 Okay, so let's start. So the last thing we did is an example of a space which is connected, but not path-connected, right? And this is the ordered square. You said something, let's talk after this, okay? This, this is one point which is not correct. I will give another example now, okay? More familiar, so this was the ordered square. The ordered square is connected but not path-connected, okay? And here's another example, more familiar maybe. This is an example. I will, this, okay, so what? We take the graph of the function sinus one over x, okay? Starting at a certain point. So this is a graph, g, graph. So what is it? It's x cross sinus one over x, x bigger than zero, obviously. And smaller than some constant, okay? Which I don't care, some constant. So this is a graph of the function sinus one over x, no? That's what it is, the graph. It starts somewhere, it depends on the constant. And then it goes in this way, and now it comes oscillating, okay? Sinus one over x, it's in some sense, you change sides, no? They're all these which come here, come now in this path, okay? So this is a graph of this function. And g, obviously, g is path-connected, okay? This is an exercise, g is path-connected. Why is g path-connected? Well, it's a graph, you take two points and just follow the graph, right? You follow the graph, so it's path-connected. That's no problem. So that means it's connected also, okay? Path-connected is easy, no, in general, as I said. So it's also connected, g is connected. And now I take, so here maybe it starts somewhere, that doesn't matter, okay? On this side, it doesn't matter, very interesting, isn't it? So then also g closure, so what is this? Now I take the closure, so sorry, this is the subspace of R2, okay? This is obviously, we are in R2, so this is the subspace of R2. That's auto-polar, clearly. So we are in R2, standard analysis R2, okay? So what is the closure of g? And then it's clear what you have to add is this interval, right? These are points in the closure. If you take a neighborhood here of any point here, then you always intersect this. Now it comes arbitrary close, okay? So the closure of g is g union, what is this? This is 0 times minus 1, 1, okay? This is this interval here, right? That's the closure, clearly, no other points in the closure. These points are in the closure, no other. So this is also connected, it's connected. The closure of a connected set is connected, that's a general fact, okay? The closure of a connected set is connected again, so this is connected. And this is our example, this now does not look very path connected, okay? g is okay, but g bar is connected, but g bar is not path connected, okay? That's a fact, so it's connected, but g bar is not path connected. Intuitively, this might be quite clear, no? Because what you don't see is a path which comes from this part to the other part, okay? In some sense, you have to jump to go over. Well, I will prove. One has to prove, I will indicate the proof, okay? That it's not path connected, no. But it's intuitively, it's strange to have a path from here, from this part to this part, no? That's the idea, so we prove there's no path, which goes from here to here, okay? And then it's not path connected. That's the second example, more familiar, no? Because this is a standard R2 graph of a standard function times one of x and so on and so on. So it's not path connected, so why not? So this is, so suppose we have a path which goes from this part to this part, okay? Suppose there is a path, f from the interval to g bar, such that which goes from one path to the other, such that where for zero is in this part, okay? Zero times minus one one, and f of one is in the other part, is element, so this is the graph, it's okay? So we have a, suppose we have a path which goes from some point on this here to some point in this here, right? And I will prove that that is not possible, okay? Now we have to, we need to contradict, okay? There's no path going from this to the other, from this path to this path, no path, okay? That's what we want to prove, it's okay? So what, let me give the idea now. We can assume that f of t is in g for t bigger than zero, okay? So maybe we come here, then we come back, come back, but we take the last point where we are here. And after this we are here, it takes a soup, okay? That's the maximum where we are here, of all points, and after this we are here. The path finishes in some point here, no? I don't know which one, and starts in some point here. So we take the last point where we are here and then we are always here, okay? Taking the soup, but with the maximum, okay? So that means, we can assume that the path starts here sometimes. This is p, a point p, okay? F for t bigger than, it's only f of zero. That's, so f of zero, this is our point p. This, of course, is in the other part, okay? So this is zero times minus one, one. And after this, we are on this side, that's no problem, okay? And now we want to contradiction, f of zero is p, so this is f of zero. Now, f is continuous in this point, zero, okay? F is continuous, so that's f is continuous in zero. Maybe the f is continuous in each point, okay? So we give a neighborhood u, what kind of neighborhood? So I give a neighborhood, which is I intersect with an open ball with a radius which is not too large, okay? So this one, okay? So this is the ball around p with some radius epsilon, and then we have u equal bp epsilon intersection with g bar. So what is this? How does it look like? This is this one, no? This part? And then we have all these pieces here, okay? Right? Infinitely many approaching this. They look, no, it's even here. It depends where this point is. They are not connected even here now, okay? If we take this point as p, then we have this. But in any case, whatever p is, it doesn't look connected, what we see here, okay? These are single pieces, right, of one or another form. Either intervals, here's intervals even, okay? Or intervals of this type, which are connected, okay? Well, okay. So there's a neighborhood of zero, which we take an interval, okay? So then there is a neighborhood. So I take zero, we are in the closed interval here, no? Zero, epsilon kind of takes zero delta open, okay? Here closed, because that's the first point. Of zero in the interval i, such that the image is in u. Okay? Such that the image, f of zero delta is contained in u, by continuity, u is this, okay? The ball, okay? So that we get by continuity. But now we need a contradiction, okay? This is connected. That's the point. This is connected. These intervals are connected, okay? The image of a connected is connected. That remains connected. So this is connected, okay? But the image here, so this has the point p, no? And then all the rest are on the other side, here, okay? That's the last point. f of zero is p, this is this, okay? This would be the center of this, it doesn't look like this. This is this, and then after, if t is bigger than this, we are on the other side, okay? So we are in u, somewhere here, no? This is also our neighborhood u. And then, but this is not connected, clearly, okay? This is not connected. This cannot be connected, okay? So because you see that it's not connected. So there's some, you have this point, and you have some other point here in the image, okay? But you can separate all this here, some there, okay? But clearly, so that's the last where I should write that. Clearly, but clearly, f of, what is it? Zero, f delta is not connected in u, okay? It's not connected in u. And so we have a contradiction, so it's not possible to go from a path from here to here. That's the point. That's the second example, okay? That's called, in some sense, this is called topology sine curve, okay? This, I don't know what exactly, but this is a well-known example in topology. Also an analysis, I suppose. So this is the second example, okay? Of a space which is connected, but not path connected. This is, okay, now, the point is that this is not locally connected in this point. So I give the definition of locally connected. We will not do much with this, so definition. A space X is locally connected. Locally connected in a point X and X. If for every neighborhood u of X, there's a smaller neighborhood v, there is a neighborhood v of X. A v of X contained in u, so a smaller neighborhood which is connected. So for each neighborhood, you find a smaller neighborhood which is connected. That's locally connected, okay? And the example is, and the problem in our example is that, so the example is this, sine curve and so on, plus this here, no? That's this v bar. V bar is connected, no? That's, we wrote, but not locally connected in certain points, obviously, no? In all these points, it's not locally connected, okay? But not locally connected. That's a point, but not locally connected. In fact, in all points of this part, it's not locally connected, okay? It's not low connected in all points of this part, which is zero times minus one, one. If you take any point here, in fact, any point of this does not have a single neighborhood which is connected, okay? So it gets very wrong, okay? There's not even one neighborhood which is of, well, one neighborhood of this point, one small neighborhood, okay, I should say, okay? I mean, what is the neighborhood of this point? You can take everything and this is connected, okay? But in fact, so if you take a small neighborhood here, if you take a small neighborhood here, it's never connected, okay? In fact, so this, I wanted to write. What did you write? But not locally connected in all points of zero, yeah. It's clear, okay? So this is an example of a space which is connected, but not locally connected, okay? And you see what goes wrong here because this falls apart in many things, okay? You can separate something. You see the line, you say, this is left, right, okay? And you separate this, okay? So this is locally connected and this is probably it. Then the last thing which I want to do for connected things is another definition. So you define also locally pass-connected, okay? But I will not write that. Locally pass-connected is the same. You say it's locally pass-connected in a point if for every neighborhood of X, there is a neighborhood V of X contained in U which is pass-connected, okay? You just connect it, replace by pass-connected. That's locally pass-connected. And it's clear that locally connected, connected implies pass, sorry. Pass-connected implies connected. So locally pass-connected implies locally connected, right? I will not write. So again, so locally connected. And then the last definition, definition, component, connected component. So X is a space, topological space. A component of X, sometimes a connected component, but component of X is a maximal connected subset. So, and then we have also pass-component. A pass-component is a maximal pass-connected subset of X. A pass-component of X is a maximal pass-connected subset of X. So this is a definition, component, pass-component. And now, well, it's well defined, this definition. If, so that requires a little proof, okay? If we have two components and they have a point in common, then they are equal, okay? If two components, so in fact, two components are either equal or disjoint. Because if you have two connected subsets and they have a point in common, then the union is connected, okay? That's exactly this, right? If you have two components connected and they have a point in common, then the union is also connected. So two components are either equal or disjoint. Two pass-components are either equal or disjoint. Why that? Well, I make a picture here. So let's say we have here a pass-component, okay? With this pass-connected, pass-connected, not connected, pass-connected, and this is pass-connected, and they have a point in common. And then I say the union is also pass-connected, same situation as here, okay? If it has a point in common, the union remains pass-connected. Why is that? So take any two points. If they are in one of these two, it's anyway pass-connected, no? This is pass-connected, this is pass-connected. So an interesting case is where you take one here and one here, okay? And we have to find the path. How do we do that? Well, we have a point in the intersection. That's okay, intersection is not empty. Now we can find the path here from this to this, because this path is pass-connected, okay? But we can also find the path from here to here. So another path, G, okay? And now we can easily construct this path, first F and then G, from this point to this point. That we will do anyway later, okay? But it's clear that we have a path. In the first half of the path, we take F in the second half G. And by the pasting lemma, it's continuous also. This product, it's called the product of passes, okay? The product of F and G. And so this means the same for path components. If they have a point in common, then they have to be equal, okay? Because otherwise, it was not maximum, no? So we have components, path components. And now we give example. So we have here, this example, no? This is G bar, which I call G bar. So connected means one component, of course, right? That's the same, no? Connected means one component. The whole is connected, okay? One component. So this is connected, how many path components here? Path components. You cannot go from a point here to a point here, no? But G is path connected. Two, yes, two. So G bar, G bar has two path components. Which are G, no? The graph itself. And zero times minus one one, right? Because this is path, here you have no problem with the path, no? Here you have no problem with this, follow the graph. But you have a problem to go from one to the other. That's what we indicated, okay? So there are two path components. Now take our other example. So we ordered square. This is connected. So it's one component, okay? Connected is the same as one component, clear? Clearly, all right? It's not path connected. That's what we proved, okay? So now the question is, what are the path components of the ordered square? Yes, exactly. The path components, X times, so I should have written this. X times the, this is the ordered square, no? And these vertical lines, okay? So this is X times I, no? X and A, I. Obviously this is an interval, right? So this is connected, path connected. Interval is path connected. But you cannot make it larger, path connected. Because if you make it larger, then you have two points with different first coordinates. And there's no path, also in the hole here, to a point, so the fact is, there are no, let me write here, if two points have different first coordinates, there's no path from one. If two points in I ordered square have different first coordinates, there's no path between them from one to the other, no? Why not? So why not? So that means this is maximal path connected, okay? We cannot enlarge, we cannot take this point because then there's no path. So why? If you have different points here, why is there no path from here to here? Sorry? By the same proof, yeah, there's no point from here to here, no? But now we have these two points, there's the same proof. Okay, if there would be a path from here to here, let me make a picture. No, I don't, yeah, so here's the same picture. So let's suppose we have a path from here to here, no? What I mean, so this gives me an interval, right? These two points, okay? If I have a path from here to here by the intermediate value theory, I get all points between, as before, okay? By intermediate value theory, as yesterday, okay? I get all points between, because this is the value, this is the value, these are all points between these two, okay? I get everything, everything in this interval here, okay? So then the same argument applies. Now I take the pre-image of all these here, okay? This is open, pre-image is open. You have uncountably many, so we get uncountably many rational numbers. So it's the same argument. If it's this, yesterday it was this and this point, okay? And now it's this and this point, and we have this interval from here to here already, okay? Yesterday we had the interval from here to here, we just had everything, okay? But it's the same argument, exactly the same argument. So that proves that if it's path-connected something, this is path-connected, it's an interval, but it cannot be larger. So these are the path components. So this means they are uncountably many path components here, okay? It's a very strange example. Here we have two path components, no? Here in this example, they are uncountably many path components. It's connected, but uncountably many path components. So uncountably many. So these are some concepts, no? Locally-connected, locally-path-connected components, path-component, which are sometimes useful, but we will not go into this. With this, I finish now the chapter on connected, and we have to go to the next one. It's sort of nice to play around with this, no? Another question, which I don't want to do now, is you take the ordered square and ask, so in which points of the ordered square, the ordered square is locally-path-connected, locally-path-connected, okay? Or path-connected, anyway it's connected, or connected, it's connected anyway. Connected is not the problem, but because intervals are connected, okay? But path-connected, locally-path-connected. What are the points in which it is locally, so you can give exact, I will not, maybe I give this as an exercise, okay? Then you can write. So what did you say? Second coordinate, so we are here or here? What are these points? Where it's, yeah, it's not locally in this part, because if you take a neighborhood of this point, then you can, of course, you can go here on this side, okay? But on the other side, you have to go through this, and then you get all this stuff, and maybe the other side. This, okay? And then you get all this stuff, and then you finish somewhere here, no? So get all this stuff here. And again, this is not path-connected, but always the same argument, okay? So it's clear, in this point, it's not locally path-connected. In this point, it's the same argument, it's not locally path-connected. Sorry? Yeah, it seems anyway, so that's already the answer then, yes. Well, let's, no, no, no, no, no, no. Yeah, so here it is. So, and probably here, we have to check, so here you have this neighborhood, okay? So this is locally path-connected. In this point, yes, in this point, yes. In this point, no. In this point, no, we have to go here. So it's all these points, all these points with the exception of this and this. That's the final answer, okay? Yes. So this wasn't extra, I didn't want to do that, but no, he gave the answer, so it's similar to the arguments, okay? All of this. Okay, now we have compact spaces. It's nice, we start from the beginning again, all of us, compact spaces. So we need to introduce a language also here, okay? So the language is as follows. A covering, an open covering, you need what is an open covering of a space X. It's a collection of open subsets of X. Well, it's clear what it is. It's a collection, collection, family. Maybe not set because it's a collection of open subsets of X whose union is X. So they cover X. So definition, that's a very short definition of compactness. The shortest possible. A space X is compact. The less technical also, a space X is compact. If every open covering of X contains a finite sub-collection which covers. Has, contains a finite sub-collection. So we will have examples, then this is words. It's compact if every open covering of X contains a finite sub-collection which covers X. Finite sub-collection which covers X. And I will use here, and this is not a book, but it's very convenient. I will say contains a finite sub-covering. It's not clear what is the sub-covering in general, okay? One has to define, okay? But I will say finite sub-covering for this. You have a covering, okay? You can find it in many of these. You don't change the sets, okay? And then you call it sub-covering, okay? So this is convenient. Which contains a finite sub-covering. That's what I will use, a finite sub-covering. Sometimes you make sub-covering, you change the sets, makes them small or something. That may happen, but we don't want you. So examples. But before examples, I have to prove, I will prove a lemma, which is useful here. And it's very boring, but it's useful. Lemma. So this condition of compactness, you know, probably. It's not the typical from analysis. They are very, various others. I mean, somebody who wants to define compactness would not come up with such a definition. You would say compact means, in RN, it should mean bounded closed, maybe, okay? Haneborel. Haneborel. In, all you say, any infinite subset has a point of accumulation, a limit point. Or any sequence has a convergent sub-sequence or something like that, right? That seems to be more natural, but it's more technical. Each of these definitions is more technical and less general. So finally, this was the right definition of compactness, except house stuff, okay? Maybe house stuff, maybe not house stuff, okay? That we will, here we don't have house stuff, okay? Just this very short definition, okay? We come back to the other versions, anyway, okay? So lemma, lemma. Let Y be a subspace of X. Y and X be a subspace. Then Y is compact. And I introduce a language here, find a sub-covering. I introduce also here a language. Then Y is compact, if and only if, every open covering of Y in X, what does it mean? So this is a language, open covering of Y in X. Here it's not in something, it's just open covering, okay? Now say in X, that it means in X and Y is contained in the union, not equal to Y, may not be open, okay? By open sets, that is, by open sets in X, okay? Whose union contains Y, not equal to Y. Whose union contains Y. Because Y may not be open, okay? So it's not equal to the union of open sets in general, okay? So contains only. So that's open covering of Y in X, okay? That's the language, okay? It's compact, if and only if, contains a finite sub-covering. By open sets, again, by open sets in X, okay? By, so finally many open sets in X again. Whose union contains Y, okay? So this is a lemma. That means if you have a subspace or a larger space, we can either work with open sets in Y, or with open sets in X, and it doesn't matter, okay? Sometimes it's convenient to use open sets in X, okay? So we don't have to, okay. This is very, the proof is very easy. I mean, you just pass from open sets in Y to open sets in X, or you pass from open sets in X and go back to open sets. You want me to prove it or not? To get used to it? Should I write the proof? What? Yes? Yes. Yes, I always did that, but if you say, that's boring, then I will not do it, okay? So to get used to the language, okay? Of course it's useful. So prove. So if and only if. If and only if, why? So we have two directions, no? Y is compact. So let's start with this one. This means now Y is compact, okay? That's your positive. Y is compact. And what we have to prove? Every open covering of Y in X, okay? So let U equal, yeah, it's some language, no? U equal U alpha, alpha and J, be an open covering of Y in X. What does that mean? That means that, so this is a written U, okay? And this is stamped U as usual, no? Base is B and then element stamped. Be an open covering of Y and Y. This means these are open in X, okay? U alpha, open in X. And the Y is contained in the union U alpha, alpha and J. Not equal, okay? And now it's clear, it's easy of course, then U prime. This means U prime, what do you do? You take U alpha, intersection Y, alpha and J. So this is open covering of Y in X, no? What is this now? Yeah, that's what you're gonna say. Open covering of Y, yes, in Y, exactly. Open covering of Y, and that means in Y. If you don't say anything. This is an open covering of Y. So Y is compact, okay? This of course means there's a final sub-covering, okay? There's a final sub-covering. What is it? U alpha one, intersection Y. U alpha n, intersection Y. Of Y, no? What do we have to prove? Of this first one, okay? And of course then you have U alpha one, U alpha. So this implies that U alpha one, U alpha n. It's a final sub-covering of Y of this first one, okay? Now we go back to the first one. Covers. So these are in U and cover. And cover Y, okay? Cover. So this is one direction. And the other direction is obviously similarly now. Shall I write all to the other? Yes? To get used? Yes. Yes. So the other one is this one, no? So what do we have now? Every open covering of Y in X contains a final sub-covering of Y in X. And we have to prove Y is compact. So what means Y is compact? Every open covering of Y has a final sub-covering. So let, well, U in the first case was different, no? So U prime. I'll start with U prime. Equal U alpha prime alpha nj be a final cover, be an open covering of Y. That means U alpha prime is open in Y, okay? So U alpha prime is open in Y. But what means open in Y? That means U alpha prime is equal U alpha intersection Y where U alpha is open in X now. That's a sub-space topology, right? And then of course, U, which is U alpha, alpha nj, right? This is an open covering of Y in X now. Open covering is an open covering of Y in X. Now you have to say in X, open covering of Y in X. Here's just open covering, Y in X. So the hypothesis, so U has a final sub-covering, okay? My hypothesis, every open covering of Y in X contains a final sub-covering, okay? So this implies U has a final sub-covering. U alpha one, U alpha n. And this of course, now we go back. This of course implies that U alpha one prime, U alpha n prime, you intersect now, okay, with Y. So U alpha one prime, U alpha n prime is a final sub-covering of U prime, of the original, okay? It's a final sub-covering of the original U prime. This one, open covering of Y. And this means this is Y is compact. So this implies Y is compact. Each open covering contains the final sub-covering, okay? So Y is compact. So that's the proof. And now we can do examples. We just have to choice now, okay? You want to work in open sets in the subspace or in the large space, okay? And very often in the large space is better. Less technical sometimes, okay? So, okay, examples. So first two easy examples from analysis, okay? One over n, n in n. So this is the subspace of the reals. So is compact or not? Not compact. Why not? Well, but that's not our definition, no? So we have to, we cannot use closed and bounded, okay? It's bounded but not closed. But that's not our definition. Then we have to prove that it's the same in R, no? It's not so clear. So we have to give our definition, okay? We have to use our definition. By the way, let me call this Y. So one question is, what space is Y? What kind of space? What is the topology of Y? Yeah, Y has a discrete topology. That's exactly the discrete topology. The discrete space is compact, if and only if it's finite. No, no, finite, finite. Points are open, finitely many points. The discrete space, so this is compact, if and only if it is finite. Because you're covering by one point sets, no? And you cannot forget anything, okay? So it must be finite. Otherwise, so it's not compact, okay? So then we know Y equals zero, union 101, N, N, N. So in general, proving non-compact is easy again, no? And not useful, maybe. Because you just need one open covering which does not have a finite sub-cover, okay? You have to find just one, which does not have a finite sub-cover. Proving that something is compact may be more difficult. But this is useful, okay? Because you have to prove that every open covering has a finite sub-cover, no? This is the other, okay? It's like connected, no? Connected unless you prove, not collected. Just one example of separation, okay? Here's just one example of covering which has no finite sub-cover. So this one, now you say it's closed and bounded, but that we cannot use in this moment. Yeah, well, that's the same, why? The theorem? No, no, so it's compact. Yeah, that's a nice, that's the same. You say you have a sequence and it's limit point. That's compact, always. Yeah, that's okay, so let's prove it more general, okay? So this is compact. So XN converges to X, okay? In the space X and in X and then they take Y equal X, union XN, N in N, this is compact. This is special case, no? But we have to use our definition, okay? So what is our definition? Every open covering has a finite sub-cover. So we start with an open covering. So let U equal whatever, U alpha, I always run alpha and J, okay? Be an open covering of Y, Y, this is Y. In X, I prefer to work with open sets in so much space, okay? Open covering of Y and X. So we have the point X, no? The limit point, there's a crucial point obviously, no? Isn't one of these, no? Let X be in U alpha zero, alpha zero and J. It covers everything, so X is somewhere. So this is the neighborhood of X, no? So XN converges to X, what does it mean? So XN converges to X implies there exists N in N, no? Such that XN is in U alpha zero, for N bigger or equal, for N bigger than N or bigger equal, I don't know, whatever, okay? For N bigger than N, right? So that means in U alpha zero, we have almost everything already, okay? We have all points XN here, okay? So missing finitely many. So let XN be in U alpha N for and now choose the missing for N, one second, for N, one second, for N from one to N. These are the missing points. The first N points in the sequence, okay? These are missing steps. And this implies that in U alpha zero, U alpha one, U alpha N is a finite sub-covering, so it's compact. It's always the same proof, no? You have to prove compact here, and you start with an open-covering, okay? And then you, almost, what, we will see, for example. This is an analysis, now I'd like to give another example, which is strange, maybe. But the proof is similar to this. So example, let X be a set with a final complement topology. So we take a set and we take a final complement. You recall that, okay? That's one of our examples also. So what are open sets? Open sets are those sets which have final complement and the empty sets, yeah? And X, of course. But X has final complement, no? Because the complement is empty, yes. With the final complement. Then any subset of X is compact. Any subset Y of X is compact. What are the closed subsets? Final sets, no? The final sets. The closed subsets of X are the final subsets. Not only, and, and X itself, that's maybe not fine. So here's a big difference between, without definition, between compact and closed, okay? If X is infinite, there are many, everything is compact. But the closed subsets are only the final subsets, okay? So you say something is wrong with the definition, baby, no? So this means the compact subset is not closed with the definition, okay? Then we need house stuff, okay? If we add house stuff, so we will see in a moment, okay? Then we have to add house stuff. So sometimes in some books, compact means our condition plus house stuff, okay? To avoid this strange phenomenon, okay? So any subset Y of X is compact proof. And you see, it is a similar proof as this, okay? So let Y, let U be an open covering of, and so. Let U, well, it's always U alpha, alpha in J. Let U be an open covering of Y, and I always prefer an X. That's why I proved the lemma first, okay? Go through X, open covering of Y and X. And we have to prove there's a finite sub-covering, okay? What was I saying? So Y is not empty. We can assume Y is not empty. If it's empty, well, the empty set is compact now. Because you don't need any covering as an empty sub-covering, okay? We can assume Y is not, so let, so we choose one point. Let Y zero be in Y, fixed. It's not empty, we can assume, so let's take a fixed point. So Y zero is in one of these, okay? Let Y zero be in U alpha zero. It's a covering, so it must be in some of these, okay? But the complement of U alpha zero is finite. X minus U alpha zero is finite. And that means I am not only interested in the point in Y now, okay? So Y intersection X minus U alpha zero, okay? It's finite also, of course. So there are finitely many points also. So that is finitely many, Y1, Yn, right? Finitely many. And now the same as before. Now, the structure is the same as before. Let, how did I write before, Xn I didn't use here. Xn, so let Yn to alpha n for n between 1 and capital N. These are the missing points, right? Everything else is, and this implies then that U alpha zero, which contains almost everything already, okay? And then I add these, U alpha one, U alpha n. It's a finite sub-cover, finite sub-cover. And so it's compact. This is a strange example, but it's sort of the same proof as this one, okay? We concentrate on this one point and then we have a neighbor which contains almost everything and then we have finitely many missing and we take finitely many. So in this case, in this case the same proof, okay? So this means, so observation, attention, so observation. This shows that a compact subset Y of X need not be closed, okay? It's in general, it's not clear. This is an example, okay? Need not be closed. In general, okay? Compact subset means we have a subset and you take a subspace, okay? The subspace topology and this need not be closed, okay? So here's the example here, we have the compact, everything is compact, okay? But the clothes are very restricted, only the finite substance. So there's a big difference between clothes and compact here, okay? Which one? Well, let's forget metrizable in this moment. Why metrizable? Yeah, it's not metrizable. Why is it not metrizable? Say again, mm-hm. Yeah, that's true, yes. However, metrizable is not the good word here. There's another word which we used already, which I used already. Yeah, how so? So that is the next one. So this is the proposition, first the other direction, okay? So proposition, a closed subset of a compact space is compact. This is no problem, a closed subset of a compact space is compact. If it's something compact and something closed in compact, then it's also compact. So proof, can think of the proof. This is very easy. So what we have, a closed subset of a compact space. So let y be a closed subset. So this means this is compact, and this is closed. That's our situation, okay? And we have to prove that y is compact, right? So let, all is the same, without thinking. Let u be an open covering of y in x. Let u be an open covering. I don't even need names, as it seems. Be an open covering of y in x. I also prefer to work in the large space. Then, well, u, we get y. We want to get all of x now. So what is missing, x minus union, x minus y. And what is this? This is open, no? Y is closed, so this is open. Is what an open covering of x? Now we get everything, it's an open covering of x, okay? X is compact, so this means we have a final sub-covering, okay? There's a final sub-covering. u1, un, this, maybe I need, maybe I don't need, okay? But I can add it anyway, okay? x minus y. So these are in u, and x minus, what is it, y. I can add anyway, okay? In any case, maybe this is sufficient. Because we have an open covering of y in x, might be already sufficient. So these are in u, okay? And we want y, so we just take only these now, and that's it, okay? So this implies that u1, un is an open, is a sub-covering of what? Of u, of y, of u of y, okay? That's not so, that's a proof, okay? So this is a trivial proof, no? You have a closed subset of a compact, you have an open covering, you just add x minus y. And then you have an open covering of the whole, and then you have a final sub-covering of both, okay? So this is very easy. The next one is this point. So it says that every compact subset is closed, it's not true, of a house of space is closed, okay? So, proposition. Every compact subset of a house of space is closed, so proof. So let's fix notation, let y, as before, let y be subset, so this is, what do we have? Compact, y is compact, and x is house of, okay? So this is our situation, x is house of. Without house of, it's not true, no? No, already, something. We prove that, what is it, x minus y is open, okay? That's the idea, we prove that. We have to prove that y is closed, so we prove x minus y is open. X minus y is open, that's what we want to prove. So open means, given a point, we should find a neighborhood which contains, of the point, which is contained here. So let x zero be a point in x minus y. That's a complement, right? All right, in this way is a complement, x minus y. So let x zero be in x minus y. So maybe I make a picture here. So this is x, this is y, and now we have a point in the complement, x zero, okay? And now for each point, y, small y in y, okay? So we have two different points here, it's house of. So we have these random neighborhoods, okay? Truth to show on neighborhoods, neighborhoods. That's neighborhoods, some of the shorter. U, U, so this is fixed, no? This is a fixed point for the moment, okay? It's fixed, y is not fixed. So it depends on y here. Neighborhood, so it's the index is y, Uy of x zero, and Vy of x zero, Vy of y. So here we have this neighborhood. This is Uy, and this is Vy, this one neighborhood, okay? Of these two points, this point. Then, so this we can do for any y, okay? So what you see immediately, if you vary y in y, okay? And you take all these Vy's, what you see is an open covering of y, in x, okay? So then V, all these Vs, Vy, y in y, take all. It's an open covering of y in x, by open sense in x. But y is compact, okay? So that means, finitely many of these suffices, okay? Y is contained in finitely many, y is contained in finitely many of these. Vy1, union, union Vyn. So finitely many suffices to cover. That's a finite sub-covering, okay? Finite sub-covering. What do you want to prove? What do we want to prove? We want to prove that we find the neighborhood here, which is contained here. What is the neighborhood? The intersection of the other ones, okay? So let u be intersection, u, y1, intersection, intersection, u, yn. And important is we have finitely many, okay? If we would do it immediately, then the intersection of open sets is not open general, okay? But this is the neighborhood, we have only finite, we have a finite intersection of open sets, no? And this is open. So it's a neighborhood of, what is this point? X0, it's a neighborhood of X0 contained in, yes, contained in x minus y. And this implies that x minus y is open, okay? So why is it contained in x minus y? No, no, one second. Y is contained here, no? Take a point, one second, one second. So, y is contained here. So if there would be a point here in this finite intersection, which is contained in y, okay? There's no point in y, that's what we want to prove, okay? In u, intersection y is empty, that's what we prove. That's what we want to prove. U intersection y is empty, okay? So it's in x minus y. Suppose it would be not empty, okay? That means here's a point here, which is in y, okay? But the point here is in all of these, because of the intersection. If you take a point here, it's in all of these, okay? A point here is in one of these, okay? Maybe the first one. But here it's in all of these points, okay? So, that means some intersection here is not empty. You want me to write that? Yes. Contradiction. Yeah, contradiction, right? Should I write yes? Yes. Well, maybe no. Yes, four minutes, that's not okay, the clock, okay? It's open. So suppose u intersection y is not empty. Okay, that's contradiction, no? So let x be in, what, x we don't have, no? No, x is your good name, x be in u intersection y. So x is in y in particular, no? That means x is, x is in y, right? That means it's in one of these. So x is in u, v, y, i for some i between one and n. It's in some of these because it's in here, okay? But x is in u, this implies x is in all of these, okay? So in particular it's in u, y, i also, okay? U, y, i. And this implies that v, y, i, intersection u, y, i is not empty. And that's a contradiction, no? Okay, because at the beginning, we have, where is it? It must be disappeared. Here, u, y, y, y of, Ah, disjoint, disjoint, disjoint, disjoint, okay? Disjoint neighborhood. So that's the, clear that. Okay.