 So I have to come back to this point, yes, okay? So we have infinite subset, okay? And we want to find the limit point, right? Infinite subset, limit point. So we can assume that this is countable, okay? And then we have a upper bound, that's a lemma, okay? If it's countable, it's a upper bound. And then you say, or we can say, we take the least upper bound, of course, okay? Then there's a least upper bound. But this least upper bound might be an isolated point, might be a maximum of A, okay? It might be a maximum, an isolated maximum. So it's not necessarily a limit point, this least upper bound, okay? The least upper bound may be a maximum. If it's a maximum, you cannot conclude anything, okay? However, you can choose this countable subset in a specific way, okay? And then it works. So what you do is, so this is infinite. I take the smallest element, okay? We have a well-ordered set, no? There's always the smallest element, right? So A1 is the smallest element of A, okay? Then I take A minus A1, there's the smallest element. So there's the next smallest element, okay? Right, this is next. So then A2. So this is the next smallest element, it is, no? Next smallest in A, it's the smallest of A minus A1, no? And then you go on in this way. So what you construct in this way, you construct the countable subset, a very special one, okay? So this is a countable subset A1. So what you get is a countable subset of A, no? In a very specific way. Now this countable subset has a upper bound, any countable subset has a upper bound, then it has a least upper bound, because it's well-ordered, okay? And this least upper bound now is a limit point, okay? This is, it's increasing sequence, okay? And now if you take the least upper bound, then it's a limit point. In the other case, if you take an arbitrary and you take upper bound, the least upper bound, this might be a maximum, okay? And this might be isolated, this maximum, right? I mean, you have some infinite set, and then you just add one set somewhere, okay? Some point somewhere, this is the maximum, right? But it's isolated, okay? So if there's this isolated maximum, it doesn't work, right? But if you construct in this way, then the least upper bound will be a limit point, okay? This is the sequence then which converges against this least upper bound, okay? So you can, it's more or less true, but you have to modify in some way, okay? In this way you avoid, I looked in the book now, and in the book he uses compactness, okay? He gets a closed interval with the smallest element and the upper bound, and then it's compact. So you use compactness, and then it's limit point compact, and then you find the limit point, okay? But this also works in some way. That yesterday, okay? Because if you, it's not correct if you just take the least upper bound, okay? If this is the maximum, then it doesn't work. Anyway, this in some sense, this can be. So another point is a theorem which we know more or less the proof already. If X is Hausdorff, no, not Hausdorff, if X is compact Hausdorff, yeah, compact Hausdorff implies X is normal. The last thing yesterday was metric spaces, matrizal spaces are normal, okay? And compact Hausdorff spaces are normal. So the proof we know, part of the proof, and the other, so first we prove X is regular, okay? First, we prove that X is regular, and that is something which we saw in a similar, we prove that X is regular. Well, sorry, G3 is sufficient, no, Hausdorff anywhere is G3, okay? And this implies that regular. Well, what is regular? Regular is G3 and Hausdorff, no? G3 and G2 or G3 and G1. So what is G3? G3 is, we have a point, and we have a closed subset. We have a point, and we have a closed subset which doesn't contain the point, and we have to find disjoint neighbors, right? That's T3. So what do we do? We have, this Hausdorff here is important now, because it's Hausdorff, so take any point here, okay? And then, this is fixed. Take any point here, we have disjoint neighbors, right? So we have disjoint neighbors, this and this. Hausdorff, we need Hausdorff. So we get, so this is a point, if this is a point X, this is fixed, so this is UX and VX, disjoint neighbors, okay? And then what we do, we, so this is our set. This is closed, no? But we are in a compact set now, so it's compact. Closed in a compact is compact, no? We don't even need Hausdorff here, this, okay? So it's compact. And so we have an open covering, no? Of how to call this, closed compact. So this is A, whatever, okay? So we take all VX, X and X, X in A, what is this? Open covering of A in, in the whole space, no? Which is X. And this means then that A is contained in finitely many, because A is compact, in finitely many of these. So VX1, union, union VXN. And this you call V, what is this neighborhood of A, okay? And then you take UX1, intersection, intersection UXN. And this is you. This is the neighborhood of, or point, which has no name. And these are disjoint, okay? So this implies X is T3. This argument we saw, well, did we see this? There was a lemma when we proved what, yeah, exactly. There we saw exactly this argument, okay? A compact subset of a house of space is closed. This is the same house of, this house of, yeah, exactly. So you can see this, okay? And now we want to prove not T3, but T4, no? And we do the same now. So T4, you have two close sets now, okay? Disjoint, right? And they are compact also, no? Closed, and this implies compact. This is easy, this is also, this is close in a compact. And now you call, this is A, and this is B. And this is B, and now you do exactly the same as before. For each point of A, X. We know now T3, so it's a close and the point. So they are disjoint neighborhoods. Disjoint neighborhoods now of B and of X, okay? So this is the neighborhood UX, and this is the neighborhood VX. But this is a close set, here's a point. That's the only difference, okay? But we have T3, and now we do the same. That's it, okay? So A is contained in finitely many, right? This is a neighborhood of A, and then you take U1, UX1, UXN intersection. This is U, this is neighborhood of B, capital B now, okay? And they are disjoint. So this implies X is T4, that's it, okay? So it's two steps. First, you use Hausdorff and compact space. Hausdorff implies regular, regular implies, Haus T2 implies T3, implies T4. Or if you want Hausdorff implies regular, regular implies normal, okay? What did you just say? So these are these two steps. Okay, now we prove something interesting. First a lemma, and then we prove something interesting. So a lemma, which I will not, maybe, let's see. So X, I, X is T3, well, regular if you have Hausdorff, but that's not important here, okay? Just T3. If and only if for every X in X and neighborhood U of X, for each point and each neighborhood of this point, there exists a neighborhood V of X. There is a neighborhood V of X, such that the closure of V is contained in U. So this is important condition, okay? A smaller neighborhood, but not only smaller, the closure of V is still contained in U. And then the same for T4. X is T4, if and only if, and now instead of a point, you take a closed subset, okay? For each closed subset, A of X, and neighborhood, what is it? U of A now. So this is an open set containing A, no? Neighborhood of A. There is this smaller, there is a neighborhood V of A of the set now, of the closed set, such that the same condition. The closure of V is contained in U. So this is T4, this is T3, okay? Well, I will not, the proof works in the following way, proof. So I, and two I is the same proof then. Here you have a point, here you have an open set, but the proof is exactly the same. So what do you have? We have two, just let's make a picture, okay? So we have maybe this direction. So what do you have? X is T3. For every X, then we have to prove, we have X and neighborhood U of X. So we have X and the neighborhood U of X. And then we have to find the smaller neighborhood, okay? V, such that the important condition is this one here, okay? The closure is still contained. So what we do, of course what we do is, this is U and we take the complement of U, no? So this is X minus U, and this is closed because U is open. And does not contain X support, okay? Because X is in U. So, but now X is T3, okay? What does it mean? This means a point and a closed set, which is not contained, a closed set and a point, not contained in the closed set, you can find these joint neighborhoods. So you find these joint neighborhoods of this point and this closed set. So how does this look like? It looks like this, disjoint neighborhoods. And, well, this goes outside, no? Okay, X minus U is everything here, okay? So these are disjoint neighborhoods. And this smaller neighborhood, this is our V, okay? This, the closure of this V is contained in this one and then it's contained in U. So, this works, okay? This is V. So V is this one, no? And the other neighborhood goes forward. This is this one and the other proof is similar, of course. Now we have to prove that X is T3, right? So what? T3 means you have a point, you have a closed subset, right? And we have to find these joint neighborhoods. But now we have a neighborhood, of course, of the all point, no? This is A closed, what? X minus A. So we have a neighborhood, well, I don't know how to make the picture in the best way. I should A, let me try another. So what? We have a closed set. So the closed set A will then be this, okay? So now X minus A. So this is the neighborhood of this point here, okay? X minus A, what is X minus A? X minus A is this here, okay? That's the neighborhood of this point, right? And now we have the condition. If you find a smaller neighborhood, this one, how is it called? V, yeah, right? Such that the closure of V is still contained in X minus A. So the closure of V is still contained in this larger, okay? But we have to prove, I forgot. We have to find, which direction are we? So we have to define X and A. What are these two open sets? One segment, we are in this direction, okay? So there's a neighborhood V such that this is contained in U and we have to prove X is T3. So we have this point and we have this A and we have to find the joint neighborhood of X and A, right? And V is the neighborhood of X and what is the neighborhood of X minus A? X minus V closure, okay? So V is the neighborhood of X and X minus V closure, so this is still in U, okay? That's the point. It's the neighborhood of what? Of X, of A, yes, of A, right? That's what we're looking for, X and A, okay? And this is schematically the proof, okay? This works. There's some small details, but I don't want to insist on this. Because now I want to prove something else. So this is equivalent to T3 and T4, okay? Given a point, given any neighborhood, there is a smaller neighborhood, such that the closure is still contained in U, okay? This is equivalent to T3 and T4, the same, we don't have a point, but we have a closed set, we have a neighborhood. Okay, so now it's theory. This is, this is, this is orison, orison lemma. So X is normal, T4 is sufficient. Let X be, X is T4 or normal, which is stronger, but we don't never need house stuff here. So we need T4. Let X, sorry, be T4 and A and B design close subsets of X. And A, B disjoint close subsets of X. So X, T4 means they have disjoint neighborhoods, okay? But that's not, that's the definition of T4, okay? Then they exist, so they exist. A continuous map, F, from X to 0, 1, so this is a real interval, okay? Standard, it's a real interval 0, 1, such that F of A is 0 and F of B is 1, F of X is F of, what? F of A is 0, yes, F of A is 0. And F of B is 1, in this interval here, okay? A goes to 0, B goes to 1. So we say A and B can be separated by a continuous function. Real function, real valued function, whatever. That means exactly this. And B can be separated by a continuous value function, okay? That means there's a function X from 0, 1, such that A goes to 0 and B goes to 1. Then we say they can be separated by a continuous function to the real, okay? That's a way of talking. In fact, observation, X is T4, if and only if, that's equivalent to T4. So A and B disjoint closed subsets. And any disjoint, any two disjoint closed subsets of A, of X can be separated in this way, okay? But continuous real valued function. So one direction here, if X is T4, this is the level of over zone, okay? And the other direction is previous, okay? One direction is this theory, and the other direction is previous. If two disjoint closed, if you can separate, so T4, what means T4, you have A, B disjoint closed subsets. And you have to find disjoint neighbors, okay? So how do you find this? So then you can separate by this function. And what are the disjoint neighbors? Yes, that's good, yes, okay? If X is, what, which direction here? So proving, which direction? If and only, proving this direction now, okay? So this is equivalent to T4, this condition, okay? But one direction is trivial, and the other one is the lemma of over zone, and now we prove the lemma of over zone, okay? So this is a way of talking, can be separated by a continuous real valued function, that's the way of talking. Okay, so I prove of over zone, lemma. One thing which we may notice immediately is the following. If, for example, X, if X is, so X is T4, okay? If X is connected, now that might happen, of course. X is T4, normal space, connected, okay? Regional spaces are connected in general. What happens? If we want this condition here, okay, to zero one. If X is connected, what can we say about F? What? What? I don't, no, no, no, let's say it's connected X. What is this function then? F of A is zero, F of B is one. If X is connected, what is the, what? Yes, that's in general, so F is, no, F is not connected. The image is connected, yes, you have zero and you have one. It's subjective, but this is the intermediate value, if you want, okay? The intermediate value, so if it happens that X is connected, then by the intermediate value theorem, F must be subjective. That means you have value zero and one, but you get any real value in between, okay? That means that we have to construct this function, okay, F. It's not an existence proof, okay? We construct the function F, but we have to, in some way, get every real value in between zero and one, okay? If X is connected, we cannot avoid anything, okay? So this shows in some way, not the difficulty, but why it's interesting, no? We have to construct F. If you construct this function F, we need all real values in this interval, okay? In some sense, we need all real values. That's an interesting point, okay? So it means that it's not completely trivial, no? I mean, okay, that's a remark, so proof. So we prefer to work with rational first, okay? And then the reals we can approximate with rational, okay? We need something countable. So let P be the set Q, these are the rationales, intersection zero, one. So all rationales in zero, one. These are called P, okay? All rationales. This is countable, of course, okay? Countable. That's important for the construction. That is countable, because it will recursively, okay? So what we want to define is, so I say what is the plan. We want to define for each P and P, so the rationales in the interval. For each rational in the interval, small P in capital P, and open set UP in X, UP in X, such that whenever P is smaller than Q, and they are in P, okay? So the rationales, P, Q, also Q is in P, then the closure of P, now that looks vaguely familiar, is contained in UQ. So this is an increasing set, okay? They are getting bigger and bigger, but that's not sufficient. We need this condition, the closure of something smaller is still in this bigger. So this is important condition. And then we have to start in the right way, okay? And finish in the right way, but that will be seen immediately. How to start to define the sets, okay? Anyway, this condition is the important condition. This will give continuity of F at the end, okay? If you don't have this, then we have nothing, okay? Continuity is, we need this. So this is the, so we define, we define this, so P is countable, no? This is countable, that's an important point, oops. P is countable, so we define recursively, okay? But for recursively, we have to count P, okay? That's not the natural order, of course, okay? So we have to arrange P in the sequence, okay? We have to arrange P in the sequence. We arrange, well, P is countable, no? So we can arrange P. So we arrange P in the sequence, starting with, so we need a good start, starting with that. So we start with one, we have one and zero, no? And we start in with one is the first, okay? Starting with one, then the next one is zero. These are obviously rational numbers here, point P, okay? And then we don't care, okay? And so on. And this is arbitrary, we just choose, okay? The third one. So this is completely arbitrary. We just, the first two, we fix. And now we define UP following the sequence, okay? We define UP recursively, inductively, following the sequence. This is not the natural order of these numbers, no? This is well-ordered, okay? The rationals, it's not so well-ordered, no? This is arbitrary, recursively, what do we have to say? We define UP recursively, following the sequence. Following this sequence, okay? So we have to start with one, that's the first one, okay? We start with one, what is U1? No, how is it called, U, they are called U, okay? So let U1, so we define the first one, that's one. Let U1 be equal to X minus B. A and B are these two sets, okay? At the end, yeah, okay, X minus B. On B, which value we want, one, okay? Now I have to define U0, that's the next one, U0, okay? So U0, and we need, of course, zero is smaller than one, so we need U0 closure contained in U1, okay? So what, X is, so X minus U1, what is this? This is the neighborhood of what? F, A, that's right, this is the neighborhood of A. U1 is the neighborhood of A, so X is T4. Then we have the lemma, which I didn't, which I just made a picture, okay? So we have a closed set, we have a closed set, and we have a neighborhood of this closed set. Then we find the smaller neighborhood, such as the closure is still contained, okay? So lemma implies there is a neighborhood, so how should I call this? U0, that's our U0. U0 of A, such that the closure, U0, is still contained in U1, okay? That's exactly the lemma, that's equivalent to T4. This is this condition, no, in this special case. U0 closure is contained in U1, okay? That's it. In U0, the value will be, well, how do we find the function, no, these are all these numbers, so at the end we will look at a point, where is it, in which of these sets, U2 is it, where's the first, where it is, okay? And that's our value, more or less. We have to take least lower bounds, okay? Because exactly values, we need real numbers, it's not sufficient to. Anyway, this is, and now suppose inductively, inductively that f, that Up is already defined, for what, for the first, how many do we take? It's already fine for the first n numbers, rational numbers in the sequence, okay? For the first n numbers, rational numbers of the sequence, which we fixed, okay? This is the sequence, and here is n of the sequence. The first n rational numbers of the sequence, and then we have to define it for the next one, okay? For the n plus first. For the n rational numbers of the sequence, and I call this Pn, these are the first n, these are this second, third up to n, so this is Pn, okay, the first n. So Pn plus one, so let R be the next rational number in the sequence, the n plus first, okay? Let R be the next rational number in the sequence. And we have to define it for R, okay? There we have to define all the values. For the first n rational numbers of the sequence, such that this condition holds, okay? With the condition of the sequence, the first n rational numbers of the sequence Pn, such that the condition holds, obviously. We want this condition. Let R be the next rational number in the sequence, okay? So important to see this condition, okay? So when we are leaving this, so this is this condition. If P is smaller than q, then u, p, but that's what we want. Let R be the next rational number in the sequence. So we have, make a picture, now we have, here we have zero, here we have, this is the first one, the second one, and then we have Pn, now this is Pn, the element of Pn, okay, n. And then we have the next one, R, somewhere. I don't know where, it's not in the order, no? It's arbitrary, it's the order, okay? So here somewhere we have R, that's the next one, right? And now we have an immediate, because this Pn is finite, finite, obviously, no? So we have an immediate predecessor, immediate successor of R in Pn, in the natural order of the reals, okay? Now in the natural order of the, not the sequence, no? So let, so I call this, I give this a name, I call this, whatever it is called, P and Q. So this is P and this is Q. So let P be the immediate predecessor, how to write that? C, no? In P of R, the predecessor of R, in P, with respect to the natural order of the reals, okay? This might, this is the first one, this is the second one, this is the third, this is the fourth, this is the fifth, this is the sixth, so this is an arbitrary order, okay? But now we have the order which we see, okay? In the natural order of what? Of this interval, okay? We are in this interval. Is that okay? One second, Pn are the first n elements of the sequence. The first one is one, the second one is zero, and then you go up to n, okay? The n's one, and this is Pn. So these are n numbers, C and the interval, okay? They're n numbers, and now I have the next one, the n plus first, this is R, okay? And then in the natural order, you have a P in Pn which is the predecessor, and the Q which comes after, okay? In P, in P, no? Let P be the immediate predecessor of R in P. Those are n elements, which I may, one, two, three, four, five, six, seven, okay? Yeah, but this is Pn, Pn, sorry. Yeah, P are all rational numbers. This is Pn, Pn, yeah, I said P maybe, okay? So this is Pn, up to this n is, oops, it's Pn, okay? These are the first n, these are only n. P are all rational numbers, yeah. I said P sometimes, maybe, instead of Pn, I don't know. It's clear what Pn is, okay? The first n, so I find it, then the next one, and then this P and Q, no? You see, which comes, okay? Nothing to do with the sequence. The sequence is jumping and doing, is arbitrary, the sequence, okay? Let P be the immediate predecessor of R in Pn, sorry. That's the point, okay? I have to write Pn, of course, yeah. Pn, no, no, no, that's capital Pn. That's R, no, no, there are no points here. I didn't give name to points. These are all capital P's, these are sets. P is a set, capital P, Pn are the first. The point is this R, okay? That's the point, I didn't give names. You don't see if this is capital or small, okay? These are all capital, these are sets, okay? This is a set. What is Pn? Where did I define Pn? Define for the first n, rational numbers of the sequence. This I call Pn, okay? It's a set of the first n, okay? This is capital Pn, that's not the point, it's a set. Which is difficult to see because small P and capital P, if you cannot compare, you don't see, okay? Let Q, so what is Q? Be the immediate, the next one, the next largest successor of R in Pn, again, in this finite set, okay? In the natural order. So here you see, here you have, what this picture is in some sense is P7, okay? You have one, two, three, four, five, six, seven, now P7, right? The first seven, and this is the eighth, okay? R, and this is P and this is Q in the natural order, okay? That's it. So UP and UQ are already defined, okay? Inductively, UP and UQ are already defined. With P smaller than Q, no, P is smaller than Q, and then we have this condition immediately, okay? By induction. So we have this condition for P and Q, where it's already defined, okay? And now I need UR, the next one, okay? For this, where's R? This one, I have to define now for this one. That's the next. But again, what is UQ? UQ is the neighborhood of? Where to write? Let's not, UQ is a neighborhood of, of what? Of UP bar. Obviously, this is closed, no? It's a closure. So X is T4, what does it say? If you have a neighborhood of a closed set, the lemma, okay? Again, the lemma, all this is lemma. If you have a neighborhood of a closed set, then we find a smaller neighborhood, such as the closure is still contained in the larger neighborhood, okay? The natural order of this, yes? Let me forget the immediate, maybe it's not the best word. Now with respect to the, these are finitely many points. PN, that's the final set. So you have this real number, this rational number R, and you have finitely many points, okay? And then you say, I take this, which comes in the natural order before this, and this is the next one, okay? The smallest to this point in the natural order. The sequence is jumping, maybe, okay? It's jumping from here, to here, to here, to here, to here, to here, to here, okay? The sequence is arbitrary, but the natural order of the reals you see, okay? So maybe immediate, the predecessors of R in the natural order, okay? Immediate, maybe it's not necessary. Yeah, in PN, of R in PN, which is finite. So there's a neighborhood U0 of A, no, what, where are we? Lemma here. So the lemma says, so we have a neighborhood of a closed set, so we find there's a neighborhood UR, and this is what we're looking for, of UP closure of this closed set, such that the closure of UR is still contained in UP, no, no, one second. There's a neighborhood UR of UP bar in UQ, no? Yes, UQ. This is a larger one, okay? So this defines UR, okay? By induction or recursive definition, UP is defined for, no? You need more blackboards, you don't see, okay? By induction, recursion, UP is defined for all P in P. Now it's kept a small P, a point in the set, okay? Yeah, that's a disadvantage. I mean, this is a small P, okay? This is a small P, this is a small Q, okay? These are points, the sets are capital. And then we define also for all rationals, so define, well, this is not so important. Define U, define, if, sorry, use this notation. So define UP for all, but this is for all P in Q, no, for all rationals, okay? We define it for the zero one, okay? By setting, let's say UP equal empty, empty set, if, so this is increasing, okay? This UPs is increasing, okay? So UP is empty if P is smaller than zero, then we just, and so we extend in a trivial way. And UP is all everything, X, if P is bigger than one. So we extend in a trivial way, okay? If you have smaller than zero, it's empty. If it's bigger than one, it's the whole set, okay? And the rest we define, okay? In an increasing way, that's the first step, okay? We have defined now, for all rationals, with this condition, okay? This whole set, that's the first step. And now we define our function, okay? So we define our function, so second step. I don't know really where to write it here. Second step, so I didn't write first step, but it's clear that up to this point was first step. Second step, definition of F, the function F. So what we do, we have to define F of X, no? And we're looking for this X. This X is not here, okay? Because this is empty. So maybe it's not in the first, okay? But in a certain point, it is there, okay? Because then we have everything. And so we are looking where we have it for the first time, okay? More or less. Of course, we have to take, we need real numbers, no? The rationals are not sufficient. So we need the least lower bound, okay? Of these. So define, it's a definition of this. So for X in X, let QX be, all P rational, we look for all rational numbers, such that X is in UP. So we're looking for all UP, such that X is in UP, okay? And then we define F. F of X, so this is a definition. So what do you suggest? Greatest lower bound, no? Okay, we have, where, so the greatest lower bound, greatest lower bound of QX of this set. Well, we have to say something about this set. What is this set QX or P rational? So anyway, QX is clear that it contains no rational which is smaller than zero, okay? This is empty set, okay? Smaller than zero, we don't have anything. And it contains everything which is larger than one. Well, this is a definition here, no? UP is empty anyway, okay? So we don't have these points, P in, okay, in QX. Because then we have the empty set, no points at all. And here we have everything, if P is bigger than one, okay? So this is a definition and then immediately it follows that F of X, this is between zero and one. As I said, we need all real numbers, okay? This is lower bound, okay, of rationales. So we get all real numbers. And then the point is, what is F of A? So now we defined in the right way, so F of A. So what was A? A is contained in U zero, it's the second one here, okay? This is the neighborhood of A. So zero is in this set, okay? For points in A. So I say this is zero. All points in A are in U zero. All points of A, U zero is the neighborhood of A, okay? All points of A are in U zero. So the least upper bound is zero. Because other, if smaller than zero, there's nothing, it's empty, okay? And F of B, U one, that's the largest, okay? This is X minus B. So we don't have any point of B so far. But after this, we have the whole, okay? So we get everything, which is larger than one. So the least upper bound is one. So here we have one, so this is okay. F of A is zero, by construction, by these choices, okay? Of zero and one, or one on zero. We have these conditions, we are in this interval, and this is, so what is remaining? Continuity, okay? We have only to prove that F is continuous. So this is the remaining claim, F is continuous. So claim, F is continuous. And you need this condition, which I don't see anymore, okay? Since, I mean, one can make a picture, try to make a picture here. Sorry, I should have left this. F of X, this is the definition, this is Q of X. And this is the definition, this is. So F is continuous. I mean, you can make a picture, no? But I don't know what is the best way to make a picture. So this maybe is A, okay? Here we have, how do you say in English, height zero, okay? Above, height zero, value zero. And A, we have value zero, no? And now, so the B is maybe this. But to make a good picture, I go to the outside, okay? The close set, the other close set, okay? And now, we have all these in between, these UPs, okay? We have all these, so we have height zero here. And this B finally has height one, okay? Here we are at level one, okay? We have to go from zero to one. And here we have all our UPs, okay? And if we get a point here, we look where it is, this point, X, okay? It's maybe not in A, it's here. So at a certain point, we, X is in this, okay? And then we look at this P. But we have to talk the lower bound, okay? We need, this is a real number, no? And then we need this condition, UP in UQ, why? I mean, this has to be continuous, no? So what may happen is that these sets go in this way. So it gets large, higher and higher, okay? From zero to one, okay? Like a crater, right? This is a bottom, and this is the, okay? And it goes from the bottom, and we need all reals, okay? To go from bottom, we cannot, right? But it has to be in a continuous way, no? If it happens something like this, this is bad. Because in some sense, if you go, so you have other levels here. If you go here to this point from this direction, right? Then if they are all touch here, these sets, okay? On the bottom, then you fall down here, okay? This is not continuous, and this is excluded by this condition. That's something like this happens, okay? Where you go to some level from ups, okay? And then you come to such a, here you go through all levels, okay? In a nice way. This happens at a certain point, no? Then you fall down. That's not continuous, right? But this contradicts this condition here, okay? If the closure of the smaller one here is not contained in the larger one, okay? It becomes, you see, the closure of the smaller one here is not in the larger one, okay? So this is the condition. This is continuity. Maybe I don't, well, we have 25 minutes. And you want to see continuity or not? I mean, I give the idea here, okay? That this is the important condition for continuity, okay? This implies continuity. It's not so difficult. I don't know. Maybe I don't prove continuity. Or you want to see? I mean, it's a long proof, right? It seems, it's not. The whole is a long proof, not continuity, okay? So I prove, okay? I don't know, I mean, if you are, yes? Yes. Okay, yes. That's good too. So this is the most interesting proof of the whole course, okay? Well, the most difficult, most interesting, most. So let's prove, okay? So this is a picture, right? I told you why this is important. If you don't have this, if you have only this, then this has nothing, okay? This has nothing. You can take all equal, okay, almost, okay? And then you don't have continuity or nothing. But that's a good idea. So you have level zero, and then you have a crater. You have level one, and you have to go down in a continuous way from level one to level zero, okay? That's exactly what is happening, okay? In a continuous way. And not continuous means that you come to some level, and then it goes down just, okay? That's not continuous, right? That's exactly what continuity is. You fall down. So I prove that it's, well, sorry. Probably I need the definition of F, of course, no? So F is continuous, claim, F is continuous, okay? So what? So we, I make a remark, observation. And this helps to X in UR bar implies F of X. So this, again, is, well, you see here, but I write, greatest lower bound of Q of R, and this is smaller equal to R, I say. Then if X is in UR bar, UR closure, then FX is smaller equal to R. Why? Because it's UR closures in all which come after this, no? UR closure is contained in all UQ, R smaller than Q. It comes in all, okay? Because this is contained, it's contained here, it's contained in all these. And you take the greatest lower bound. So you finish with R, or sorry, QX, thank you. Okay? That's the first, and the second one is similar. X, if X is not in UR, result closure, no? Then F of X is, I write again, so it's easier to read. Greatest lower bound of Q of X, again, okay? And this is bigger or equal to R, no? Larger equal to R. Well, if it's not in UR, it comes later, okay? And then you take the greatest lower bound, so maybe it's R, or larger, okay? So it's larger, or bigger or equal to R, okay? X is not in UR, okay? R is not in the set, QX, but maybe everything which comes after R is in the set, okay? That might happen, so this means it's, maybe R or larger, okay? So this is also, so this is the observation, these two things, okay? And now I prove continuity. So let X0 be in X, such that, so we prove it's continuous in this point, okay? In every point. And so we give a neighborhood of, we want a neighborhood of F of X0, so let F of X0 be an element of, and I choose an interval, open interval, okay? Which is our neighborhood. So this is CD, CD, content in R. So this is a neighborhood of F of X0. What we have to find is a neighborhood of X0, such that U, such that F of U is contained in this, right? We want to find a neighborhood of X0, such that F of U is contained in this neighborhood, okay? That's continuity, local definition of continuity. So what is this neighborhood? So again, we have C and D. We have F of X0, we have C and D. So we choose rational, it's P and Q, okay? P and Q, and these are rationales now. Let Q be in rational numbers, such that this situation happens here, okay? Such that C is smaller than P, is smaller than F of X0, is smaller than Q, is smaller than D. And our neighborhood, I can write immediately, so let U, the neighborhood which we are looking for, is then UQ, so here's defined, this is a rational number, minus UP bar, this is open, okay? This is open, okay? This is open because this is closed. So this is open. This implies that, first of all, X0 is in, so let's see, X0 is in UQ, and X0 is not in U, X0 is not in UP bar. So this means this is the neighborhood, U is the neighborhood of, this would imply that, this says that U is the neighborhood of X0. So let's see this, X0 is in UQ. This is the observation, okay? Observation. So let's see, if X0 would not be in UQ, X0 is not in UQ, then F of X would be, what? If X0 is not in UQ, if X0 is not in UQ, then F of X0 is larger or equal to Q, say again. So X0 is in UQ, okay? If, I say, if X0 would not be in UQ, okay? X0 not in UQ means that F of X0 is larger or equal to Q, but F of X0 is smaller than Q, okay? So that's okay. And the same, X0 is not in UP bar. If X0 is in UP bar, then F of X0 would be smaller or equal to P, but it is larger than P. So this is okay, okay? By the observation, you assume that this is not happening, so the observation implies that this is the neighborhood of U is enabled, okay? And now I have to prove that F of U is contained in CD, okay? That's the last point, and this is also, also F of U is contained in CD. Actually, it's contained, sorry, it's contained in PQ, in the closed interval PQ. That's better. And this is contained in CD, open interval. And this, of course, implies F is continuous. This sign implies F is continuous in the point X0, no? So we find the neighborhood. F of U contained in this. So F of U is in the closed interval PQ, which is contained in CD. Why is that? What is U? U is this. So if we take a point in U, okay? Then it's in UQ, right? If a point is in UQ, then it's obviously here also, okay? Then F of the point is small equal to Q, okay? It is not in UP bar, okay? It is not in UP bar. And you can, if it is, so it's not in UP even, okay? If it's not in UP, F of X is smaller equal to P. So we are between P and Q, okay? So we are between P and Q, and this is in CD. So F is continuous. That's the end of the proof, okay? So I fix, this is, you just look at this and formally, you can look at this and you check, okay? So it is continuous. That's a nice proof, no? I will give maybe today, I will give one application, okay? But today not because, well, I will write, but I will not prove today, I'll prove next time. This has many applications. And one which is interesting for us, I will write now, but not proof because then we have another proof but it's too concentrated too much. So this is a Eurozone. So first there's a proposition and this, I'm not, this I will not prove anyway. So we need a Eurozone, okay, a lemma of Eurozone. But we need normal, okay? But in the hypothesis, it's only a regular, but so there's a proposition which is not too difficult. It's in the book, so proposition. Every regular space, every regular space with a countable basis is normal. So regular plus second countable is plus normal, okay? This is reasonable, not too difficult, but I will not prove it now, okay? Instead, what I will give the proof next time is a Eurozone metrization theorem. This is needed for this formulation. Metrization theorem. We never found a reasonable, we have necessary conditions for metrizable, no? House of first countable normal, no, okay? Normal is necessary. We never found a nice sufficient condition, no? Except trivial ones, discrete space. Who is on metrization theorem? So what does it say? It's very short and it says that if X is, what? X, a regular with a countable basis, okay? Implies X is metrizable. So finally, there's the first nice sufficient condition in the history of anthropology, okay? For metrizability. Regular with a countable basis implies metrizable. The main tool is urizone, as we will see, okay, not today, okay? We apply urizone, so we need normal. Regular is not sufficient for urizone. We need normal. But regular with a countable basis is normal anyway, okay? So I will not in five minutes prove this now, okay? I give the idea of this proof, it's not too difficult. The main tool is lemma of urizone for this, okay? It's a nice proof, so I will sketch the proof next time. But now we have, so X metrizable. Now I make a scheme, metrizable. Then we have sufficient conditions, necessary conditions, no? So X must be, what are the most interesting ones? Hausdorff, yes, but, Hausdorff, no. First countable, that's better to write countable basis. No, you don't have to think, this is second countable, okay? So first countable, and now we have normal, okay? Metrizable space is normal, so these are, and not sufficient, well, we have this urizone, okay? What do I have? X second countable, and then regular, or normal, then it's equivalent by this, okay? Whatever, regular, or normal, implies X is metroid. This is the urizone metrization. Urizone metrization, okay? So what you see, you have, let me write normal here, okay? In some sense, normal is, you have, you see on this side, on this side anyway, okay? But here you have second countable, and here you have first countable, no? So normal is okay, no? This is, but it must be between second and first countable somewhere, okay? Then you say maybe it's one of this or this, okay? But this is not true, it's not, so here this does not hold, and this does not hold. So we need, that's the last, we need examples, that is, that's not hold. So if X is metrizable, it is normal, of course, no? Normal is on both sides, okay? If X is metrizable, it is normal, that we know, okay? Second countable is a problem. What you, that's an easy, here we have an easy example. That metrizable, but not second countable. Then you take any discrete uncountable space, discrete, so we take any discrete uncountable space, okay? Then it has no, it's not second countable. Uncountable, because you need points, no? For a basis, you need all points, and if it's not countable, you have no countable basis, no? You need all points, which are open, points are open. So any, take any discrete uncountable space. This is metrizable, no? This is a discrete metric. However, it is not second countable, okay? So this direction, this is necessary. This is sufficient, but not necessary, okay? And now we want an example here. So we want a space, which is first countable, first countable normal, here we have many examples, but all strange examples. And it's not metrizable, right? So what are our favorite examples? RL, but we have to, well RL is first countable, no? It's normal, that one has to prove, it is normal, okay? RL is normal, I didn't prove that, but it is normal. What else? Ordered square, let's try this. Ordered square, even easier, it's normal, why it's normal? Why is it normal, the ordered square? It's compact, house top. So it's normal, it is first countable. So this we proved everything, okay? This is also countable. Then our last is S omega, let's try S omega, what is S omega? S omega is first countable, okay? It's also normal, that we didn't prove, okay? But it's not metrizable, okay? So here, in this direction, we have all our favorite examples. Also, the list of strange spaces says that this is necessary, but it's not sufficient, okay? Well, this is sufficient, but it's not necessary. So it's between second countable and first countable. We need a condition between second countable and first countable, okay? And that is another chapter of the book, which we will not do. It's a subtle condition. It took about, I don't know. I mean, the story is interesting. It took about, well, I mean, the story of Uzon is also interesting, or sad, whatever you want, no? So he wrote this paper with the Urizon lemma, and with the Urizon matrization theory, when he was 26 about. He was traveling in Europe with Aleksandrov, another Russian mathematician, no? And they were in France and swimming, and he drowned in the sea. So it was a rough sea, Britannia in France, no? And so he had finished his paper, okay? And then that was the end of his life, okay? Some week before he had finished his paper, no? And then he was 26, then he did, okay? And that was it. And then, of course, he knew this, no? And then it took about, I don't know how many, this was 19. Well, I should be careful. I would say about 20 more years to find the right condition, okay? That took a long time, at least 10. I forgot exactly when that was, but maybe 20 years, no? And then the right conditions in between these two, they're found by two different approaches, not one by a Japanese Nagata, no? And one by Russian again, which was Smirnov, I think, okay? So independently, they found the, so this must be 1940 or 50, 40 maybe, but anyway, I mean, there's a gap between 10, 15 years, no? So maybe Urizon would have been faster, but there was this incidence, no? Aleksandrov is also famous, no? Aleksandrov Compactification, okay? This is Aleksandrov from General Topology, no? And he was traveling with Urizon. They were young both, no? And then they had this, everywhere they went, they wanted to swim to take my sport in the sea, no? And this is dangerous in France, in some places, no? There are currents, and so that's what happened. Aleksandrov became very old, no? He was 85 or something, okay? So that's also interesting history here, okay? The history of the... It's really interesting. Yeah. Well, he was 26 and he proved these two beautiful things, the Urizon matrization and this application, the Urizon lemma and this application, and that's the main result of this paper, okay? 19, I don't know, somebody, 1915? I forgot, I have to look, okay?