 I'll give myself a cube, but in this case maybe the cube, okay, so let's say the cube is two dimensional, and you know, I'll try to do the same story again on the faces of a cube, okay? And here, for instance, there is something that I can do, is pick a new center on this face, like here, okay? And whatever, so the image was contained in the boundary, and now in this face of a boundary, what I will do is I will push points, I will project radially in this direction, and in this direction I will get one point. Here, I cannot really do that, so you know I'll keep, usually when I continue, I continue on all the faces, or not on all the faces, but in my picture, I do it step by step. Again, here there's not much I can do, but here, again I can pick any point here, project on that point, and forget about the point there. Okay, so let me try to read my slide and see whether I'm missing important information. So you start from your initial cube, you start with the inside of a cube, you try to find a center, which is not on the set, you project from the center, and you get a new image, so I guess this is what I call pi m here, and I get a new image which is in the union of the faces of a cube, which I call this colleton of dimension one less, okay? Once I'm finished with this, I have a new set, which is lying on the faces of dimension, in this case, n minus one. If n minus one is large enough compared, for instance, with the dimension d that I'm thinking about all the time, then I know that in each face, what I didn't say before is that the image is still d-dimensional, with host of measure finite, so you never cover a face, so if you pick a point near the center of each face, you can re-project on faces of dimension n minus two, and you get a new set, so a new projection which is just composing two projections, and a new set which is the image, which still has a host of measure under control, and which is contained in the colleton of dimension n minus two, and you continue like this. At some point of time, you hit the dimension d, okay? And then two things can happen. In general, you hit the dimension d, and you have a projection on the faces of dimension d, and you're stuck. You're not going to be able to find anything better. In some cases, you're lucky. It just turns out that each of the face, that the measure of a set is still so small, but it never covers half a face, so you will be able to find a red point like here in every small face of dimension d, and then you can project once more of a set onto a colleton of dimension d minus one, and for me, it's essentially as if I made the set disappear. So I'm even especially happy in this case. Any, okay, and so this is probably one of the cases where you shouldn't look at the slide, because, yeah, okay. So that's the basic federal flaming projection iterated up to the faces of dimension d, maybe d minus one if you're lucky, and you have holes. Step number three, we do that on a bunch of cubes. So why would you care doing it on a bunch of cubes? It's because here on the boundary of this cube, we've been moving points around. But remember, we'll have to construct competitors of deformations of sets, and the deformations of sets, there is a place where the thing has to be the identity, so it's not so clear to me that I should be able to move points on the boundary. So I'll put a caution between what I like to do, which is do the full federal flaming projection, and what I can actually do, which is a full federal flaming projection most of the time and change it a little bit near the boundary, okay? So I still give myself a big cube, okay? Which I think I'll call q. I give myself a large number n. Taking the number n large will allow me to make some errors small, okay? And I cut my cube into n to the power little n, the dimension cubes. I will leave it as an exercise to find out whether my horizontal n is the same as my vertical n. It's improbable, but anyway, let's imagine it's like this. So I cut this cube into equal cubes and to the power little n of m, okay? And what do I do? In each of these cubes here, I do the complete federal flaming projection on each of those cubes here, okay? One thing that I didn't say or I mean it was written maybe but I didn't say is that once you do the federal flaming projection on the face, it is the identity on the boundary of a face, so we can extend it to be the identity everywhere else, right? Which means that my federal flaming projections glue to each other very nicely, right? So here when I'm saying I'm doing the federal flaming projection here, so let me try to see what happens. So here I'm taking risks, okay? There are cubes which don't contain anything and for those things things are easier. Let's say for this one, which is easier so I can see somewhere in the middle and I will project my green thing on the face and I will get something like this, okay? And let's say here I will have another point and I will project on something like this and so on and so forth. Here I would be probably projecting here but I could project on the other side. Let's say here I will take a center like this and I will project there. I would project there, okay, yeah, and so on. Okay, I don't do the rest, you imagine what happens, okay? And then here the set was of dimension one so in principle I should stop and again if I'm lucky like in this face, if I was lucky in all the faces like this I would be able to push it to a point and I would be even happier, okay? Now, the reason why I decided to, so on the inside here you do the full feather of flaming projection but when you're here you are a little bit more cautious because for instance on the boundary here you want the identity, okay? So I'll tell you exactly what I do in this cube here. So on this cube here it is not contained in the boundary of Q so I still feather of flaming project and I will get this part here. This face here is still not contained in the boundary so I'm allowed to project on it except that I can't but if there was a hole I would project on it so for instance this face here I can maybe there was a tiny bit here so I would be able to project in this direction that would be fine, okay? But on this face here which is contained in the boundary I decide I cannot touch on things anymore because I want the identity anyway so I will not move points in the middle, okay? That's the whole story. And again it's designed so that I mean this is the danger of a region which I will be a little bit afraid of. On this region I don't know exactly what I'm doing but I hope it's not gonna be too important in the estimates and inside I do really what I like, okay? Still no problem about this? Okay. I mean since we're using this feather of flaming projection all the time it's better. And again I did my, I mean usually I try to do pictures in 3D but it's not clear. It's always working better. All right. So let's try to use it and let's try to do Alpha's regularity the upper bound on the set of large co-dimension, okay? Where we stopped last time, okay? So we have an almost minimal set. We start instead of working on the ball we work on the cube. It doesn't matter so much for the estimate so we take this large cube Q zero. The cube Q zero we cut into and we suppose that there is a lot of mass here on Q zero. We cut a Q zero into pieces. I don't recognize what I was supposed to do. Okay, well anyway it will come back. So I draw this picture with a large N to be chosen later. I suppose that I have a lot of mass in there and again in this part here I project up to the end and there I project not up to the end. And since it's deformations I get a competitor for my initial set in Q zero. It's the identity so it stays the same outside of Q zero. And then I try to estimate the measure of a competitor in Q zero. All right. Okay, so let's try the competitor is composed of so what I wrote here still I mean so before I make the slide disappear is I have this new competitor I call it G. F was the inside of E in the intersection of E with Q zero and the almost minimality of my set says exactly the thing that I numbered 13 which is cost of measure of a set inside Q zero is less than cost of measure of a competitor G inside Q zero plus a small error. What do we remember? So and again F is more or less the same as E and G is my competitor. And I go to the next slide, okay? So again G is composed of two pieces the piece that lies in here and the piece that lies on the dangerous annulus there and we'll have to estimate their measure separately. And I start with the thing that I really wanted which is this part here. So this part what I know is that the set is contained in the skeleton of dimension D of the union of all my cubes and I have a very brutal estimate up there is less than the measure is less than C times the number of little cubes times the measure of the skeleton of each little cube. So it looks the scaling is delta so delta is the size of a cube, okay? Of a small cube. It sounds bad because there is an N to the power N minus D but don't worry about that, it's a constant. I will choose N, it's going to be a very large number but essentially I get a bounded contribution from inside. Okay? What's happening outside? So what's happening outside is I have to estimate the measure of the image, okay? My federal flaming projections have a property that's once you're in a cube you can move inside the cube in the boundary of a cube but you stay in the cube. So if I want to estimate the measure inside this outer layer the only thing that I have to do is I take each of the cubes here I look at what happens to my federal flaming projection. I find out that at each stage I multiply maybe the measure by constant C there is a certain number of stages so in each cube I multiply the measure by some constant, okay? So essentially what I'm saying is that the measure that I have here in the image is less than C times the measure that I had before in the points that appear. And just to be on the safe side there is a second layer that I remove. Okay? I remove it to be on the safe side because maybe points of this cube will end up on this boundary and I wanted to include the measure of the boundary, okay? So something like this. So this is what's happening here. I have Q2 is Q0 minus the two outside layers it's still very small. A, the annulus will be the difference between the big cube and the smaller cube. So it's the annular region which has just width two, okay? Which is very small. And the estimate for the part of the competitor that lives in here is less than C times the measure that I had in the annulus. Okay? And in principle I have this estimate over there on the slide. Let me check. So G minus Q1 is less than C times sort of measure of what I had in the annulus. Okay? So I put all my estimates together. I remember I have that the measure of E was less than the measure of a competitor plus something small. The measure of a competitor has two pieces. The one inside the grid which is bounded. The one on the exterior which is bounded in terms of what happened on the annulus. And I get something like this which is host of measure inside the cube is less than large number but controlled. Host of measure inside the annulus plus smaller. Okay? I assumed that the, oh sorry, yes. This plus this is very small compared to the total amount of mass that I had before because this was my assumption. I said start assuming that there is a lot of mass inside. So I write it like this, like one half of this plus that. And what I get at the end is that this quantity here is controlled by the quantity in the annulus. Okay? That's what I have here. So that's what I get. Okay? It's not, I mean what would be beautiful if you didn't have this measure of the annulus. The measure of the annulus comes from a fact that we have to glue something until it's unpleasant. However, N is very large. So here I am in a situation where I have host of measure in the whole cube less than some constant times host of measure in a very, very, very thin annulus. And I'm saying this sounds fishy. Okay? Especially since I can't control who is my cube, cube zero. So the picture that I should draw is more like this, right? This is my annulus. And I get the conclusion that if I had lots of mass in there then in fact something like 10% of the mass was on the inside annulus. Okay? The standard way to continue the argument is to say, okay, this looks like a differential inequality saying that the mass inside here is less than, is less than some amount times the derivative of that mass getting a differential inequality and proving that actually the mass inside decays extremely fast and in fact there is nothing in the middle. Another way to say this is that once you have this, so 10% are gone and you can start the same argument with a square which is and a new annulus here. Okay? And again, 10% more percent will go away and you iterate. You find out that when you iterate you can even take annulus, annulus that are smaller and smaller and at the end you get that this part here was empty or had some control. Okay? So the truth is that I'm saying all those things to convince you that once we have the estimate up there we're in business. But I don't want to be more precise than that because it's just playing games and the right way I claim to play the game is to choose your initial cube Q zero correctly at the first place and get the contradiction immediately. And again the hint for choosing the best Q zero at the beginning would be take the cube which is I think is as large as possible with a huge density and saying if it was as large as possible you get the contradiction immediately, okay? I don't want to do this because first I always get confused and again I think the important information is here, right? You should not expect that most of the mass is near the boundary. Okay? Sorry about, so this is the, but the rest of the proof is more or less. Okay. One more argument. So lower alpha's regularity which means the lower bound. So the lower bound I announced to you that it looked like it was harder because you'd see less why the set should not be allowed to be too thin. At the same time now here is the reason why the set should not be too thin, okay? So again, start with this thing here and imagine that actually the host of measure of the set was very small inside this cube. You do the feather flaming projection and you find out that in each cube the measure that you started with was small. It's multiplied by at most a constant. So at the end it's still going to be small. If it's very small it means that no face of dimension D of my grid will be completely filled by my set, okay? By the image that I construct. That's good, that's exactly the condition where I said that I could iterate the construction once more, okay? And in fact project on a skeleton of dimension D minus one. Okay, let me repeat what I've been doing. I've been assuming this time, so again N is given to me. I assume that the measure of a set in the cube is extremely small depending on N. I do my initial feather flaming projection the usual way and I find out that the measure of the image is still extremely small, so small that it cannot contain the full face of dimension D. Then I do it once more and I can project. So whatever I project here will be on a skeleton of dimension D minus one, which means I essentially made the set vanish, okay? That's the inside estimate. As usual there is an annulus here. In this annulus here what I do is I essentially multiply measure by constant and we'll have to do an estimate about that, okay? Let me see if the estimates that I have here sort of fit. So in principle I explained the story about this. The cost of measure of a projection is always very small so you can do one more feather flaming projection. This time we get this competitor G and the competitor G is contained inside in a D minus one dimensional skeleton which doesn't count plus the part that leaves an annulus and the part that leaves an annulus will control the same way as we did before, okay? So what we get is cost of measure of a set in Q zero less than cost of measure of a competitor plus smaller. So that's the same thing as the last time. And this time this measure here, there's nothing from the grid inside so you're just left with the part that was leaving an annulus plus something small. And on the next page I have, so it's the same estimate as I had before. It's simpler because there is one less term, okay? So you get this thing here. And again I claim this looks fishy because again you have something like 10% of a mass of a cube which has to be lying on a thin annulus, okay? So now the way we say this cannot happen is by reverse engineering. I think I will not even read the whole story. So let me try to say how you do it. So you start from a cube where the density is extremely small, okay? And instead of trying to see what happens on that cube and continuing, what you'll do is you say, okay, let's replace this cube where the density is very small by the first cube, so you try the half of a cube, half of a cube, and so on and so forth. And you stop the first time the density starts becoming a little larger, okay? So this way you get a cube where the density is very, very small, but on half a cube the density is a little bit larger. In fact, now the second thing that you can do is by pigeon-holing or something like this, you find out that you're in the middle between this cube and that cube, I mean they're very small and they're not so small. You will find another cube for which the annulus here has very, very small measure by averaging, okay? Maybe on this one I was unlucky but here I can arrange that the measure is large, which means that this thing here does not happen, okay? I'm just missing one thing in the argument. You could say yes, but in fact, what this proves is essentially I start with this cube with very, very low density. I try to find the first cube where the density gets larger and in fact I don't because all the cubes have low densities starting from this point, okay? And then I appeal to a little bit of a geometric measure theory, not so hard because at almost every point of a set, the upper density of a set is strictly positive. It's even larger than some constant that we know of, okay? So this is one thing that's, okay, so anyway, you take, again, you take a set with strictly positive cost of measure at almost every point of this set, the measure of small balls, you can find small balls centered at this point where the density is larger than some constant that depends on the dimension. Maybe it's even one if you would normalize things correctly, okay? So here it means that if I started from a cube here that I just centered on one of those points of densities, okay, when I do my attempt of trying to find a cube where the density is large, I will manage, right? Because there are cubes that are as small as I want, where the density is large. So if I start from a very low density, there will be a point where I get density which is very small but not so small, okay? And that's the way it can be. Again, so in this case, I made a small attempt to tell you, again, the story is that as long as you have a thing up there, you're in good shape. And here I try to make an attempt to, yeah, to tell you how it works. So this was, okay. So this was Alphor's regularity. And I think it better stop if I don't want, yeah. Bravo. Bravo.