 Okay, so let's start. So we are discussing various examples of topologies. Last was product topology of two spaces, another one subspace topology. So the notation, we have all those x topological space. So it's set with a topology, topological space, and there's a subset which is called y. Y subset of x, okay? We have a subset of a topological space. We want to define a topology on y. So t, y, the topology on y, these are all intersections. Y intersection with u, such that u is open in x, okay? U open in x, u in t. So this is topology of x. So this is a topology, and this is a subspace topology. So this is a definition. Well, one has to prove that the topology, this is very easy because what is the three conditions? The empty set is, of course, the intersection of the empty set. The whole set, y is intersection with the whole space, x. And then we have arbitrary unions. Well, for arbitrary unions, what do you have? Arbitrary unions, no? Union alpha and j of open sets. So the open sets are y intersection, intersection u alpha, no? Y intersection u alpha. Where these, so these are the open sets. These are open in the whole space, no? These are typical open sets. And of course, this is set theory again. So it's y intersection union u alpha, alpha and j. And so this is also open in x. This is open in x, okay? This is for the union. And the same for the intersection, y from 1 to n, finite intersections, no? Y intersection of ui, I change index, no? This is a finite, finitely many, u1, un, okay? Yeah, I don't know, u alpha, so I cannot. So this, of course, is equal. So these are again open in ui. So this is again y intersection ui, i from 1 to n. And again, this is open in x, finite intersection of open sets. So this is more or less proof, okay? Which is easy set theory, all those. So this is the topology on y. And this is a subspace topology. That's the definition. Okay, so as for the product topology, if you have a basis, so there's a little lemma observation, more lemma. If b is the basis for x, for the topology of x, then I should use the notion then by. So what we take is, we take not all intersections, but only with the basis. So y intersection b, b in b is the basis for y, for the subspace topology. It's the basis. So how do we prove that something's the basis? I will not write the proof, but I'll make a picture, no? But this is proved by picture. We apply some lemma as the basis. For each open set, for each open set, each point in this open set. We should be able to find the basis element that contains the point and it's contained in the open set. Then the lemma says, this is the basis, okay? And it generates a topology, which we have. That's the lemma, okay? Which we always use in this situation. So what we have? We have x and we have a subspace y, a subset y. And we have an open set, okay, and a point. So what are the open sets in y? The intersections, so this is an open set in y. This is the u, okay? So this is u intersection y, u is open in x, no? And we have a point, we take a point, okay? Then we should be able to find, so this is x, small x, okay? So we should be able to find the basis element which contains the point and it's contained in this open set, no? And then we apply the lemma. However, this point, of course, is in u also, okay? And we have a basis for this topology of x. So we find the basis element which contains x is contained in u. The basis element in the basis b, okay? This is b in b, this basis element, no? And then it's clear that we intersect b with y, okay? And then, of course, x is contained, so we get here this part here, no? In b intersection y, oops, b intersection y. And this is contained in u intersection y. So this is a proof, okay? Trivial proof, okay? You make a picture and look at the picture and then you write down what you see. Okay, so now I give one example here of subspace. And this is a little bit strange, but it's important. It will be important, example in the book. So here's an exercise, or example, better example. So i is the interval, 0, 1 in R. So this is a real interval, okay, i. That's why we don't call an index at i, because in general i is this interval. That's the notion of the book. And then we take i cross i. Of course, i cross i, we have here the subspace topology, okay? This is the standard topology, and we have the standard topology on the interval. Then we have the standard topology here, which is the product topology. That's not very interesting, okay? So what you take now, so this is contained in r cross r, that's clear, and we have the standard topology if you want. But that's not what I want now. What I want is, so this is ordered, okay? So we have the, what did we say? Dictionary order on i cross i, dictionary order. It takes a dictionary order on i cross i, okay? And with this topology, this is called the ordered square in the book, okay? Then we get the ordered square. That's the definition of the ordered square. And it's written in the book, i squared ordered, that's an o, okay? The ordered square. So this is a notation. This is one of the many examples of the book, which we will see again and again. I will discuss it more and more. So we take here the dictionary order on i cross i, okay? First coordinate, second coordinate. And then we take this topology, the ordered topology, and this is the ordered square. The ordered topology, okay? Dictionary, we take, of course, the ordered topology here. And then we get the ordered square, okay? That's an important example and that's a notation. Of course, you can do something else here, okay? You can take here the ordered topology, okay? We can take here the ordered topology of the dictionary order, okay? That's the only we see of the dictionary order. You can take here the ordered topology. And then we can take the subspace topology, okay? Then take the subspace topology. Then take the subspace topology, as we defined, on i cross i. So we have the subspace topology of this ordered. And we have directly the ordered topology, okay? And the observation is they are not equal. One is an interesting example. The other is not so interesting, okay? So what happens is, there's an exercise. One of the exercises in the book, but it's important so I do. So observation, you would maybe guess they are equal, okay? They are not equal. So the exercise is the subspace topology is strictly final. Then the dictionary, okay? The subspace topology, in this sense on i cross i is strictly final. They are not equal, these two topologies, but they are comparable. It's strictly final than the order topology of the order square, okay? Then the order topology. So this is the situation. So we make, how to see this, okay? So we have the order square, we make p cross. We have the order square, i cross i. That's i cross i, okay, the square. And then we have to prove, we have to prove two things, final, no? The subspace topology is finer than the order topology, okay? So we're taking open sets in the subspace topology. Well, we can work with the basis, okay? The basis for the subspace topology. First, we have the basis here, which are intervals, okay? These strange intervals. There are two types of intervals, but we need only one type of interval, okay? So intervals and we intersect. So what are intervals? Intervals are, for example, these types of intervals, no? This, this, and we have this type of interval, okay? In R2, in R cross R. And then we intersect, and what do we get? We get, we get this one, okay? So this is open. That's a closed interval, the intersection, okay? That's a closed one. This is open in, in which, in the subspace, in the subspace topology. This is open. That's a closed interval, which we get, okay? I, I started with the wrong, okay? I should, if you prove finer, I should start with the other topology. This is smaller topology, okay? But this is open in the subspace topology, but I can continue here. But, so they're not equal to topology, but not open, but not open in the, what is the other one? Order topology, okay? This is open in the subspace, that's clear, no? This closed interval, you get this intersection with this larger open interval in R2, okay? So it's open, by definition of subspace topology. That's an example of subspace topology, okay? But I say it's not open in the other topology, in the order topology, why not? Well, of course, in these points, we have no problems, okay? If we take such a point and we take an open set in the order topology, the problem is this point and this point. These two points are problem, no? Because if this would be open also in the order topology, then given this point, we should be able to find the basis element of the order topology which contains this point and is contained in this interval, right? That's always the same condition, no? So again an interval, we're in the order topology, but on this small, we don't see R2 now. We have only this space, the order topology, okay? These two points are not there now, okay? So I need an interval which contains this, right? Well, I have no problem with the lower bound, obviously, no? As a lower bound, I take anything here, right? That's clear, but I have a problem with an upper bound. This point I cannot take, that's not an element of I cross I, right? I need an upper bound. So what I have to do, I have to go to the right. So with the upper bound, I have to go to the right. And what does it mean? I have to take some point here, I don't know which one. But I have to go to the right, okay? But then I get the interval between this point and this point, okay, as an upper bound. I get everything, all these intervals here and up to this point, okay? The interval between this and this contains all these lines in between. So no basis element contains this point and it's contained in this part here, okay? So they're not equal to the topology, so this, that's what I write. This is open, so again, this was the example. This is open in the subspace topology, that's trivial. But not open in the order topology. So they're not equal to the topology. I should have started with the other one, okay? So now I show, this was not final. This was strictly final, if you prove final, okay? So now I prove final, okay, again. But this is clear, so now I have to start with an element of what, of, no, I lost. I have to, with the topology, which is course, okay? That's the order topology. So I take an open set of the order topology. How do they look like? Well, I take two points here and I take such an open interval, for example, no? And it's clear that this is also open. That's the basis element, okay? For the order square, this is open also in R2, okay? This is also an interval in R2. Or I have the same, of course, if I take a point here and maybe a point here now, then it's this one, no problem, open in R2, okay? So this is no problem. So then I can take points with different coordinates, intervals, okay? So this goes up, then you have all this stuff here and this is the interval, right? And it's clear, this is open R2, given any point here now. For example, this point, we should be able to find a basis element of the subspace topology, which contains this point and this content here. Well, if this takes this point, there's no problem again, okay? This is clear. Here we can take again this, but also for the other points. You may think, the only points, maybe look at this point, okay? This is here and now, but this interval here is intersection of something like this, okay, in R2. So the intersection of this with this is this, okay? So this is also, okay, we find the basis element. So the difference is the basis elements, maybe you find this easy. The basis elements for the subspace topology. So what are they? They are open intervals like this. There are also these intervals, okay? Which is intersection with the larger interval here. And of course, we have also, for example, these, okay? And these are sufficient because these are, we take intersections in R2, of intervals in R2 with this. And we get this kind of stuff. And for the order topology, it's different, no? The basis, this is not this. So we get again, of course, these intervals, no problem. But we don't have this kind of stuff, okay? What we have is this, then, if you don't want to stay on one line, then we have to go and forget immediately such a kind of construction, okay? Because the upper point of the interval, well, we have to go to the right, okay? If you want this point here, we have to go to the right. We cannot go up because we are not in R2, we are in I cross I, okay? So these are two topologies in I cross I. By the way, this topology is not interesting, why not? Which is this topology for the subspace topology? What is this topology? I mean, you see all these. Of course, you have also these, no? These are open also, no? So what is this topology? You can give, describe in a different way. Yes, discrete times standard. So this is subspace topology, so this is equal, okay? To I discrete topology times I standard, okay? That's exactly this topology. I discrete with a discrete topology. So this is a discrete topology, and this is, if you don't say anything, it's standard topology, standard. And you get this topology. Well, you get basis, the basis of the product is basis times by basis element, okay? Here you have points, okay? And then times basis elements of I, what are basis elements of I? We need also these intervals. Now, because it's bounded, we need open intervals. Half open intervals, if you have the largest or smallest element, no? And that we have here, okay? We have the largest or smallest element. So this is standard, this is just this product, okay? And it's not so interesting. What is interesting is this one. This one is interesting, okay? This is many open sets, okay? Too many to be interesting, okay? Just discrete times standard and many open sets. So you can forget more or less this, it's just an exercise. But this one you will see, okay? And this is I squared, ordered square, okay? That's the notion, that's the ordered square, ordered square. And the basis looks like this. And this has much more open sets than this, and this we will see, okay? So this is an example for subspace topology, order topology. This is an exercise, okay? So you should under, this comes back, you should remember this, okay? This was an exercise. The exercises are very often more interesting than the proofs. I will give tomorrow some exercises for home tomorrow, okay? So this is a subspace topology and some example. And now we have closed sets and limit points. This is very formal again. So closed sets and limit points. This is like real analysis, limit points. Maybe not limit points, it's not so clear. So what is a closed set? So X is a topological space. So the closed sets are complement of open sets, okay? So A in X is closed if, that's the definition. X minus A, the complement, that's the complement is open. Well, it's equivalent in some sense. You may work with open sets or with closed sets, no? The axioms for closed sets is empty set and X are closed. This is the complement of X, this is the complement of the empty set. Then arbitrary unions of open sets are open. Finite intersections of, finite unions, sorry, finite unions. It's just the opposite, no? Finite unions, that's the morgan of closed sets are closed. And arbitrary intersections of open sets, of closed sets are closed. So this is the morgan, okay, again. I mean, this first is trivial, this is trivial. Finite unions of closed sets, so union, finite union of closed sets. So what are closed sets? Closed sets are complements of open sets. So it's X minus, no, Ui, I from 1 to n, no? So this is a closed set here, this, I mean, this is open, and this is closed. So finite union of closed, and then we have the morgan, what does it say? This intersection of X minus intersection of Ui, is that okay? That's what it is, okay? And this is open, but we need to find that intersection. Otherwise, we have a problem, okay? And the same, of course, for the other morgan. So we have arbitrary unions, X minus U alpha, alpha in J, no? We have an arbitrary intersection, arbitrary intersection. So this is again X minus union U alpha, alpha in J. And here we have no problem because arbitrary unions are open. So this is open. And that's the morgan, two times, okay? So these are the axioms for closed sets. So we have, we prove two lemmas, which are almost trivial, but used often. So the first lemma is, so if, why, we have subspaces, no? We define subspace, subspace topology. So Y in X is subspace, let Y in X be a subspace. This means that X is a topological space, Y is a subset with a subspace topology, okay? So this is a short way to write that. And then a subset A in Y is closed. So what you suspect is, open sets are intersections with open sets from the big space, no? And closed sets are same. So if and only if, then I is closed, if and only if A is intersection, C, C closed in X, yes. So the open sets in the subspace are intersections with open sets in the large space. And the closed sets are intersections with open sets, okay? So maybe, the proof is very easy, but so I will write the proof. Well, you take compliments, no? I might take compliments, okay? So let's write the proof, okay? So we have two directions anyway, so we have to be careful. So suppose A is this one. Suppose A is Y intersection C, where C is closed in. So what do we have to prove? We have to prove that A is closed in, yeah, A is closed in Y. So Y minus A is open in Y. That's what we have to prove, okay? So this is closed in X, what does it mean? That means X minus C is open in X, no? That's the definition, compliment. And this implies that X minus C intersection Y is open in Y, yes. That's the definition, no? This is open intersection Y, these are exactly the open sets in the subspace, intersections with open sets in the large space, okay? However, what is this? This is Y minus C intersection Y, if you look at this now. You have to throw away C, okay? So Y minus C, C intersection Y, okay? Because we are in Y, so that's clear that this is the same. And this is Y minus A, okay? So this Y minus A, and this means of course that Y minus A is open in Y. So the compliment is closed in Y, no? So this implies that A is closed in Y, since its compliment is open, okay? So this is one direction, okay? That's one direction, and conversely, maybe we did this, so conversely. One has to concentrate here, no? What to write, conversely is very depends on logic. So which direction did we prove? We prove this direction. Now suppose A is closed in Y, okay? So conversely, suppose A is closed, and we have to prove that it is intersection of a closed set in the whole space with Y. No? That's what we have to prove for the other direction. So A is closed in Y. This means Y minus A is open in Y. And that means that the open sets we know, so this means Y minus A is equal to U intersection Y where U is open in X. The open sets are intersections. So what have you proved? No, no, I don't see anything. So this means, what is U? U is open in X, no? So X minus U is closed in X. I have to write no, closed where, no? Closed in Y, closed in X, otherwise we have problems, no? And Y intersection X minus U is equal to Y intersection. No, sorry, Y minus Y intersection U, no? This is what we have to throw away, Y intersection, avoid U, okay? So Y minus Y intersection U, so I want, this is A, no? Of course, this should be A, that's clear. So this is Y minus, maybe I'm exaggerating, intersection A, so this is A. I'm saying, that's what I want, this is A, what is U? U intersection Y is, sorry, I wrote something wrong, no? Minus A, U intersection Y, no? Y minus A, so it's okay. It's two times the complement. Okay, so here one has to concentrate, okay? On taking complements, open where, no? So one, it's not so pleasant to write, okay? You have to concentrate a lot on these kinds of proofs, okay? The same is true for the next lemma, which is also. So this finish is a proof, no? So the next lemma, but they are used often, so, in particular, in particular the next one, so maybe I prove that also. So what's the next lemma? The next lemma says, I call everything lemma here. The book is proposition, sometimes, whatever, I don't know. So Y subspace of X, well, I mean right, subspace of X. Again, as before, sorry. I have to define the closure before, so definition. Before I can write this lemma. So the closure, what is the closure? So I is a subset of X, X is a topological space. So this is a subset and X is always a topological space now. So we have the interior, which is interior, interior of A, also written in this way. This is a union of all open sets, which are contained in A. So U, open in X, U contained in A. So arbitrary union of open sets is open. So this is open, okay? Open in X, this is open in X, clearly. Arbitrary union of open sets, no, it's open in X. In fact, it's the largest open set, obviously. It's the largest open set contained in A. The largest open set contained in A, that's what it is. That's the interior, I didn't write, the interior of A. And you write interior of A or A, interior, okay. More interesting is the closure. So then we write sometimes closure of A, more often this notation, the closure of A. So this will be the closure of A, so what is this? So now we take the intersection, we take dual to this intersection of all closed sets. So C closed in X, such that A is contained in C now. So arbitrary intersection of closed sets is closed, so this is closed, okay? So closure of A is closed, it is closed in X, certainly. It is closed in X, and in fact, it is, what is it? Yeah, exactly, the smallest, because it takes intersection of all closed sets which contain A. The smallest closed set containing A, that's what it is, okay? The largest open which is contained, the smallest open which contains. And if you change roles here, it makes no, it's in general not well defined, okay? Because arbitrary intersection of closed sets will not be open, and arbitrary union of closed sets will not be closed in general, okay? So you can define these two, but you cannot change, what is not well defined in general is, so this is the largest open set contained in A, the largest closed set contained in A, it's not well defined in general, okay? The largest closed set contained in A, you cannot define, okay? Because you, and what is also in general not defined is the smallest open set containing A, that's also not well defined, okay? Well, you can try to write, but then you get, have to take arbitrary union of closed sets and arbitrary intersection of open sets. That's in general not open and not closed, okay? It's defined, but it's not open and not closed. So it's not, okay. So now comes the lemma, which I wanted to write immediately for getting to define. So this is the definition, and now this lemma, which I start again, so what does it say? So why is the subspace of X again? Why subspace of X? So that means X is a topological space, Y is a subset, and we have the subspace topology, okay? Which we defined. A subset of Y, and A is a subset of X, of Y, sorry, A subset of Y, subspace, but we are a subset now, okay? A subset of Y, then the closure of A in Y is, so what, you take the closure in the whole space and intersect with Y. So it's A closure intersection where this is the closure of Y in X. So this symbol is also in reference to the large space, okay, not where. So Y is closure of A in the large space, okay, always in X. Okay, so let me prove this also. It's also concentration, again, one has to concentrate, okay? It's not so pleasant, always, because one is tired, one has to apply definitions, okay? What is the definition of this? What's the definition of this, and then apply, apply, no? And then it's almost canonical. In any case, it's useful, so we have two closures. We have the closure of A in Y, and we have the closure of A in X, no? So it's better to give a name for the proof, to the closure of A. We have a name for the closure of A, so let B be the closure of A in Y, and we have to prove that B is equal to this, okay? So a claim is, of course, that B is nothing else than A bar, so this is the same. And so we have two inclusions again, no? As usual, here we have these two inclusions, maybe not. Okay, let's see. So A bar is closed in X, by definition. That's the closure and the whole space. So the intersection, this means, of course, that A bar intersection Y. So I don't write then, okay, we go, this means this and so on, implies, implies, no? So A bar intersection Y is closed in Y. That's a proceeding lemma, no? Proceeding lemma, that's what you're proceeding. Of course, A is contained in A bar. This is the largest, the smallest closed set containing A and it's also in Y. By the definition of closure, exactly, of closure now in Y, okay? Because now we are in Y. So the closure of A is B in Y. B is contained in A bar intersection Y. Because this is closed, no? And B is intersection over all closed set with contain A, over all, okay? So B is, okay, so this is one inclusion, which one? Well, this one, okay? This one, so we need the other one. Conversely, conversely, it's not so clear. For the other inclusion, no? That means conversely. B is closed in Y. That's the closure. B is closed in Y. So by the proceeding lemma again, okay? The lemma before precedes with one here with two here. So I'm not completely sure, no? Purchheader with one, I suppose. In Italiano, in Italian, it's this one. Yeah, it should be one also here, then I suppose. By the preceding lemma, maybe not, I'm not sure. One is sufficient, it seems, no? By preceding lemma. So we have a closed set in Y, so what is that? B is equal to, the closed sets in Y are C intersection Y where C is closed in X. That's what we proved before, no? I'm writing two large, it seems. So what we know is that A is contained in C. This clear A, of course, is contained in B, you know? That's a closure. So A is contained in C. C is closed in X. This is closed in X, which I'll write again. This is closed in X. And this implies that A bar is contained in C again, no? Because this is closed in X. A bar is the smallest closed set. And then finally, we have A bar intersection Y is contained in C, intersection Y. And that was B, and that's it. That's the other inclusion, so they're equal. So one has to concentrate on notation, no? If one is tired, it's not so, but everything is canonical, okay? No ideas, but applying definitions, definitions. But we have to use the notation, and we have to think all those, at which point are we, okay? So this is this lemma. So now, so I start again here. I mean, this definition is somewhat awkward, no? Of the closure. The largest, the smallest closed set containing the intersection over all closed sets, no? You don't want to find all closed sets containing and then take the intersection, no? So what we apply is the following proposition. And so A is a subset of X. X is the usual topological space. And X is the point in X. X is in X. So I, then, the following. X is in the closure of A. You want to decide if this point is in the closure of A, no? If and only if every open set containing X intersects A. So now I introduce some notation, which is very useful, okay? If and only if every, and now I write neighborhood. I call this neighborhood. But one has to see what is the neighborhood, okay? They have not one, two possibilities at least. Every neighborhood, it's a long word, but we will abbreviate later this. Every neighborhood, U of X, and what is the neighborhood? So I give a definition of neighborhood. That's an open set containing X. Open set, that's a neighborhood for us. An open set containing X. That's a definition. There's another possibility of neighborhood, more general. It might be a set containing X, and X is in the interior. So X, there's an open set, which contains X as in, contains in the, maybe the larger is not open itself, okay? But for us, neighborhood is open. Every neighborhood of X intersects, what means intersects? Intersects means has non-empty intersection with X, no? Has non-empty intersection with A. So that's the shortest way to say it. Every neighborhood of X intersects A. Then X is in the closure. And the second part is, we can take the second part is, if B is a basis, B a basis of X, yes, X, of the topology of X, B a basis of X, then X is in the closure of A. If and only if every B in B, which contains X, we don't have to check it for all neighborhoods. Every B in B, which contains X, intersects A, okay? So we have to check only for basis elements. Intersects A, as usual. One can always restrict to basis elements. That's useful because basis is easy to describe in general. We find the nice basis for topology, but the topology is not nice to describe in general. So it's good to work with the basis. So proof. And two, I will not prove. I mean, one direction follows from this anyway. If any neighborhood intersects, then every B in B, which is open intersects. So one direction is, if you prove this, one direction is trivial, and the other is very easy. Again, the property of the basis is given open set, given a point, there's a basis element, which contains a point, is contained in the open set now. So you apply just this, and then it is the same as this. So I, if, let's suppose X is not in the closure of A. If X is not in the closure of A, then of course, U, which is a complement, X minus closure of A, this is open, no? And it contains X, because X is not here. So it's a neighborhood of X. It's a neighborhood of X. And now I abbreviate neighborhood, okay? Because this is too long, one time, okay? But this is neighborhood, it's clear, no one can understand. It's a neighborhood of X, which does not intersect A. Because it's X minus a larger set than A, okay? Which does not intersect A. Since A is contained in A bar, no? Since A is contained in, this is the smallest close set which contains A, okay? The smallest close set which contains A. So this proves, what does it prove? This proves? Yes, it's something. So if X is not in the closure, then there is a neighborhood which does not intersect. So if every neighborhood, if X is not in the closure, then there is a neighborhood which does not intersect. If every neighborhood intersects, then X is in the closure. If every neighborhood intersects, then X is in the closure. So this proves this direction, okay? This proves this direction. Well, you have to just, it's the language, no? If X is not in the closure, then there's a neighborhood which does not intersect A. If every neighborhood intersects A, And then x must be in the closure. So this is the solution. OK. And the other one, conversely, again. Well, I will not prove this, OK? That's a very easy exercise. And so conversely, for the other inclusion. Conversely means for the other, not inclusion, for the other implication. Inclusion, implication, whatever is the case. Conversely, let u be a neighborhood of x. Neighborhood of x, which does not intersect. Suppose there's a neighborhood, u, which does not intersect a. This means that x minus u is closed. Let u be a neighborhood of x. And it does not contain x, no? And x is closed, and it contains a. Neighborhood of x, which does not intersect a. u does not intersect a. So a is contained in x minus u. Yes. And this means that's the closure of a. So this is closed. And then a bar. Because this is the smallest closed a bar, no? It's contained in x minus u. But u is enabled of x. So x is not here. It's not here. So x is not in. It's not here, so it's not here. It's not in a bar. So if x is in a bar, then every neighborhood intersects. So this is the other direction, OK? This proves this direction. So we have both. And now we can distinguish between two types of points in the closure. So this is a definition. Important for us, definition. So a is subset of x. Again, the same situation as before. So x in x, we have a point x, a small x in capital x. So x is the limit point of a. So we won't define limit point. A is the limit point of a. If every neighborhood of x, neighborhood, again, of x, maybe u of x, it's not important for the definition. But neighborhood u of x intersects a. That would be boundary in the boundary, no? That's the proposition. So these are exactly the points in the, sorry, in the boundary, in the closure of a, OK? These are the points in the closure. And now I add, for limit point, intersects a in a point, in a point different from x, in another point. Well, it's not clear why this is called limit point here, OK? That doesn't look. So we have this neighborhood. And it has to intersect a. But not in the same, maybe in x also, OK? But that's not sufficient. We need another, one other point, no? In a. And that doesn't look like a limit, OK? So we have to comment on this a little bit. But it's called limit point. That's the definition of limit point. And so there's a lemma, then, that's the closure of a. It's a and union a prime. So this is not a standard definition, but sometimes, OK? We will not, we use it only here. So a prime are all limit points of a. So this is the closure is a itself, a is contained in a closure. And view at all limit points. Proof is easy. So two inclusions again, no? This and this, as usual. So a is contained in a bar, by definition. The, what? Yeah, there's a smallest closed set containing a. And a prime is contained in a bar by the proposition, OK? So this is by the proposition. Intersects, every, every neighbor of x intersects this a, OK? In a point different, but in the closures anyway. So this implies that a bar, what do you want? A union a prime certainly is contained in it. And we have to prove the other inclusion. Conversely, again, conversely, for the other inclusion. Suppose, well, the other inclusion, which one is it? That's a bar is contained in this one. A bar is contained here. So let x be in a bar. X, and let's add x is not in a. Suppose x is in a bar, but not in a. Then we have to prove it's in a prime, of course, OK? Same proposition, no? So this is then every neighborhood of a of x intersects. That's a proposition, no? By the proposition. In a point different from a, x. Because x is not in a, OK? In a point different x, since x is not in a. So then x is a limit point. So x is in a prime. So x is in a prime. Sorry, x is in the limit point. So it's in a prime. And that proves the other direction, OK? It's not in a. Things, yes? Since x is not in a, x must, so then x is in a prime. So this implies, so this implies then that a bar is contained in a union a prime. If it's not here, it's here. If it's not here, a point here. If it's here, it's good. If it's not here, it's here. Sorry? Yes? No, no, no, no, no, no, no. Sorry, sorry. So you are asking if points of a can be limit points, OK? So for example, q, the closure. So q is in r, OK? The closure of q is r y. So every point is in the closure of r, you know? So given a point, given a neighborhood, this neighborhood contains an interval. This is the basis. And each interval contains rationals, OK? Since each interval contains rationals. In each open interval, we have a, interval is in general, if you don't say open interval. Well, I should say. Since each open interval contains rationals, what are the limit points of q? There's also r, no? So each point in q is the limit point of q. Because each interval contains this point, but it contains other rationals, OK? Not just one rationals, no? So also the limit points of q, every r, OK? So a intersection, a prime is not in general disjoint or something, OK? The points of a can be limit points, itself, OK? This is not the disjoint union. I don't know. What's the question more or less, or not more or less? So you can ask again. Take the integers, no? Then the closure is this, and the limit points. These are all isolated points, OK? OK, so this was the example. I will give a nice example as an exercise, maybe. It's the ordered square, no? And you have certain subsets, and you want to find the closure. And that's very surprising, because it's a strange topology. But I will not do it now. Tomorrow, I'll give you some exercises. Of course, why is it called limit point? That's not so clear here, no? One other point, limit seems limit. There's some limits, OK? Limit means some kind of convergence in general, no? Limit point. But here, you have one other point only, OK? Which is very weak. So there's a condition, which is we, this is Hausdorff. That's the first condition for topological space. So topological space X is Hausdorff. So that's the property. That's a name, and that's a, OK? X is Hausdorff. It's a Hausdorff space. It's a Hausdorff space. What is Hausdorff? That's an adjective, and it's a Hausdorff space. And maybe that's a, but I'm trying to, it's a Hausdorff space, OK? So what? If every two points, X distinct points, X, Y in X, have disjoint neighbors. If every two distinct points, X distinct from Y in X, have disjoint neighbors. Neighborhoods, now this is now neighborhoods. N, B, H, D, S, OK? So the picture is this. We have two disjoint points, X, Y. And then we have disjoint neighborhoods. So these are the disjoint neighborhoods. That's Hausdorff. So why introduce Hausdorff at this point? Because then we have lemma proposition. I come back to limit points in three minutes, but here's a proposition. So what is the impression of the standard topology R? R2 in analysis. Now points are closed. Points are closed. I mean points can be open if you have the discrete topology or some strange topology. If there's a standard topology on the reals or on R2 or on Rn, then points are closed. Finded sets are closed, no? Finded unions of closed sets, no? And this is the property of Hausdorff spaces. So if X is a Hausdorff space, so every finite set, every finite set, every point, in particular, every point in a Hausdorff space, so in particular, every point, in particular, sorry, every point in a Hausdorff space is closed. It suffices to prove that every point is closed, no? It suffices because finite unions of closed sets are closed, no? Finite unions of closed sets are closed, not arbitrary. It suffices to prove that every point, every point is closed. Well, every one point set is closed. Every one point set, singleton, yes, x, this is small x, is closed, x and x, of course, OK? So this is a property here, Hausdorff space. It suffices to prove that every one point set, x is closed, x and x. So what does it mean? That means the closure of this is again this, no? The closure of this is this again, no? If you take the closure here, we remain in this set, OK? So take any other point. So let y be in x, y different from x. So these are all the points, no? x, and we have no different point y, OK? So we're getting tired. So you want to prove that y is not in the closure, OK? That is, you have to find the neighborhood of y which does not intersect x. So we have to find the neighborhood here which does not intersect x. So x is Hausdorff, OK? So this implies that y has a neighborhood, u. So this is u, well, maybe not u. Don't give a name, u, v, two neighborhoods. y has a neighborhood which does not x. Well, we have two neighborhoods which are disjoint, even, OK? So it says a neighborhood which does not contain, because we have disjoint neighborhoods, no? If here we have a second neighborhood, if you want, OK? We don't care too much about the second one. y has a neighborhood which does not contain x. So this implies that y is not in the closure, OK? So y is not in the closure. But that means the closure of x, there's no other point. So the closure of x is x. So x is closed. So x and x, this is closed, no? The closure is closed. x is closed. So that's the proof. But you see, we don't need the full strength, OK? So that's nice to remember. So I will, this is not correct, OK? We have three minutes, right? I have three minutes here of my clock. Two, two and a half. This is OK, this is, I checked. So this is right. I don't know why they have this. So what, so we have, this is a house door, no? Every two different points have to show a neighborhood. Now there's this house door. And this is also T2. This is called T2 sometimes, no? And we don't need T2. What we need is, even given two points, each one has a neighborhood which doesn't contain the other. So the picture is this now, OK? We have disjoint, we have neighborhoods. Each one, each point has a neighborhood which doesn't contain the other. But maybe they are not disjoint. That's T1, yeah? So these are separation axioms. We will see much later more. But these are the separation axioms. And the nice thing is then, T2 is important, OK? The house door, house door we will see, and T1 we will not see. Because it's just, well, it's also important sometimes. But so we have a proposition if you want. That's an exercise. So X is T1, OK? It satisfies this separation axiom. Even though the points are closed. That's the equivalent to the fact that points are closed. That's exactly T1. T2, yeah, it would be enough to have T1 for points are closed. So proof, proof. So one, this we proved already, no? As before, right? Well, we use only T1 in the proof before. We use that Y as a neighborhood which does not contain X, right? In the proof before. That T1 is sufficient. We don't need T2. We just have T1, right? And the other direction is an exercise. Well, this is trivial. This is trivial. So what we have to prove is T1. Points are closed. So let, given two points again, X, Y, distinct points, no? X different from Y. So we have to find the neighborhood of Y which does not contain X. But points are closed. So Y minus X, because this is closed. Points are closed. So U minus a point is a neighborhood of Y. So this is open. This is a neighborhood of Y. What do we want to prove? Which does not contain X, of course. Which does not contain X. So this means it's T1, OK? So this implies X is T1. It's symmetric in X and Y here, obviously, OK? So Y has a neighborhood which does not contain X. And this is symmetric completely. So X has a neighborhood which does not contain Y, OK? It's symmetric. This condition is symmetric in X and Y. So T1 is just points are closed, OK? Of course, now the general game, so I finish. Two minutes over. Sorry? So this is, points are closed, no? And so this is open. The complement of closed is open. Ah, what is U? Yes, thank you. There's no U, no? Careful with the X. Yes, yes, of course. You are right. No U. So we take X, the whole space, no? X is closed. X minus X is open. So this is the neighborhood of X, of Y, of the other point, which, yes, thanks. So tomorrow I will ask you an example. So this game in point set topology now is, well, maybe it's the same T1 and T2, no? If it would be the same, it would not be called T1 and T2. So they are different, no? So then we have an example. We want to see an example which is T1 but not T2. It's not the same condition, no? So you can guess what this example, we know an example already, which is T1 and not T2. This is a finite complement topology here. Finite complement topology of an infinite set, OK? So maybe tomorrow we start with this. So it's easy to recall T1, no? Because its points are closed. And T2, it's a little bit stronger. But in many cases, T1 is sufficient, OK? In almost all of the cases, it's T1. I don't know. I don't remember in this moment a single situation where you need T2 and T1 is not sufficient. Let's see, let's see. So we have a sequence which converges to this point. And you say it converges maybe also to this point, OK? But so T1 means if you have these joint neighborhoods, it's clear this sequence has to decide where to go, OK? If you have this side of neighborhood, this type of neighborhood, we have some problem, yeah, right? Yes. Yeah, it might be in the intersection always, OK? Yeah, that's a good remark, OK? So we have for uniqueness of limit, yeah, that's why we should have talked about that anyway, no? In house-toff spaces, if you have a convergent sequence, then the limit is unique. That's an important property of house-toff spaces, OK? And as you say here, it seems that T1 is not sufficient, maybe, OK? Exactly. So that's the situation, OK? Where you want house-toff, T2 and not T1. Well, we discussed tomorrow, two minutes, if we have an example, OK? Yes. So for today, five minutes over.