for each x, we find the Bx, OK? And so you have to choose for all. Yeah, I suppose you cannot avoid X of choice, OK? You have to use it. But as usual, we never think in mathematics, except in this week or last week, because we are talking about these logical stuff. So it's injective. This is injective, this method, OK? Then this is not countable. This is not countable, right? This implies B is uncountable. It's not countable. It's uncountable. Why? Since S omega is uncountable, of course. Then you have to prove it's injective. Of course, that requires. No, no, no, no, no, no. So the condition is this one. And if you choose Bx in this way for all x, then this function is injective. Phi. From S, phi is injective, OK? So I have to prove that. And if this is an exercise in the test, then you have to prove it, of course. So suppose x is different from y, OK? Let x different y, xy in S omega, right? Take two points, different points. So we have to prove that Bx and By are different, no? Suppose it's ordered, no? Linearly ordered. So suppose x is smaller than y. Doesn't matter which one, OK? So we have Bx and we have By. And we have to prove they are not equal, right? So Bx is contained where? Alpha x prime. By is contained where? Alpha y prime, right? So now x is smaller than y, means that y. So it's clear that y is here, no? OK, y is here because y is an element that's enabled of y, OK? And now y is not here. Why not? Because y is not here. y is bigger than y is larger than x. So it may be equal to x prime. That's the next one. But then it's not here. Or it's larger, still larger, OK? So in any case, it's not here. Is it OK? Yes? Because this is the next element, OK? x, sorry. y is larger than x. The next element is x prime, OK? So y might be equal to x prime. But then it's still not here, no? Because x prime, OK? So this implies that they are different. But it's similar as for RL. The same proof, no? You should see the proof for RL also. It's similar, very similar. So that's it for S omega. So our main spaces and R, which I wrote that, main examples where there's RL, then there's R something, Rj, with various topologies, no, R omega. Then there's the ordered square. And finally, we have S omega now also, OK? These are the four main examples of. And they are all a little bit. This is the most natural with the product, OK? This here, the others, F. But you can read the introduction of the book. And then he comments on that, OK? And he uses this S omega. This S omega and S omega bar can be used for many counter examples in general topology, OK? So not only what we did here, OK? Here's an example of a space, but it can be used for many things, OK? This S omega. Sorry? It is? It's four years. S omega? Well, it's there. I mean, it's an object, which is there. You cannot, you can ignore. And you can say, I don't like that, ignore it. But it's there anyway, OK? It exists. And it's canonical. In some sense, it's natural. In some other sense, it's not so natural. But I mean, all this was introduced into mathematics or logic by whom? Who's the name? This kind of cardinalities and canto, yes. Canto. What? The first name is canto, OK? And when he introduced, he proved. I don't know. He proved that the real numbers are uncountable. That's the irrational number, the transcendental, the algebraic numbers are countable, OK? Algebraic numbers are countable. So non-algebraic numbers are uncountable again, OK? Because it's a complement of a countable set, and so on, and so on. But when he introduced these concepts, OK, the other mathematicians were not, they said he's a little bit, but that's a typical thing, OK, in science, no? If there's a school, and all people doing some kind of mathematics, and that somebody comes and does something strange, no? Something new, something, and then the others don't like that, of course, no? Because the real mathematics or whatever physics, that's what they are doing, not somebody who comes. And very often, there are people who do strange things, OK? And they go away, and people will forget again. But not always, no? They are always these people who have new ideas, no? And then the people never assess, no? He's not, he's out of mind, OK? He's not too serious, but, ah? Kantor? No, Kantor, I'm not so sure. Girdel, yes. Many, in logic, it's a difficult field, Girdel. Kantor, I'm not so sure. Girdel, yeah, he had some problems, I guess. So it's also interesting. So that's it. But you should know a little bit S omega, no? Because maybe I give some easy thing in a test, OK? Prove that S omega is not second count, for example, no? That's not so difficult, OK? And also for RL, the same, OK? So you should have, you know, the proper, some of the main properties of these four spaces, right? OK, so now we finish the concept of apology. I'm going to give some definitions now and prove something tomorrow. So this is the last what we will doing, the separation axioms. So now I guess this is now more standard than S omega. So we know already two. If you have two points, then there is T1, right? What do you mean T1? T1 each has a neighborhood which does not contain the other. That's T1. Each has a neighborhood which does not contain the other. This is the equivalent that points are closed, OK? So if and only if points are closed, that's T1. Then you have the famous T2, what is this? This is Hausdorff, no? So you say Hausdorff in general. That means any two points have disjoint neighborhoods. And now you have T3. So now you take a point and a closed subset. You replace by closed subset, please. So you take a point and a closed subset. So closed subset, not containing the point. Then we have disjoint neighborhoods. This is T3. So this T3 says that points and closed subsets, a point and a closed subset, not containing the point, have disjoint neighborhoods. Well, maybe I don't write, OK? So you have a point, you have a closed subset which does not contain the point. Then they have disjoint neighborhoods. And then we have finally two closed subsets, disjoint closed subsets, disjoint closed subsets, have disjoint neighborhoods. This is T4. Well, one second, I can't immediately do a regular normal. So we want a hierarchy, no? But then this does not imply this. Well, it's not, of course, you have to find an example. Because if you want to apply, one second, you have an example. If you want to apply this to points, you have to know that points are closed, OK? You have to know that points are closed. And this is T1. So in general, T4 does not imply T3, does not imply T2. Well, T2 implies T1, of course. Yes, but not now. We will see tomorrow. So this is not so good, OK? So what means a regular? A regular means T3 and, well, whatever you want, T1 or T2. That doesn't matter. And T1. Then we have to do also, OK? And normal is T4 and T1. And now we have a hierarchy, OK? Then if we define this, so a regular normal, then a normal implies a regular, implies our stuff, implies T1, OK? So we have this T1, T4. And then we have normal is T3 and T1, a regular, and normal is T4 and T1, OK? And then we have this hierarchy. Normal is the strongest, then regular, then, but we cannot write it for T4 and T3. That doesn't work, OK? And now to convince you that these are very, this is an important condition, OK? Why? So I give you a theorem. And that makes immediately clear that this is an important condition, theorem. X metric implies X normal. A metric space is normal, OK? Well, then we get metrizability, no? Metrizability. We need first count of it, OK? But we need normal. So normal is what's necessary, condition for metrizability. The proof is easy. So we have the normal. House of is OK. That's what you have to know, OK? So house of or T2, T1, well, or T2, right? The house of we have, OK? House of is OK. You take two different points. You take balls, no? What radius? Half the distance, right? You have two different points. You have the distance. You take half the distance, and the balls. You have half the distance at the short, around these two points, OK? As you see in the picture immediately. So that's, I don't know, and then we need T4. So for T4, we need disjoint open subsets. Let A and B be disjoint closed subsets of what? What is our span, x? Let A and B be disjoint closed subsets, OK? Maybe I should use the notion here. For each A and A, yes, A and B. For each A and A. So A is not in B. B is closed. So there's an open ball which contains A and does not intersect B, OK? There's an open ball B, D, A, with some radius, epsilon A. If it depends on A, OK? There's an open ball disjoint from B. B is closed. So the complement is open. The complement contains A. So there's an open ball which contains A and is contained in the complement of B, OK? For each A, there is an open ball disjoint from B. For each B, in B, there is an open ball B, D, B, epsilon B, disjoint from A, because also A is closed. So the complement is open. And now it takes a union of all these. So A is the union of all these B, D, A, epsilon A, A and A. And B, I take the union B in B, B, D, B, epsilon B. This is the neighborhood of A, obviously, no? Neighborhood of? This is the neighborhood of B, clearly, right? We have all points of A and B. Sorry? Ah, sorry, sorry, sorry, sorry. Thank you. So what are the disjoint neighborhoods? I don't know, U and V. So U is the neighborhood of A and V is the neighborhood of B, OK? Yes, U and V, not A and B. But we need disjoint. They're not disjoint, maybe, these. So I make a little trick, I write over 2, OK? And then they are disjoint, now they are disjoint. So this implies that U intersection V is empty. So these are the disjoint neighborhoods, OK? So we have to prove this. If there is a point x in U intersection V, suppose there is a point in the intersection, OK? Then x is in sum, then x is in sum B, D, A, epsilon A over 2. It's a union of, it must be somewhere here, no? And also x is in sum B, D, B, epsilon B over 2. So suppose let, we have to compare these. Suppose epsilon A is bigger than equal to epsilon B. I don't know which one, but it doesn't matter, OK? Suppose epsilon A is bigger than epsilon B. So I'm tired, and so the distance between, the distance between A and B, the distance between A and B. So these are fixed points now. If you want A0 and B0, OK? Epsilon A0, but there are too many indices here. One second, one second. What? We have here B, B, A, epsilon over 2 is a complement of B. Say again, B? B, B. This one? Yes. Of B? Yes. Yes, of B, yes. And B, but the company has made it a set. Let me finish the proof first. So DAB is smaller than equal to. So we have a common point here, x, OK? DAX plus DXB, no? This is smaller than AX. What is AX? Epsilon A0 over 2 plus XB epsilon B0 over 2, no? This is smaller or equal to epsilon A0. And what does it mean now? This implies that B is in BD0. So this means B, intersection BD, A, epsilon A0 is not empty, no? This is a contradiction. We have through this, this point, OK? So that's the proof that the intersection is empty. I don't know. You say there's this is complement of B, yes? Yes. The complement of A. What do you say again? So this is clear. BD, A, epsilon A, is contained in the complement of B, no? That's true, yes. B is contained in the complement of A. Yeah, that's by construction, right? Yes? Here we take over 2. Yes. And now? Intersection is empty. Which intersection is empty? These two is empty? That's not clear, no? No, this one is very clear. It would be empty, yes, but it's not clear. You have to go to the half radius, OK? That's the trick. It's not clear, OK? That's the proof, OK? The time is finished. Yes. So the next thing is also compact spaces are normal, OK? So matrix spaces are normal, compact spaces are normal. So it's clear that normal is something important, OK? Interesting condition, no? Compact spaces are normal. Compact household spaces are normal, OK? And matrix spaces are normal. So it's an important condition, normal, being normal. So tomorrow we discuss normal spaces. And then it's finished, the first part. Concept of positive, OK? Tomorrow we finish with the first part.