 OK, so the proof that we still have to complete is this. It's the fact that small l2 is complete. So yesterday we remained here. We have a Cauchy sequence xn. Hence, this is the definition of being a Cauchy sequence. And we proved that there is convergence of all components. So for any k, we proved that there is convergence of the components. So what remains? So we have something in the limit, xk. For any k, we have a number. Then we define the sequence xk. This is a definition. So what remains to show is that x is in l2 and that the convergence actually is not of all components, but is the convergence in l2. So this was what remains. Now, for any l in n, I have that this, for any l, for any epsilon positive that exists, such that xk n minus xk square m less than epsilon square for any n and m. So in particular, since this infinite sum is less than epsilon square, a forciori, this is less than epsilon square for any fixed l. So letting m going to plus infinity here and using star, it follows that for any n bigger than n bar. Now, this is true for any l. And epsilon is independent of l. Therefore, for any n bigger than n bar. So we have at least something similar to the convergence. You see, we are very close to say that xn converges to x, actually. We are almost there. The only thing is to observe that we have this, but then let me write this. So I want to show that x is in l2. So I have to control the sum of square. I have to control this, right? Once I control this, then this is the definition of convergence in l2. So let me write it xk square. I add and subtract x, OK? I add and subtract this. And so this is less than or equal than xk by the triangular probability of the absolute value xkn square, OK? And this is less than or equal than. So there is also another inequality which is, say, a plus b a plus b square less than or equal than 2 plus square plus b square, OK? This is clear. So apply this inequality with this choice of a and this choice of b. So this is less than or equal than 2 times xk minus xkn square plus 2 xkn square. And so we see that when I sum this over k, when I sum this over k, the sum over k of xk square is less than or equal than a number, even a small number, 2 epsilon square, plus 2 times the infinite sum of xkn square, which is finite, because each xn for any fixed n, each xn is in L2, because this shows that x is in L2. So we have proven 0.1. Hence this finite, which is, like to say, x in L2, which implies 0.1. So to conclude the proof, we have to show that this convergence takes place actually in L2. But this is exactly what is written here. Now let me, for any epsilon positive, there exists n bar such that this is less than this. So this sentence here in the box is exactly as to say that this converges to x in L2. So this shows the theorem. So our small L2 is an interesting, reasonable, infinite dimensional Hilbert space. I left also another exercise yesterday on the semi-norm, P. Did you try to do that? Is it OK? Everybody solved the problem on the semi-norm? Yes? So P of x1 minus x2 is larger than or equal than P of x1 minus P of x2. This is because P is sub-additive. So remember, P of x plus y is less than or equal than P of x plus P of y. And therefore, this is simply obtained by writing this like this. So from this, we have this. But then we can exchange the role of x1 and x2. So P of x1 minus x2 is, remember also that P of lambda x is equal to P of x. And so P of x1 of x2 is also equal to P of x2 minus x1. OK? So therefore, this is larger or equal than P of x2 minus P of x1. We obtained that this, say, P of x2, P of x1 minus P of x2 is less than or equal than P of x1 minus x2, but also larger than or equal than minus P of x1 minus P. So this means that P of x1 minus x2 is larger than or equal than the absolute value. So this gives us a little bit more surprising that this happens. OK? Now, finally, the last exercise that I left was to see whether inside the Hilbert cube product minus 1 over k, 1 over k, we can insert a ball centered at the origin. And the answer was yes, that it was no, we cannot. So this cannot contain a ball centered at the origin. Do you have the proof of this? Which is the proof? Yes? So let's assume that I have a ball contained here. OK. So now we will obtain, yes, it's true. If there is a point here, say, a point, actually, 0, 0, 0, a point epsilon over 2, because this is open, actually. So if there is a point with one component, like one component equal to epsilon over 2, say, then I take k, so that 1 over k is less than epsilon over 2. And then I take the point 0, 0, 0, 0, and I take k component, I put epsilon over 2. This is a point here, but it's not a point here. So there is a sort of isotropic object. And this isotropic object cannot be contained in this object here. However, if you have not followed this proof, we will obtain, once more, this fact as a consequence of the next theorem that I will say. However, the proof is direct, as he said. OK. Let me show you this theorem. So let x be a point in L2, and let r be a positive number. OK. And so Br of x, as usual, is the set of all point y in L2 such that the distance from x to y is less than r. This is a usual notation, open bold. And this is the distance in L2. So whenever I d, d means distance in L2. OK, so this object here is not, sequentially, relatively, it's not relatively compact. OK. OK, this is maybe, so you see, what it is saying is that neighborhoods in the topology of the distance are not local. This essentially says that if you have a neighborhood in the, if you take, say, the closure of this neighborhood in the topology of the distance, then this cannot be compact. Or if you want, if you have a compact set, then it has empty interior. And so this says that this topology has, the number of compact sets for this topology is not enough, in some sense. There are many open sets, but not so many compact sets. And this is maybe not good. So one would like to have something, some new topology that has more compact sets. And this is the origin of the so-called weak topology. So at the end, one needs to put on this space another notion of topology, weaker than the topology of the norm, but with more compactness properties. So behind, after this result, there is the problem of choosing another topology in A2. You don't see this distinction in finite dimension. In finite dimension, weak topology and strong topology are the same. You don't see anything. But in infinite dimension, of course, in finite dimension, this is relatively compact. If you close it, then take a sequence here, then there is a limit, some sequence converging to, OK? So I mean, this is, I put in parentheses sequentially because for metric spaces, we can always use sequences, OK? So if you want, you can put sequentially, or you don't put sequentially, it's the same because we are in metric space, OK? So proof. So you have to keep in mind this result because this is maybe one of the starting points of functional analysis, meaning that compact sets have empty interior. Compact sets have an empty interior. And this is a sort of origin, one of the starting points. So proof idea, find a sequence. So let me call it sequence Yn, find the sequence of Yn in Br of X having no converging subsequence. Try to find the sequence. So if you want to make the proof more transparent, so X is a point in L2. So X is a sequence, is a point in L2, OK? Now, if you want to make the proof more transparent, maybe you can take X equal to 0. If you take X equal to 0, then you show that this holds at the origin, but then essentially you have the result. So if you don't like now the proof that I will do, take X equal to 0, which makes it more transparent, OK? So I have X at the center of the ball. And I want to find Yn sequence of sequences. So for any n, we define Yn as follows. Yn as follows. So the component K of Yn is equal to the component. So if K is equal to n, or if K is different from n, I split this, and then this is equal to the n component, plus say r over 2 here, and then Xk. So if X is equal to 0, just to understand a little bit, if X is equal to 0, then Yn K is equal to 0, if K is different from n, and r over 2, if K is equal to n. This is just to understand a little bit better. Notice that any element then Yn is in L2, clearly. And also, the distance of Yn from X is less than r, as you can see. It is immediate, because it is actually r over 2. Therefore, Yn belongs to be for any n, for any n, OK? Now we can consider the distance between two points of the sequence. So take now n different from m, and consider the distance between Yn and Yn. It is clear what is the distance between these two points. I mean, because the distance, so the distance between Yn and Ym squared, if you want, it is by definition this infinite sum of Yn K minus Ym K squared by definition. Now, I remember now the definition of 1. So this is equal. So this reduces just two contributions, only when K is equal to m or n. Otherwise, this is 0. So this is equal to Yn n minus Ym n squared plus Yn m minus Ym m squared. I mean, the infinite sum reduces only for those index K equal to either n or m. And this is r over 2 squared. This is r over 2 squared, right? So this is 2 r over 4 r squared over 4, OK? And so this is equal to r squared over 2. Therefore, the distance is always equal to r over square root of 2. And therefore, Yn cannot have any converging subsequence. If you prefer, you can just think about this case. x equal to 0, it is more transparent. So Yn cannot subsequence. So the fact that you have at your disposal an infinite number of dimensions allows you to make some kind of tricks. In any direction, a constant distance from the origin, in any direction. And then this sequence cannot have converging subsequences. So exercise, let C be a subset over 2. If the interior of C is non-empty, then C is not compact. Or in other words, K, K compact, K sequentially compact, OK? So let us try to do the exercise. Assume by contradiction, C is compact. And assume that its interior is non-empty. And assume its interior is non-empty. So take a point x. So if the interior is non-empty, there is an interior point. So there is a point in the set, and there is a radius. The topology is the one induced by the norm. And so the ball is contained in C, OK? Now, a compact set in a house of topological space is closed. And this is a metric space. Therefore, it is a house of space, right? So we also have this inclusion at C is closed. So this shows that this contradicts exactly the previous, because this shows that then this is compact. This is the closure. This is contained in a compact set. So this is compact. Yeah. This is a closed subset of a compact set. Therefore, this is compact itself. So I am using just this. So then we are x is compact. Which is, in contradiction, what we had just proven before. Is it OK? So this is something that should create some difficulties to our intuition on what is smaller 2. Because now smaller 2 is complete, is separable, but compact subsets have empty interior. So this is something that is not so easy to imagine, I believe, when one study topology. So this is a very good interesting example of topological space with this strange somehow property. And this is also the starting form, as I said, of introducing a new topology, having more compact sets. However, as I said, from this you can start to understand how difficult the function analysis is, because infinite dimensions probably something outside our intuition, in some sense. For instance, let us do this exercise. Let f be a function continuous with compact support. f is identically 0. So there are no continuous functions on L2, no non-trivial continuous functions on L2 with compact support. And so when you see measure theory, for instance, or the space of continuous functions with compact support has, in general, a lot of importance. Well, in this case, you have to replace continuous functions with compact support with something else, because there are no non-trivial continuous functions on L2 with compact support. So let us try to see why. Do you help me? Can you help me? Do you have some? Yes, per image. And where do you use compactness of support in this argument? Yeah, but then you have to use compactness of support. Yes, take a point exactly. Take x in L2, such that, say, f r. So assume by contradiction that f is not identically 0, OK? So assume by contradiction not identically 0. Hence, there is at least one point in L2 where f is not 0, OK? OK, without loss of generality, we can assume that f of x, say, is positive. By continuity, since f is continuous, there exists a neighborhood called y, neighborhood of x, such that f of y is positive for any y in u. Now consider the closure of u-bar. Now the closure of u-bar is necessarily compact, because it's contained in the support of f. Because u is contained in support of f. So that, as before, u-bar is contained in the closure of the support of f, which is compact, which is equal to the support of f. And therefore, this shows that there is a compact set with non-empt interior. The support of f has non-empt interior contradiction. So all of this is in order to say, pay attention to infinite dimensional spaces. Something that you think to be extremely natural, like this, does not exist. And so if you want to take some test function over L2, you have to change the class. Theorem L2 is not sigma-sequentially compact. It's not sigma-compact. This means that it's not the union of a countable number of compact sets. Proof, assumed by contradiction, that you can write L2 as the union over n of kn with kn-compact. Then we know. So I assume that you can cover L2 with a countable union of compact sets. So not only L2 is not locally compact, meaning you cannot be covered by the union of a countable number of compact sets. Because this we know, then for any n, since it is compact, it has empty interior. Do you see the contradiction? Now, the contradiction is the contradiction with bare theorem. You know, bare theorem. That says that under this condition, at least one of the compact sets has no empty interior. Because this is complete. The completeness over 2. If you have a complete metric space, which can be covered as a union of compact sets, at least one has no empty interior. At least one. So we know that there are not so many compact sets in L2. So after this discussion, this discussion, we deduce that there are not too many subset sets in L2. And this will be a problem for us. But we also know, because we know, that they have empty interior. However, we started the lecture by observing that there is a set, strange set here, contained in L2. So let me call it C contained in L2. That could be compact. This could be. Why it could be? Could be compact. Well, we don't know exactly if it is. The previous was minus over k, 1 over k, because it was symmetric. Now, this is exactly the same. So the fact that there is zero, it is because this is more standard. Maybe this is really called the Hilbert Q. So we have already observed that it is in L2, because the extremum is made by a sequence convergence in L2, the right extremum, the right extrema. We know this. We know that this is closed. This is closed, convex, closed. We don't know it is compact, actually. But we know that it has empty interior, at least, because this was observed at the beginning. So that is why, I mean, we know that interior C is empty. And this is why we hope to have at least one example of non-trivial compact set. So maybe this is a candidate. This is a candidate of non-trivial subset. So now, this is called the Hilbert Cube. And let us try to prove that it is compact. So we have a theorem here. Are there other questions, problems for the moment? Is it OK? Do you see some proof of this? It's not easy. You already know. So is there somebody else that see the proof? What about the following proof? This is a product of compact sets. Yes, what about using the Theon of theorem? Theon of theorem says that the product of compact sets is compact. This is general, big, extremely strong. That is the point. So I assume that I want to prove this using Theon of theorem. This is the product, countable, even countable. Not more than countable. Actually, just only countable product of compact sets. Theon of says it is compact. Yeah, it is compact, but in the product topology. And we know that the product topology has more or less compact set than the topology over 2. More, because it has less open sets. Is it clear? No? I mean, there is a theorem in topology, very big, which says, in general, the product of compact set is compact for the product topology, however. So one could invoke this theorem and say, well, this is compact. The problem is that you have proven it is compact in the product topology, not in the topology over 2. And the topologies are different. In particular, it is more easy to prove compactness in the product topology that compact is in the 2. So we have to change. By the way, in this countable, I mean, there is also no need of so big theorem like Theon of for proving compactness in the product topology of this, because this is a countable product. And therefore, one can do this by hence taking a sequence and the diagonal argument. So actually, maybe Theon of is not necessary, because you take a sequence of points here, countable sequence. Then you take the first component. The first component lies in 0, 1. So x, 1, n. This is a sequence, first component. Then you can extract a subsequence such that you can extract, say, a subsequence indecised by n1 such that the first component is converging. So that, say, x1 n1 converges as n1 goes to infinity to a point n1 is a sequence. So n1 is a sequence of indices, sequence of natural numbers. Maybe the notation is not perfect. Just to say that this is converging to a point at x1. And since 0, 1 is compact, actually, this belongs to 0, 1. Then I consider x2 n1. This is a sequence. It belongs to 0, 1, 1, 1, because I'm taking the second component. Therefore, it has a subsequence converges to a point in 0, 1, 1. So let me call n2 this subsequence extracted from the sequence n1. So n2 is a sequence extracted from n1. Therefore, x2 and 2 converges to x2 in 0, 1, 1, 1, 1, et cetera, et cetera. I mean, this procedure can be repeated the countable number of times to extract just one sequence such that one sequence n prime, I don't know, n prime, sequence extracted using a diagonal that sequence of indices such that xk n converges to xk as n goes to infinity for nk. And this belongs to 0, 1 over k. So this shows by hands using, however, a diagonal argument, which is not so trivial, show by hands compactness in the product topology. So you don't need, actually, for this countable product, you don't need invoking, maybe, Ticanoff theorem. So now, well, let us prove that this Hilbert cube is actually compact in a topology of L2. So there is a theorem of topology which says the following, which actually is an interesting characterization of compactness. Theorem of topology is that compact if and only if. It relates compactness with completeness. If it are if, complete and not totally bounded. This is an interesting result of topology. Now I'll tell you what does it mean. Maybe I recall you what does it mean, totally bounded. For any epsilon positive, there exists, say, an epsilon. OK, c contained in a topological space x is so definition. Let x be a topological space. Capital X, a subset of a topological space xc is called totally bounded. If for any positive epsilon, there exists an index and epsilon. And there exist points x1, xn epsilon points of x such that c is covered, the union of pi xi contains c. Ah, sorry. This is a ball of radius epsilon, totally bounded. It's called totally bounded. Remark, maybe home matrix space. So x matrix space, because we are using the balls. So since we are using the balls, we have a matrix space. Home, totally bounded, implies bounded. But bounded does not in general implies totally bounded. But the converse is not always true. Well, this implication is very easy. Because say, take epsilon equal 1. Then you can cover your set with the finite number of balls of radius 1. Therefore, there is a big ball of radius, I don't know. You can cover your set with the number capital N of balls of radius 1. So you can cover your set with the number of balls of fixed radius, and your set is here. Now by the triangular property, there will be a big ball containing all these finite number of balls. So you cover your ball with finite number of balls of radius 1, say 100 of balls. And then there will be a big ball of radius 100 plus 1. I don't know, whatever, covering all the balls, containing all the balls. And therefore, this means that it is bounded. Bounded means exactly that it is contained in a big ball. So the proof of this implication is easy. And indeed, the word totally says totally stronger than without totally. But the converse is not always true. It's not always true. And therefore, you cannot just prove completeness and boundedness in order to have compactness. You have to prove completeness and total boundedness in order to prove equivalence to compactness. Now, do you see why converse is not always true? Well, but now you are using a lot of things, the existence for an orthonormal base, for instance. I mean, only one point is one. The rest is the many. They are called the many. Yes. And we take each of them some small ethylene over which is joined each other. So let us try to see if I have understood. So you take L2. And then you take balls centered, say, at the point 0, i, 0. You take the ball bi. You take? 0, centered at 0, radius equal to 1. The ball bi is centered at the origin with the radius 1. Yes. Yes. Then? Yes. 1, 0, 0, 0. Oh. You take one. So points in this ball. Yes. Take the closure. 1, 0. Yes, closed. OK. This 1, 0, 0, 1, 0, 0, 0. OK. These are elements of this ball. So let me call ei. Ei at this point. Where were? So ei is this point. And then I take the neighborhood 1, third centered ei. These are disjoint. Yes. They cannot cover all b0. And b0 is bounded. Well, there is another maybe example that we can do. Let me see whether it works. Take on r, the distance between two points, to real numbers, is equal to 1. If x is different from y and 0, if x is equal to y. Take the following distance. So all points of r have distance 1. If x is different from 0. And 0 if x is equal to 0. So all points that are distance 1 from the origin. Discrete space. So this is bounded space. But if I take a ball with less than 1, I cannot cover it with a finite number of balls. More or less the same. OK. Well, he claims that this is a bounded set. This take is x1h. Maybe it is better to take the discrete space. So the ball covers the sequences so that the sequences is bounded. And if it is any number less than 1, then we only cover that if we need to choose the ball that covers, finite ball that covers the other sequences, it cannot be because the sequences is finite. And not definitively constant also. Well, I think that these kind of examples can be constructed simply taking this degenerate distance in which all points have distance 1, unless they are equal essentially from the origin. So you really don't need an infinite dimensional space. You just can work this in the real, but just take a distance very, very, very singular. Now, of course, if we prove that C is compact, then this will imply that it's interior is empty. And so we will have another proof of what we have said at the beginning of the lecture. Beginning of the lecture was have you proved that this is an empty interior? Yes, that's right. So this will give another proof of the fact that the interior of C is empty. Now, when you said epsilon net, maybe you were referring to total boundedness, I think, something like that. So let us try to do the proof. Because if you want to work, I mean, we have something. I can take a sequence here. And I know there is a subsequence such that all components converge. This is the Ticanov type argument that I made before. So what would remain is that to prove, actually, that subsequence not only converges in the sense of product topology, but it converges also in the sense of L2. What I'm saying is that we have already half of the proof without epsilon nets. So maybe you can try by yourself. Take a sequence here. We know there is a subsequence such that all components converge to a point of a 2, diagonal argument. This is not enough. That subsequence must converge also in L2. Try to prove this by yourself. Now we do another proof. OK. So, well, we have to show essentially that C is totally bounded. Because we already know C is closed. L2 is complete. Therefore, C is also complete. OK. So C L2 is complete. C is closed. C is contained in L2. This we know. And therefore, C is complete. So what remains to prove to show C is totally bound. Bound. OK, so fix epsilon positive. Take n bar, a number n bar in n such that the tail k from n bar plus 1 and infinity 1 over k squared say less epsilon over epsilon over 2. I don't know. Epsilon over 2, epsilon over 2. OK. Now I fix an n bar. I don't claim that this n bar is then n epsilon I'm looking for, for the total boundedness. I fix just an n bar. I can always do this. OK. Now let me define a sort of projection of C on a finite dimensional space. So I take the following projection of C, all point of C having components after n bar, which are 0. So xk equal to 0 for any k larger than n bar. So I am projecting the Hilbert cube on a finite dimensional space. This is contained in a, this is contained. I mean, it's a subset. So a can be considered as a subset of r and bar. And a is also, therefore, a, just the product from 1 to n bar or 0, 1 over k is totally bounded because it is compact in this finite dimensional space. OK. So a is totally bounded. So I can apply now the definition of total boundedness. So any point z in a belongs to at least one of the balls of radius. So let me just take. So there exists points a1, a now, and epsilon in a. So there exists an epsilon, comma, there exists point in a. Such that n z in a belongs to one of the balls of radius, say epsilon over 2 centered at aj, one of the ball. So maybe when I wrote total boundedness, I don't remember if I took the point x1, xn epsilon in c or in x, maybe in c. OK. So any point z belongs to one of the balls which is equivalent to say that the minimum of the distance the minimum j1 and epsilon, the distance in l2, normal over 2, z minus aj is less than epsilon over 2 for any z in a. So this is equivalent to say that there is at least one index j such that z is in the ball centered at aj of radius epsilon over 2, this minimum way of writing. This is true for any z. So actually now I have a candidate, an epsilon, corresponding to the previous choice of epsilon. So now the idea is to show that this and epsilon is OK. So OK, now take any x in the cube and write x bar as the projection of x, x1, xn bar, 0, 0, 0. So this x bar is an element of a. Now I consider the distance from one of the aj's. And the distance now is just this plus the sum. Now any aj is in this finite dimensional space. Therefore the components of aj after n bar are all 0. Therefore here there is no aj. So there is just the sum from, let me use the index k from n bar plus 1 to infinity of xk square. OK, now this is square. Now any xk is an element of 0, 1 over k. So this is less than 1 over k square. And so this is less than bar plus 1 infinity 1 over k square. OK, so by our choice of n bar, this is less than or equal than epsilon over 2. Ah, sorry, there is a mistake here. So let me take the square here. So let me take this, it's of course the same. Maybe I should take epsilon to the square, it doesn't matter. OK, so now for any j, for any j, for any j, for any j from 1 to n epsilon, this is true for any j 1 n epsilon. So I now take the minimum over j. From this inequality I find that the minimum, this is a minimum because it's a finite number of indices. So the minimum for any epsilon, the minimum over j, x minus aj square is less than or equal than the minimum from 1 over n epsilon x bar minus aj square plus the tail plus epsilon over 2. But I know from the finite dimensional argument that my projection, the finite dimensional space, is less than epsilon over 2. So this is actually less than or equal than epsilon. So that should be an epsilon square, I'm sorry, because then I want bolts of radius epsilon. So if you adjust everything by putting here epsilon square, epsilon square, epsilon square, epsilon square, and epsilon square. OK, sorry for that. So what I've shown is the following. So conclusion is, the conclusion is, min j from 1 to n epsilon x minus aj square less than or equal than epsilon square for any x in the cube, which is exactly like to say that I'm covering with a finite given epsilon, I'm covering with a finite number of bolts of radius epsilon centered at these projections, a1, an epsilon. So this shows, hence, given epsilon, I found n epsilon and points in c, actually in a, but a is contained in c, obviously, such that this, which is equivalent to say that it is totally bounded. And you see, the proof is not easy. There is always the idea, so that the difficult part of function in this kind of argument is that you always have the idea that infinite dimensions count less and less and less. As the dimension increases, they are not so important. So what we are doing here is that we are covering a big part of c, finite dimensional, using totally bounded. And then we have something which remains, something which remains that, however, is not so important. It's the tail. So the point is to make this argument concrete. It is never so easy to do this. In this case, it works. And so there is at least one proof that c is compact. Maybe you could try by yourself another proof using sequences. We already know that the sequence converges in the product topology. Then it remains to prove that that converging subsequence actually converges also in the stronger topology of L2.