 So, so the host of condition is the following, no? We have two different points. Any two different points have these joint neighborhoods. That's host of, or T2, okay? And T1 is, any two different points have, each one has a neighborhood, which doesn't contain the other one. That's T1. So these are two separation axioms, okay? There's also T0 and we will see T3, T4. They are not the same. So for example, if X is, X has a final complement topology, a set with final complement topology, is T1. This is T1. Why is it T1? Well, if you have any point here, okay? If you have any two points here, no? You want the neighborhood of this, which does not contain this. And what is the neighborhood? That's the whole space minus the other point, okay? A neighbor in the final complement topology, final complement. You take the whole space minus this point. That's an open set, okay? Which contains this point, obviously, right? Final complement topology. X minus one point is open, okay? So X minus this point is open and contains this point. So this is T1. Each point has a neighborhood which does not contain the other point, okay? So it's T1. If X is infinite, it's not T2. But not T2. If X is infinite, if X is finite, and has a final complement topology, what is the topology, another name? We have a finite set and we have the finite complement topology. There's no condition, right? It's a discrete topology, okay? There's this house stuff, obviously, no? The discrete topology is house stuff. The points are open, right? So this is again, if you want, so this is an exercise, okay? So you have two points. Suppose two points, two distinct points have, I want to prove this, okay? Suppose two distinct points have disjoint neighborhoods, okay? Any two distinct points, suppose, we have the finite complement topology, okay? Suppose X and Y have disjoint neighborhoods. So this is an exercise, U1, U2. So this would be X1, X2, no, X, well, X1, X2, okay? So it doesn't matter, but let me call it X1, X2, U1, U2, okay? X1, X2, and this is U1, this is U2, disjoint neighborhoods, so we are in this situation, right? This means U1 intersection U2 is empty, they are disjoint. That means X is equal to X minus U1 intersection U2. This is empty, no? And we have again the Morgan, always the same, in set theory, almost. X minus U1, union now, okay? So it goes to the union. X minus U2, right? This is the Morgan again, but these two are finite. We have the finite complement apologies. These are neighborhoods, U1 and U2, means that the complements are finite. So this means that X is finite. Then we have the discrete apology in this house of course, okay? So if X is infinite, it's not house of, right? Yesterday he made the remark, somebody of you made the remark. So what is the definition, let's give the definition of, we have to give that anyway at a certain point, okay? And you know that, of convergence. So X is a topological space and we have a sequence. A sequence, X1, X2, X3 in X converges to X in X to a point. X in X now, so what is the definition? If for every neighborhood U of X, there exists capital N, a natural number, there exists capital N in N, so a natural number, such that, such as what? XN is in U for all large N. So XN is in U for all N bigger equal to this fixed capital N. This is the notion of convergence, okay? That's the definition of convergence in general topological spaces. So we write, we will write XN converges to X, okay? If this, okay, that's a convenient way. XN converges to X and what you remarked, somebody of you remarked yesterday, so in a house of space, if a sequence converges, then the limit point is unique, okay? So observation, that's trivial almost now. If XN converges, then the limit point, the limit X is unique. So if you have this point X, this is the sequence XN, and this is unique, it cannot convert because if you have any other point, Y, okay? Then we have disjoint neighborhoods, sorry. If it is house of, if X is, that's the most important, which I forgot, I'm talking about house of spaces, no? If X converges and the limit point is unique, if X is house of, that's the most important here. If it's not, one second. So if it's house of, then for any two points, we have disjoint neighborhoods, okay? And then for large N, this sequence is in the neighborhood U of X. This is small X, okay, in this neighborhood. So it's not in this neighborhood. All these points are not in this neighborhood, which I just wrote, okay? So in a house of space, a convergent sequence converges against a unique point to a unique point, okay? If it's not house of, it's not true. For example, the most stupid example is the indiscreet topology, no? In the indiscreet topology, every sequence converges to every point. There's only one neighbor, which is the whole space, and the sequence is in the whole space, so that's a very stupid suggestion, no? In the discreet, in the indiscreet space, indiscreet space, every sequence converges to every point. In the discreet space, what are the convergent sequences? Discreet, no, no, what? Discreet, discrete, what? Yeah, what are the convergent sequences? Not exactly, what did you say? The convergent sequences are constant, almost. Not exactly, they are finally constant. So the first is not important, but up a certain point, they have to be constant. Not constant, but you say finally constant. At a certain point, they become constant. Before it's not important, okay? The first terms are not important, right? So not constant, but finally constant, okay? Another stupid question, exercise. We have, let's say, the exercises are more pleasant than the theory sometimes. We have, let's say N with a final complement topology. The natural numbers with the final, the natural topology with the natural numbers is a discrete topology, okay? The subspace of the reals, it's discrete, okay? The natural. Now we take the final complement topology, which is one of our examples, okay? The final complement topology, this is T1, no? It's T1, not T2, no? Not T2, T1, not T2. Not house stuff, that's what we, that's stuff. So then take the sequence one, two, three, four, five, and so on, no? Does it converge? And if yes, what is the limit, huh? It converges against one, is that true? Yes, it's true, it converges against one. Any neighborhood of one has final complement, so excluding finitely many, you have everything in this neighborhood, right? Every neighborhood of one is open, has final complement, so it contains all natural numbers, except finitely many, which you can forget, okay? So up a certain point, everything is in this neighborhood, okay, all natural numbers are in this neighborhood, right? So it converges against one, and then, other terms, it's not what, it converges against one, and there are other points? Everything, it converges against every point, right? Yes, right, one, two, three, so if points, okay? It converges against any point, no? Because the complement is finite, so it contains, that's a stupid example, no? But that's the case. This is T1, so T, I said yesterday, in this course, later, we always use T1 almost, okay? T1 is sufficient very often, no? But for converging, it's not sufficient, no? So it goes very wrong with T1, no, the unique limit, no? T2, it's unique, okay? But in T1, we can have everything, okay? Well, maybe your question will talk later, okay? Because I have to go on with this theory, I suppose. So why did we introduce this at this point? So there's a proposition, okay? Let me write this proposition, so proposition. So extra Hausdorff space, extra Hausdorff space, you never say T2, you say Hausdorff, that's it, okay? T2 is too technical. And then somebody forgets, what is T1, what is T2, no, Hausdorff is clear. And A subset of X and X and X, so this is the situation. X is the limit point of X, so it's the same. X and X is the limit point of X, of A, sorry. It's the limit point of A, if every neighborhood of X, if and only if, if and only if, if and only if, so I'll write this, if and only if, every neighborhood of X intersects A in a point different from X, that's the definition, no? And now we have a Hausdorff space, we say it intersects every neighborhood, abbreviate, neighborhood of X intersects in a point different from X, it's the definition of limit point, no? And now we saw in infinitely many points, okay? So it looks a little bit more of a limit, but we still have no limit, okay? But in infinitely many points. So we come a little bit closer to limit, no? At least you find infinitely many points. But limit means we have a limit, but that's still not true, okay? There we need another condition, a stronger condition, which we will see, okay? No limits, but infinitely many points. Well, okay, so sometimes I give no proofs, now I give two proofs because they are both easy and nice. So the first one, so here are two proofs. First proof is constructive, no? You make that picture. Well, let's make a picture. So we have X, we have A, and we have, this is A, this is X, and here we have our point X, okay? And now, well, one direction is trivial, no? Which one is trivial? If you have this, then X is the limit point, okay? So that you, I forgot to write, this is trivial, okay? So this direction, no, which one, the other one, no? Yes, this is trivial. So we prove the other direction, okay? So let you be enabled of X. It's, this follows immediately from the definition, okay? It's much stronger than the definition. So you enabled of X, that's U1, enabled of X. Let U be enabled of X, so here's U. So we're proving this direction, no? So X is the limit point, no? So U intersects A, okay? U intersects A, this is not empty. So let, choose a point which call A1, U1. That's U1, the first one, if you want U, equal U1. So choose a point A1, okay? So A1 is here somewhere, you make that picture, you see in the picture, A1, yes, A1, right? So X is T1, how stuff, but here T1 is sufficient, no? X is T1, what does it mean? X has an A vote, which does not contain A1. Small X has an A vote, U2, which such that, which does not contain such that, what is the first point? A1 is not an element of U2, right? And we can assume U2 is in U1, we can assume. So I prefer to write in this, it seems that U2 is contained in U1. Why can we assume that, U2 in U1? Well, if not, what, sorry? Yes, if not, you take the intersection, okay? The intersection of finally many neighbors is always a neighborhood, so you make it smaller. You just make it smaller, okay? So can apply U2, okay? And now that's honestly proof, right? So we have here the smaller neighborhood. Which is U2, so we make a picture, now it's clear, no? So the proof is clear. So now U2, intersection A is not empty, no? Because it's a limit point, oh, it's a limit point, no? So let A2, okay? So we find, sorry, that one is, so here's A2. Well, now I should make bigger this picture, but A2, okay? So recursively, we construct the sequence, okay? So recursively, define, so you go on with A3, no, A4, A5, define An, sequence. An, all distinct, okay? Because we avoid everything what was before, all distinct. In, where everything is in U intersection A, no? They're always smaller, smaller, smaller, smaller. That's the proof, okay? That's the constructed proof. That's a nice proof, okay? You make a picture, big picture, no? And then you have your neighborhood, so you have the first point, you have the smaller neighborhood, you have the second point, you have the third neighborhood, you have the third point. And so you see the points, no? What you cannot say is that this converges to X, okay? That's not true in general. Then we need something much stronger, okay? Another condition. We don't talk about conversion. We have a sequence here, and there is here, but convergence is not clear. Why not? It's not clear because there may be too many neighborhoods at this point, okay? There may be neighborhoods which have nothing to do with these neighborhoods which are somewhere in some other, okay, too many neighborhoods. This will be an important condition for the space, okay? When we want all the convergence, okay? That we will see, but here we don't have. But still we have infinitely many points, okay? So it's a little bit more limited. So there's a second proof from the book. So I give two here now, because it's more interesting. So we still, two I. So we still have this direction, okay? We're still in this direction. So this is from the book. They're the same equally easy, but... So let's suppose I'm also here. X, so what is our sequence? It's a limit point, no? We have a limit point of A. So we have a limit point, X, as before. And U is the neighborhood. And suppose U is the neighborhood. And U is the neighborhood of X, such that the intersection is finite, okay? Such that U intersection A is finite. But I don't know if X is in A, okay? I don't want X here. So I take A minus, maybe X is not in A, okay? So A minus X is A, okay? But this is finite. So I can, it's a finite set. X in A, A1, I don't know, how did you call it? X1, Xm. Well, I prefer A. We are in A anyway, no? So A1, Am. It's finite, finite, okay? That's not what we want. However, this is a finite set, no? So this is open. We have T1, T2, house stuff, okay? Here we need T1, okay? So finite sets are open, okay? Because it's T1 or T2, whatever you prefer. T2, house stuff. Final sets are open. So this is, what do we want to prove? I avoided X, right? So this is not X. So this is finite, finite sets are open. So this implies that this is the neighborhood of X. So this is the neighborhood of X. Intersection A minus X is the neighborhood of X. It's open. It doesn't contain X. So it's the neighborhood of X, right? That's our definition of neighborhood, which does not intersect A. It contains X. It's a neighborhood of X. Yes, you're right. Intersection A minus X. So what I want. One second, one second, yeah. I'm tired. Yes, yes, yes, yes. I wrote something wrong. Let me, U intersection X minus. So this is A1, Am. Not in the best form, okay? Final sets are what? In the house of space? Close. Right? So the complement is open, okay? So this is open. So this is the neighborhood of X. Which does not intersect A. Intersection, which does not intersect A. Because we avoid these points, okay? Intersection of U with A, it might be X. We don't care about X. But the other points are these points. And these we don't have here, okay? We avoid these, okay? So this is the neighborhood of X. So this means, this is also written in a stupid way. So this means X is not a limit point, right? So this implies X is not a limit point of, of what? Of A, okay? So it's written in a stupid way also, okay? Well, I prefer the first proof. It's constructive, okay? I made a mistake. I'm going to have to concentrate a little bit now. We need, the points are closed. But we have to give a proof, of course, okay? The first is constructive. This is sort of, well, you use something and you write something. One second, I don't like how it is written. This is stupid. A stupid way to write, okay? I don't know how it is written in the book. But we suppose X is the limit point, then we prove something, and then we write X is not a limit point, okay? This is a stupid contradiction, no? That's not, so I correct this, okay? To write it in a, this is correct, okay? But it's not written in a nice way. So suppose U is the neighborhood of X. Suppose, suppose there is a neighborhood U of X, okay? There is a neighborhood U of X, such that U intersects A only in finitely many points, okay? If there is such a neighborhood, then X is not a limit point. So if X is a limit point, then it intersects in infinitely many points. This is better way to write, no? The other is this kind of stupid contradiction. You write something which you want to prove, then you independently of this, you prove something, and in the end you write the contrary of that, what you wrote before, no? You can, so this is a good way to write, okay? It's clear, no? It's not a contradiction, no? We prove in some sense A plus B, not B, is equivalent to not B, not A, so that's what we prove, okay? It's not really a contradiction. Okay, then what, that's exactly the, yes? You have to prove that every neighborhood has infinitely many points, infinitely many points. So infinitely many points? Yes. That you have to prove, that you still get infinitely many points. That you have to prove, you exclude X, okay? And then you still have to prove you have infinitely many points in the intersection. That's exactly this proof. Okay. It's not, I mean, it's easy, but you have to write something. It's exactly this proof, okay? Okay. Because you have only finitely many, okay? And then X is not a limit point. Here you have to concentrate, okay? So I made some mistakes. The first one is sort of immediate, right? You see these infinitely many points, no? You see how they, you can construct them in some sense, okay? So that's why I give two proofs, okay? The first one is in some sense nicer. This one is more sophisticated, no? You use points are closed. You're not, you're, okay? You have to write something, okay? Okay? Yes. But still the interesting point is we have a sequence now, infinitely many points now, but we don't have convergence in general, okay? We will see examples. And that's interesting point. Still we cannot suppose that we get a sequence in A which converges to X, okay? That's what we want to have maybe, but that's not true. So we will see examples that this is not true. You don't get a sequence which converges. In A which converges to X. That, there you need some other condition, okay? Some stronger condition. Sorry? Yes. Yeah. In the closure, a limit point, let's say. Yes, in the closure or limit point. And if it's in A itself, you have a sequence now. Constant sequence. A is almost the same point. So interesting point if it's a limit point, okay? And then you want a sequence in A which converges to this point, okay? Maybe you want, but you don't get that, okay? No, one second. If it's an isolated point, it's a limit point only if it's in A. No, no, it's not in A point. If this is A, then this is not a limit point. It's not a limit point. So you have this. We are now two here, okay? Schematic picture. So you have here neighborhood which intersects. So A is this. A is this and this point. Isolated points are not limit points. Yeah. If it's in A, it's in the closure, of course. But it's not a limit point. Isolated points are not limit points, okay? So it's not isolated. But still, we will see that you don't get the sequence. Okay, so now let me go on because... So this is an interesting condition, hostoff. Now continuous function. We have to talk, of course, about continuous functions. Well, continuous functions. Yeah, functions, maps, whatever. So let me give the definition. The definition is very easy. So, definition. X and Y are topological spaces. A function, a map, F from X to Y is continuous. That's the definition. If pre-images of open sets are open. Pre-images. If pre-images of open sets. In other words. So this is a fast way to say that, okay? If U, if V is open in Y, then... What is a pre-image? F minus one, no? Then F minus one of V. So F minus one of V. So this is a pre-image. It means all points. So the definition, of course. F minus one is a double, no? This is not the inverse function. It might be also the inverse function. It's bijective, no? And then it's the same, okay? But in general, it's only the pre-image. It's not the inverse function. Pre-image. So these are all points X in X. Such that F of X is in V. That's the pre-image of V, no? That's the definition. So if V is open, then F minus Y is open. And pre-images of open sets are open. Well, we will say that anyway, but not now. This is the definition. The definition is also open sets, okay? No, for open for some reason. Yeah, close. You suspect maybe it's close, no? It's okay. You get some compliments, no? And you look a little bit around. And then it's true for close. But the definition is open, no? The proof that it's equivalent is easy, okay? You go to compliments, okay? Open and close are compliments. And so the pre-images of complimentary sets are complimentary or something like that, no? And then it seems to be, okay, yes. So this I like to distinguish, call it the global definition of continuity, okay? Global. You have no point here. You have, for each open set in Y, pre-images open. So this is a global definition. Because then we have a local definition, no? This is a global definition, okay? So let me call this I, okay? And now I have a second, two I. So local definition. Local means in a point, okay? So f is continuous in a point, x in x, small x in capital X. f is continuous in a point, x in x. So this is epsilon delta, except we don't have metric spaces, no? So if for every neighborhood v of f of x, if for every neighborhood v of f of x, there exists a neighborhood u of x, such that, such that, such that. f of u is continuous in v. So in analysis, in real analysis, this is given epsilon neighborhood, epsilon ball of f of x. You find a delta neighborhood, a delta ball in x such that this holds, okay? But we don't have epsilon delta here, because we have general topological space. So this is a local definition of continuity, no? In x, local definition. So this is a local definition of continuity. So this is continuous in a point. And this is just continuous, no? This is very short, no? Prematures of open sets are open. What is the relation between these two? Well, that's what you suspect, no? So proposition. f is continuous. f from x to y is the same notation, almost is continuous, if and only if. f is continuous in every point x and x. This is a little bit more general than this local, no? You can say in some point it's continuous, in some point it's not continuous, so you have the points where it is continuous, no? So it's a little bit more general, no? So f is continuous, f is continuous in every point, x and x. So that's the proof. I don't know what is the, I mean, now what in general you take as a condition, as a definition in analysis, epsilon delta, no? But also this one. Prematures of open sets are open, so both. I'm not epsilon delta. So we have two directions. This is, f is continuous. So we have to prove this one, okay? This, we have to prove this. So v is a neighborhood of f of x. So let x be an x. And v a neighborhood of f of x. Then what is u? What you take for u? This is trivial, this is real. What is u? So u, so this is a definition, no? It takes a preimage, f minus one, no? So this implies f minus one of v. So this is open, no? Enabled is open first. So the preimage is open in x. It takes a whole preimage. And of course f of u is f, f minus one of v, is contained in, not equal to v, contained in v, no? It might not be surjective for something, okay? But it's contained in v. So this is trivial, this direction, okay? A little bit more interesting is the other direction. So which is this one? This one, this one. So now if f is continuous in every point, so f is continuous in every point, yes? So let v be open in, so I use almost this notation again, open in, but we have no pointer. Let v be open in y. You have to prove that the preimage is open, okay? The preimage is open in x. No, that's continuity. To take a point x in f minus one of v. So let v be open and take any point in the preimage f minus one of v. This means, of course, f of x. Then f of x is in v. So v is the neighborhood of x, f of x, no. So v is the neighborhood. It's an open set containing f of x. It's the neighborhood of f of x. But f is continuous in this point, okay? f is continuous in x, no? In each point. f continues in x. In x, in this point, x in x. If you write, you see, this is small, this is capital, okay? Otherwise sometimes it's clear from there. f is continuous in x. So this means for the x this, there is a neighborhood u. There's a neighborhood u of x. ux. It depends on x, no? For each x. So we call ux of x such that f of ux is contained in. So what is then f minus one of v? Then the preimage f minus one of v. This should be open, no? And what is it? Just write it this way. You take union over all these, no? Each of these ux, f of ux is contained in v. So each of these is contained in. ux is contained in f minus one of v, no? Okay? Applying f minus one. ux is contained in f minus one of v, no? Right? Because f of ux is in v. So you have this inclusion. And the other inclusion, because you have all x in x here, okay? If you have some x in f minus one of v, then you have ux, okay? Which is a label of x. So this is by construction trivial also, okay? Both are trivial. So this means this is union of open set. So open is open in x. And this is the end of this proof, okay? You just formally check both. Yes, thank you. What is u? f minus one of v? u, not x. Yeah, right. Sorry, there's always some small one that's to concentrate on the notation. No, sorry. Get distracted. So x in u, okay? x in f minus, which is f minus one of v, no? It's this one. It is this one. Example. So in analysis, you have, in real analysis, you have continuous functions, differential functions, continuous, no? So you have some nice examples. So I take r. So this is a standard. And I take rl. This is a lower limit. Oh, let's, is this continuous? Ah, identity. Sorry. Yeah, I wrote id, identity. I abbreviate. I forgot to say that. So this is id, see, identity, yes, sorry. Identity map. Yeah, cool. That's, I always abbreviate identity. It's not continuous, no? Not continuous. Yeah, right. I should have started with the other way. From rl to r is continuous, no? Right? So the general principle here, not principle, the general thing is that the identity map from x, t1. No, I write the topology. This is the identity, okay? Sorry. Id is the identity map. From x with one topology to x, the same x, because I have the identity map, no? I need the same x. t2 is continuous, if and only if. Sorry, Mark. So what should I write? t1 is finer than t2, yeah? And here you have, that's rl. It's strictly finer than r, okay? So it's not finer than r. It's strictly finer than r, and r is not finer than rl, okay? It's not finer, yeah. So definition, homomorphism. Here we have one of the important problems of topology, no? Okay? So, definition. f from x to y. So these are topological spaces, okay? It's a homomorphism if f is continuous. In topology we have continuous maps, okay, in general. Homomorphism means isomorphism, okay? For groups, isomorphism, in topology, homomorphism, okay? So bijective, bijective. And that's not sufficient, okay? So what should I add? Yes, and the inverse map. So now it's the inverse map, not the pre-image. The inverse map, f minus one. This is also f minus one. From y to x is also continuous. And this is not automatic, okay? Why not automatic? Because we have an example before, no? The identity is from rl to r is continuous, bijective, but the inverse map, which is identity from rl to r, is not continuous, okay? So it's not automatic. We have to write it down here, okay? And for homomorphism we want that. And this is an interesting problem in topology. So this is a homomorphism problem, no? That's a homomorphism problem. This you can say is one of the central problems in topology, of topology, in topology. The homomorphism problem says that given two topological spaces, x and y, decide if they're homomorphic or not. Very natural, no? Given two topological spaces, x and y decide whether, if, whether, x and y are, sorry, if there's a homomorphism, then you say x and y are homomorphic, okay, of course. If there is a homomorphism, then you say x and y are homomorphic, okay? I don't write it, but it's clear, okay? Homomorphic groups, homomorphic spaces, okay? So given two topological spaces, it decides whether x and y are homomorphic. So I can write it here, are homomorphic. Homomorph, homomorphic. That means there exists a homomorphism, no? There exists. That's the central problem of topology. This one, okay, you can say. There exists a homomorphism. There are f from x to y. There are two aspects of this, okay? Two aspects. If you're given two, for example, examples. Well, I write minus one, one. x is minus one, one. The real interval, okay? In R, right? This is the real interval, okay? Standard topology, no? Standard topology, from analysis. And R, the reals itself. So these are two spaces and you say, are they homomorphic or not? What would you say? No? Not the same, they are homomorphic. Let's say, sorry. They are homomorphic. And how do you prove that? Construct the homomorphism. That's analysis, okay? The what? Yes, yes. It goes minus one, minus pi over two, or pi over two. All intervals are homomorphic, okay? Yeah, the tangent is okay, yes. So, construct the homomorphism. There are many ways to construct the homomorphism, no? Construct the homomorphism. That's the only way to prove that. The tangent is okay, no? But it's not minus one, one, it's minus pi over two, pi over two, okay? But any two intervals are homomorphic, okay? Homomorphism, by the way, is an equivalence relation, no? Homomorphism, so a remark, maybe trivial, but homomorphism is an equivalence relation and Buster. It's an equivalence relation. On what? On topological spaces, on all, okay? It's an equivalence relation. So intervals, all intervals, okay? So, and here, you can, as you say, I can take this interval, no? They are homomorphic, okay? All intervals, all open intervals are homomorphic. All open intervals are homomorphic. You can get some linear homomorphism, okay? First, you can center around zero and then you have a linear, okay, multiplication with some constant, no? Okay, we have two intervals. First, you translate to zero, right? You have an interval somewhere out in R, in the reals, no? Then you have the midpoint, okay? And then you get the translation, so the midpoint is zero, okay? That's what you do with both intervals and then you have two intervals around zero and then you just multiply with lambda, the smaller, where lambda is the ratio of... Yeah, that's analysis. It's not so much our, okay? But the idea sort of... So all intervals are homomorphic. And now comes the intriguing question. Better like, so problem, examples. Examples, that's the first example, okay? And now two are R, again, the standard. And now I write RL, yeah. RL is a strange space, no? So you say, R is a standard, R, RL, that's some strange topology. So you say not homomorphic. But then you have a problem, no? Now you have a problem, yeah? It's easy, no? Not easy, but you have two spaces. If they are homomorphic, you think and think and then you see the homomorphism. Here, you have to exclude the homomorphism, okay? That's a different order of difficulty, no? You have to exclude. So in general, how do you exclude that? The only way, you need some property, some topological property, okay? Which one has the other not, okay? So need some... One needs... That's the point. Some topological property. So this I don't define, but we give examples, okay? Topological property, okay? Which one has and the other not. A topological property is... Which one of these has and the other does not have, okay? And the other not. So we don't have many topological properties, okay? Sorry? One point in RL is not... Oh, that's not true. It has not a discrete topology, RL. Be careful. Points are not open in RL. That would mean the discrete topology. Okay? Points are not open. RL is not the discrete topology, okay? It's not... So you may say I met at RD, the discrete one here, okay? So let me write topological property. So a certain example, then you say R and I take RD, discrete, okay? And then you would say they are not on your morphic, not on your morphic. So how do you prove that? You say it's easier this one now. Why... How do you prove that R... R is not discrete, okay? It's not the discrete top... R has not the discrete topology. Now that's the point. You say points are not open, okay? In R. So this is easy, no? Since... That's what you say, okay? Since points are not open in R. Are not closed in R. Not open in R. Open, open, open. Since points... If there would be a homeomorphism, no? Then points are open in RD, discrete. And so the image of a point is a point and a homeomorphism makes open sets to open sets, okay? But this is... So this is what you say, okay? Since points are not open in R. R has not the discrete topology. Now that we know, okay? R has not the discrete. So that's the property, topological property. Points are open. That's the discrete topology. That characterizes the discrete topology, no? But that's the topological property. Since R has not the discrete topology. That's clear. Being a discrete space. Exactly, yes, right. That's the topological property. Being discrete, okay? Yeah, this is different. Because we don't have many properties so far, no? And yes? Yeah, that... So in R, L, certainly we have open sets which have a minimum, okay? But this is a notion of order. This minimum, no? And now you say this open set has a minimum in R, L. And now you apply F. And then you have an open set in R. The standard rules, right? And then they say this has not a minimum, okay? But why should a minimum be preserved by such a map? It's not order preserved or whatever. This set, 0, 1, they are open in R, L, no? Of course this is not open in R, that's clear. But the image in R, you don't know the homomorphism. The map might be very complicated. It might not preserve the order. So then you have to prove that if this has a minimum then the image should have also a minimum in R. But then I'm not convinced that you can... There's an easy proof to do that. Here, this 0. No. No, no, the map is very complicated, okay? It's the identity map. The identity map is of course not. But this map is extremely complicated. It's on rational numbers, irrational. It's a big distortion map, okay? Which makes everything complicated, okay? But it's subjective. Well, you can try to... That's an idea of course, okay? You say this is a property... Well, a property of... The point is that this notion of minimum, okay? Or maximum. It's not purely topological, no? It is order, okay? It's not purely topological. It has something to do with order, so we have more structure, okay? That's the problem, right? It's not purely topological, okay? So there's a problem. So the topological property is house stuff, no? But R is house stuff. RL is house stuff. That's the only we know. And so now I give you the property, okay? It's a good point to introduce the property, okay? To prove that. Well, what short? You have to use some property. Which one? Ah, you're talking about... Yeah, that's a good point probably. You will be talking about connectivity, okay? We will forget about connectivity at this point. Yeah, but that's... You're talking about R is... Yeah, that's true, but that you have to prove. I mean, this is connectivity. And this is... That we will do anyway, but you... That's true. You're right. So connectivity distinguishes, okay? Yes, yeah, connectivity. But then you have to prove that R is connected. Okay. Now that means R is... That's the definition of connectivity. The only empty... The only open and closed sets are... That you have to prove. If you prove that, then you are done. But that you have to prove. That means connected. That's the definition of connected. And then you have to prove R is connected. That's true, of course, okay? That's the topological analysis. And we will do connectivity, of course. Now that's the topological property. Connectivity distinguishes, okay? But then you have to prove the theorem first. R is connected. The theorem. Which is not completely trivial, okay? This is on the basis of everything in analysis, no? This is... Of course, it's... Yeah, that's a good point. The connectivity we will do later, okay? Not today, so... So we need something easier. So it's a good point to do this, because we're talking about bases. Okay? So we have bases. We know what is the basis for topology. And so the fact is, R... So has a countable basis. So this is the definition of first countable. No, no, of second countable. So R has a countable... What is the countable basis of the wheels? So what is the basis of the wheels? The bases are all open intervals. And what is the countable basis? With... Rational points, yeah? So a countable basis. All intervals... B. No, no. All intervals with rational inputs. B is equal... to... AB. A smaller B. AB and Q. This is countable. This is a countable basis. Well, you make the usual picture now. This is exercise. This is also one of the exercises. It's a book, by the way. So you have to... You have to prove this is the basis. How do you do this? You have the standard basis, all open intervals. And you have to compare. And then you apply the lemma. So you take... This is an arbitrary open interval. Okay? So what is it? CD. This is an arbitrary open interval. This is an exercise. And what... So we have... Now we have this collection here. C, okay? The lemma says... This is a basis for our topology. If, given any... open set, basis element is sufficient, given any point here, we find some of these which contains the point and this contains here. Then this is the basis for our topology. That's the lemma, okay? And so how do we do this? Yes, the density. So this interval contains... Any interval contains point of Q is dense. Q is dense. So this contains... So we have this interval, no? No, sorry, which one? And they contain rationals. Any interval contains rationals. And so we have AB, no? And rationals, and so we have X. So if you have to write this proof, then... we have this, okay? And apply a lemma, okay? So the lemma says this is the basis. Okay? It's a countable, it's clear, no? It's countable. Q is countable. So this is like... which is countable, no? Q times Q is countable. Countable, yeah. Countable set times countable set is countable, no? Well... That needs some proof also, but... RL does not have a countable basis. If this is the case, no? Then we are done. Then they are different, okay? So this implies R is not... So this implies R... So homeomorphism, you write it like this, no? Homeomorphism. But it's not homeomorphism. It's not homeomorphic to RL. Because this has a countable basis. This does not have a countable basis. But if you have homeomorphism, they are the same, no? Homeomorphism means we have... the properties are exactly the same, no? This is a bijection, and it's just other names for the points, okay? But these are the same space, the names for points, okay? That's the homeomorphism. So RL is not a countable basis. And that's a nice proof. So proof. That's in the later chapter of the book. But it has only to do with basis, so it's a good point here, okay? We define basis, so proof. It's in chapter... We are doing chapter two, okay? This is chapter four. So it gives a definition in the second of the property. So let's be... be any basis... We don't know anything. Any basis for RL. So B is not this basis, you know? Any basis. And what we have to prove is that B is not countable, okay? No countable basis. But we don't know what is this basis, no? What? So the proof is easy, but nice. So we have X... X plus one. This is open in RL, no? That's a standard basis. Not our basis, B, you know? By the way, the standard basis... This is a standard basis. This is not countable, no? Because we have too many points here, no? We have all reals here. We cannot, okay? So it's not so clear. You can try to do rationals here, now, okay? That is not the right topology. Okay? You need all reals here. So this basis you cannot restrict to a countable, okay? That doesn't work. Well, anyway, this is open in RL. And X is here, of course, no? So we have a basis now, B. This we don't know, okay? But anyway, we have an open set, we have a point. What means basis? We find the basis element which contains the point is contained in the open set, no? That's a usual stuff. So there is, there exists BX in B. So it depends on X. Such that X is in BX, which we don't know, but it exists. And this is contained in X, X plus one. This is an open set, no? Okay? So we get, for each X, we can do this. We can find this, okay? So we have, for each X choose BX for each X, okay? Choose some BX for each X. Choose a BX for each X in X. X in X is where? In RL, no? It's a set there. So such and BX for each X in RL, okay? So we have, and then we have a function from R to B X, and for each you have such BX, okay? For each X you choose this, because it exists. And now we want to prove that this function is injective, okay? That they are all different to BX. If this is the case, then R is, so we have phi, function phi. This is not countable, no? If this is injective, then this is not countable. Right? So let's see. Suppose X and Y are in R and X is different from Y and X is smaller than Y. Doesn't matter, Y is smaller than X. X smaller than Y. One is smaller, okay? They are different anyway, okay? And then we want to compare BX and BY. No? So BX and BY, right? For each X we have BX. For Y we have BY, okay? So we compare BX and BY. So BX and BY. And we want to compare this. We want to prove they are not equal, no? That's injectivity, okay? That's injectivity. They should not be equal. So BX is contained in X, X plus 1. BY is contained in Y, Y plus 1. Right? Now X is smaller than Y. X certainly is contained here. But X is smaller than Y. X is smaller than Y. So it's not here. So X is here but not here. And so this implies BX is different from BY. And this implies phi is injective. So then phi is injective. And R is uncountable, no? And that means, so R is uncountable. And this implies that B is uncountable. Not countable. So this basis was not countable. So this implies there is no countable basis. Okay? Because this is arbitrary basis, no? So RL has no countable basis. If you want to read this, this is chapter 4 of the book, okay? This proof. And the definition, so I give now the definition. So this is proof, okay? R, the standard, has a countable basis. RL intervals. RL, the strange topology, lower limit has no countable base. That's not every countable base, okay? So they are different spaces. Maybe that's another proof is of course using connectivity, okay? That we will prove later. So there are various properties, no? But we have to find one. So definition. This is an interesting property. So definition. The topological space B, X is second countable if X has a countable basis. It's a topology of X, of course. But if X has a countable basis, that's the definition. Well, this is a strange definition. I mean, we may just say countable basis, no? But we can take second countable. The point is, the second countable you suspect that there's first countable. So you have another property. There will be another property, okay? That's why it's second countable, no? If there's no first countable, then you just say countable basis. That's it, no? You say sometimes second countable because there's also first countable. It is another important condition for a topological base. But we will see much later, okay? So this is second countable. And this is a nice proof. RL is not so corollary. It's the same as before. RL is not second countable. That's the way to say it. RL is not second countable. The interesting point is that we talk what is a topological property, no? What are nice topological properties, okay? This is the main point of general topology. The topological properties, okay? What are the important or interesting topological properties of topological spaces, okay? Whereas the main point of algebraic topology is not topological properties, but topological invariance. Numbers, groups, vector spaces. For example, an algebraic invariant. Let's talk about one minute, it's finished now. One minute about groups, okay? There's a big difference here. If you have a homomorphism of groups, you will do algebra, but you know what the homomorphism of groups is, okay? For homomorphism of groups, this is bijective, okay? Then you have the inverse map. And this is a homomorphism, okay? Automatic. If you have a linear map between vector spaces, and this is bijective, then the inverse map is linear. That's automatic. You don't have to, you can easily prove that, okay? In algebra, if you have bijective and preserve the then the inverse map is okay, it's the same, okay? Isomorphism, it's a quote. Isomorphism, no? And if you want to distinguish two groups. So here you have the homomorphism problem for topological spaces, okay? Decide if two topological spaces are homomorphic or not, okay? If you want to prove that they are not homomorphic, you need something. A topological property or a topological invariant. If you want to prove that two groups are not isomorphic, you need something. You need, for groups, it's more algebra. So you need maybe a group, not topological group invariant, okay? Invariant of groups. What is the easiest invariant of a group? Yeah, for example, yes. That's... Yeah, exactly. Invariant is the order. How many elements, okay? If you have a group with 10 elements and you have a group with 20 elements, they are not isomorphic, clearly, no? Algebraic invariant, okay? The order of a group, this is more already a little bit more sophisticated, no? You have elements of order two in one, no elements of order two in the other, so they cannot be isomorphic, okay? Because isomorphism preserves order, no? Yeah. And so general topology is largely the study of topological properties, okay? Topological properties. House of first count of the, second count of the, connected, compact, the various notions of this, okay? Not just one. An algebraic topology is studying topology invariants, okay? So this will be groups. We have studied in the second part, fundamental group, okay? And then there are homology groups, co-homology groups, homotopy groups. Yeah, so there are many algebraic in there, okay? So this is one difference between these two. General topology and algebraic. That's algebraic topology, algebraic invariants, okay? Dimension, maybe. But dimension is also a number, no? So dimension, as if a group is the order, no? It's a number, just a number. The dimension, but dimension is difficult to define. For vector spaces, the dimension is the first invariant, okay? For vector spaces, the first invariant is the dimension, no? Yeah, here goes both directions, no? I'm only wondering, if they are isomorphic, they have the same dimension. And then it happens for vector space. If they have the same dimension, then they are also isomorphic. So this is a complete invariant. But that's... difficult to find. Vector spaces are easy in some sense, no? Groups are difficult, okay? There are many... You don't have anything equivalent for groups or for topological spaces, okay? So in this sense, vector spaces are easy, yes. You have to write it, okay? And comment. So I made a small list. You have to write something, okay? You have to start writing, okay, in English. And next week, I... collect... exercises. Because the final exam will be exercises like this, okay? And then you have to write in English, okay? You have to be careful and... So I give a couple of exercises. First group. These are... I comment on the exercise. What is easy, what is not easy, okay? For exercises. So page 83. And here, 6. 6. This I introduced because there is... introduced, which we can do here, but you have to... there's a new topology on the reals, okay? Which is also some nice property. So, and this is 7. So 7... only for T1 and T2. T1 is R. And T2 is... So this is the other topology, RK. It's called RK, K-topology. You have to read what is the... this topology on R, okay? And then you have to compare these two. And this is also for RK. This is also an exercise for this topology, okay? So I give this exercise so you... And here, you just give the answer, okay? Because you don't know how to write, maybe, okay? You have to compare topologies. And then we make my picture, okay? And you say what happens, okay? You just think and if you understand, then it's okay, right? Sorry? Yes, in the book, of course. So just give the answer. Because you don't know how to write that in a good way, maybe, okay? So they say this topology is not comparable or this is finer or strictly finer, okay? What happens? You want to know what happens. And here it's similar, okay? You compare RK with R, I think, okay? So you just think what happens, okay? Which is finer? Are they comparable? No? It's an exercise to think not so much to write, okay? Then there are some formal exercises and here you have to write. Give the answer for this. Then page 92 Exercise 4 Here you write. These are easy exercises. You should write, okay? In a good way. And also page 101. Page 101. So these are very easy exercises. 10. Except you have to think how to write that in a good way. That's the main point. 10, 11, 12, 13. So these exercises you write. Write the proofs, okay? Here you have to write the proofs. Write down the proofs. And they are easy. They are very easy. So it should be no problem. Except for the English, maybe, okay? And then there's the last one which is like this one where you have to think a little bit. And this is strange. Page 101 And this is exercise number 18. I hope I don't mix up. First of all, can you show me what you did? Yeah, 18 is a long exercise. Determine the closure of the order square. This is an exercise for this is an exercise for for I square orders, which is important, okay? That is just give the answer. Because it's too long to write this, okay? You have to think what happens, okay? And you just give the answer. Like this one, okay? These are the formal exercises. Just give the answer. That means you have to understand what happens, okay? You think about it and just give the answer. Because otherwise it's too long. There are five cases, no? It's interesting because you have to work with this space, okay? The order square and here you have to work with this space. Okay? So these two exercises are because you have to know what this is and what this is, okay? So I give these two exercises. The other ones are easy. Very easy. Okay, short. But you have to write in a nice way. That's the problem, no? You have to concentrate on that. Okay? So these are the exercises for next week. And you write down. You have to write down because sorry? Tuesday. Yes, yes, exactly. One week. But you have to start writing. That's very important, okay? Because the final exam is eight exercises, okay? Eight, nine exercises. If you never, if you don't do that regularly, you have a problem, no? In some sense, analysis and algebra are more computational, okay? It's not computational. You have to understand, okay? What happens? So this is it's a little bit different. Okay? The approach. The analysis you have of proofs also, until algebra, of course, but you have also computations, no? Here in topology, it's more logic or proofs, okay? Mainly proofs and constructions, okay? And then you have to write in a good way, okay? So you have to concentrate on the way how to present.