 Tako, pa je tudi, da smo nekaj izvahili, da imamo vse vse vse, da je vse vse, da je vse vse, da je vse vse, da je vse vse vse vse vse. Zelo sem odličil, da počešte, da smo vse vse praviti, da je vse, da je vse, da je vse vse vse vse vse vse, vs. jih ima x in y v ovimesti... je izgledne z klas catalogu, ki je x plus y si dvega, alternativ sem da postojimo & je v drugu dar, x plus y je veliko rekel, da je 1. Ok, imamo se, da v srečnih kajštih, je, da je supermov 2, tako, da jasno zelo vzelo v samim intervalu, ok? Ok, danes, smo videli, da smo vzelo vzelo vzelo vzelo vzelo v modulo 1, tako, da je zelo prezervovalo vzelo, in početimo, da zelo izgledamo, da zelo izgledamo zelo, da je zelo izgledamo zelo izgledamo zelo. Ok, zelo izgledamo zelo, X, Picture X, is equivalent to Y, if and only if X minus Y is a rational number. And then we introduce this function f, whose domain is k, k is the set that the set of the equivalent classes corresponded to this equivalence relation, which values in 0, one. So to any equivalent class here we can associate a small c, Znalejte to z tvoj počet, kaj je vse nekaj vse izpronu. Zelo nam je vse, da težko težko težko tježko. Težko je vse za F. Tjega je na 0 in 1. We will prove now that set t is indeed non labour causal. So the theorem is following, that set t is not variable causal. OK. Tako, zelo smo pričočati. Tukaj pričočiti, nekaj je zelo všem vsega vzela. Tako, zelo smo pričočati, da je zelo vzela vzela rešenje in v 0,1. In izgledaj smo, da je zelo vzela r, i, z 0, infinity, b and enumeration of rational number in zero one. And for instance, we choose R zero to be exactly zero. Okay so we start by considering the translation modular 1 under this enumeration. Často res difundim, da tkazbiti tkazbi, translator, model 1 v srbi. Zelo duži tko tkazba se koristite s tkazbi. Posem, da počekajzemo, ki tkazbih, taj zdaj bo zdej, nekaj je tkazbe, očer. Očer na se stavili, T i intersect tj equal to the n t set for any i, different than j. OK, the proof is quite simple, so you assume that there exist some element x in the intersection. OK, so it would have the form X would be equal, simultaneously to some t one plus some other one r, i equal to sum t2 plus rj. And we have, of course, that t1 and t2 belongs to t, by definition, with, okay? So you have that, okay, you have that t1 minus t2 belongs to, to koo. Okay, you can do the, there are, I mean, in any case, it would belong to, I think there are three options. Either it is equal to rj minus ri, or it should be rj minus ri minus plus one, or rj minus ri minus one. In any case, this difference will give you a rational number, okay? So by definition we have that t1 is in relation with t2, because the difference belongs to koo. But by construction, t has only one element from each equivalent class. Last of equivalence. And we have that t1 is equal to t2. And then you can also, if you do some algebra, you can also see that also r1 is equal to r2. So, so we get what we want. Okay, so we have that. This collection ti is, is a sequence of pair with disjoint, ay, ay, ay. Is a sequence of disjoint, okay, of pair with disjoint, okay, set. Okay, then we see that if we take the union, we have that another, another remark is the following. That we have that each x01 is in some, in some class of equivalence. And so it is equivalent to some element in t. Okay, but if x differs from an element in t by the rational number put ri. Okay, then we have that x belongs to ti. Okay, so thus we have that actually, these are a covering of our interval 01, okay. So now we, we state two properties of this set ti, okay. So we, we show that they are disjoint and the union give you 01, okay. Okay, now, so at the end we want to prove that t is, is not measurable. We argue by contradiction, okay. We assume that t is measurable. Assume t is, is measurable. And this may be answer to your question of two times of two lectures ago, no. That you ask for a counter example about the, the countable additivity property. You remember? Okay, so assume that t is measurable. Okay, so we have that since ti are defined as the translation model 01 of t. So they are measurable as well because we prove yesterday and they have the same measure. So ti are, this, this would imply that ti are measurable and the measure of t would be equal to the measure of, of ti. Okay. Okay, now we want to use the two fact that we just prove. So we want to see that the measure of, we want to compute the measure of this interval. And we know that this interval is given as the countable union of this ti, which are disjoint, is measure of ti. Okay. Okay, they have the same measure. This is the measure of t. Okay. So here we have two option for this right hand side. So either is zero or it is plus infinity. Okay. But here we know how to come. This is an interval. So the measure coincide with the length. So this is a contradiction. Okay. We get a contradiction. Okay. So basically t does not belong to the measurable, it's not measurable. Okay. From this, from this construction there are many other consequences. Okay. The first one is the following proposition. Okay. Which tells you that if you start, if you consider a measurable set E, so can you see the blackboard? Yeah. I mean, not the blackboard. What is it? If E is measurable. Okay. And E is a set, which is contained in that set, that nonmeasureable set t, then there is only one option for E. The measure of E must be zero. Okay. So the proof is easy. Okay. Again we start by an enumeration of the rational number. Let i rational number zero one. Okay. And again, this time, in analogy with what we did, we define this set E i as E translation model one of... Okay. So we have that E i at this joint. Okay. We can, you can do the same argument before, or rather you can just observe that E i are contained in T i. T i are this joint, so they are also this joint. Okay. Then... Okay. This is a quite trivial observation. You have that since... Okay. E... E is measurable. E i are measurable as well. And then, of course, also the reunion over R i is measurable. It's a countable unit. Okay. Okay. Then you have that. E i is equal to the measure of E, because it's a translation model one. And then... So we observe that. Okay. Each E i is contained in this interval. So the reunion should... The countable union will also be contained in that interval. Okay. So by monotonicity, we have that one. It's larger or equal than the measure of the union of E i R i. Okay. They are measurable. E i are measurable and this joint. So this is equal to the sum of the measure of E i, which is equal to the measure of E. And so, again, the only possibility is that the measure of E is zero. Okay. Because this is bounded from above by one. Okay. So this... Okay. Now we prove another proposition. Okay. This proposition tells you that for any... Okay. For any set E... Actually, it doesn't matter if it is measurable, whether it is measurable or not. But what is important that the outer measure of E is positive, you have that you can find that there exists a set F, which is contained in E, such that F is not measurable. Okay. So before proving, we start by a very simple remark. We have that. If you consider this set E intersection with this interval N and plus I, there must exist at least one integer N, such that E... Okay. Put it N bar. E intersection N bar and bar plus I has a positive outer measure. Okay. This comes from the fact that this has positive outer measure and these are disjoint, because if not, you can easily find, achieve a contradiction. Okay. Because you see that E... You can see E as the union over N of E intersected N and plus I. It is better to put it like this. So if M star of E intersection N and plus I is equal to zero for any N, then what you get? Then we have that M star of E. You use the countable sub-ditivity properties less or equal than N M star of E intersection N and plus I. If this is zero, you will get that M star is zero, but we started by assuming that M star of E is positive. So you could get a contradiction. Okay. So there must exist an integer N bar such that this set has positive outer measure. Without loss of generality, we can assume that N bar is equal to zero. Okay. Otherwise. Okay. So with no loss of generality, you may assume N bar is equal to... Okay. We define A, set A as E intersected zero, one. Okay. So here is N dot N plus one with any equal zero. Okay. We may... Okay. So we have that. And star of A as positive outer measure. Okay. Okay. Then we define... So let us define the set A i as the intersection between A and the set T i. Okay. A intersected T i. Okay. This is contained in A, which is contained in... Okay. So assume somehow... We argue again by contradiction. So if we assume that the sequence of set A i are measurable for any i, what we get. Okay. Then by the previous proposition. Okay. What we have? Okay. We have that. We have that. Since A i is contained in T i, then the measure of A i must be zero. Okay. We assume that they are measurable. A i is joined. We have that. And the union of this E i gives you i. It's a covering. So what we have? We have the outer measure of A. It's less or equal than the outer measure of A i, which is equal to zero. Okay. Here maybe we can... Okay. We were assuming the outer measure of A was positive. So we get a contradiction. Okay. Okay. So we proved... We proved our proposition. Here you can... Yeah. Under our hypothesis we can... I mean you can also get rid here under our hypothesis. Okay. Okay. Now... Okay. We will change a bit argument now. And we... We will try to construct the counter set. Maybe you already know the counter set according to your face. Okay. The counter set is... It's worth to study it because you can define, for instance, has a very nice property. And it is somehow a building block to construct the counter function that will provide you many counter examples. We will see later on when we will introduce... We will study the function of bounded variation, the function, the absolute continuous function. We will see that the counter function will come in that year. Okay. So we will construct it by some drawing, basically. Okay. So you consider first this interval. Okay. Yeah. Consider the interval zero, one. Okay. So we want to construct a set. This by approximation. Okay. The counter set I will call... I will denote it with letter k. Okay. And now I will start... I will show you the construction by drawing, which is maybe the simple things. And I will call each approximation of the counter set kn. Okay. So here we are at the step zero of the construction. So the first approximation of the counter set is k zero, which is the entire interval. Okay. Then this is the approximation zero. Then the approximation one is the following. Again, you consider this interval. Huh? Yeah. Yeah, yeah, yeah. So you consider one over third, one over three, and two over three. Consider the open interval, this open interval in the middle. You define k one as zero one minus this open interval in the middle. So one over three, and two over three. Okay. Oh, okay. You can let me adjust, because it will be useful. Then this is also one over three, a union of two closed interval, two over three, and one. Okay. So this is a two riser to the power one, because we are at the step one, closed intervals. Okay. Here you understand how the thing is going on. The approximation two will be the following. You, again, start by zero one. Actually, you start by the approximation one, where this interval, open interval in the middle, have been removed. Okay. You remove this. Then you focus in this two interval, and you took this time an interval in the middle here, and an interval in the middle here, and you remove it. So in that case will be one over nine, and two over nine. So you remove it. And again here you will have seven over eight, and eight over, ja ja ja, over nine, sorry, and eight over nine, and you remove it. It's a kind of fractal procedure somehow. So you have that the step two, the approximation two is what? Is zero one minus this collection, you have one over nine, two over nine, then the big interval in the middle, so one over three, two over three, and seven over nine, and eight over nine. Okay. So here you removed, you have that, no, you removed, it's made up of two to the two, four, because you have one, two, three, four, closed intervals. Okay. So now we understand how the things are going on at the step n. Okay. Cannot draw here, but you will end up with an approximation Kn, which is made of two to the times two to the n closed intervals. Now you will define the contour set as the countable intersection of all this approximation Kn. Okay. So the contour set that Kn is what? Is the approximation of this Kn. Okay. Moreover, you observe that Kn is decreasing sequence by construction. And, okay, by this, you know that it is closed, because it's a countable intersection of closed set. And you can also, from this, you can also say that Kn we are not talking about the empty set because otherwise it would be useless. No, it's different from the empty set. This is, you can see it in two way. One thing is that this is an intersection of compact set which are contained one in another. Or in a very easy way you can just say that, for instance, the element zero, one and all the end points are contained. Okay. Because you can see this the point one, one, zero are contained. Okay. So another thing that we can observe is that, okay, since k contains, we just remarked that k contains all this endpoint, this means that the set k is at least countable, okay, because it contains a countable sequence of points. But we shall see that k is more than countable, is not countable. Okay. We will prove it later on. And indeed k, we will show that k has the same cardinality of the interval zero, one. Okay. Okay. Then now we prove that the contour set is measure zero. Okay. Tour set. Okay. The fact that it is measurable comes from the fact that it is constructed as an intersection of closed set. Okay. Okay. So we have that. First of all focus on the measure of the approximation, k n. Okay. This is easy. Somehow they are, we saw that they are made of two to the power n, one of three n, which is two over three. We have that measure of k. Now we use a proposition that we show some lecture ago. So the measure of k is, by definition, is the measure of the intersection of this set, of this approximation. This is by a former proposition. A former proposition, but we also need to apply this former proposition, in fact that the measure of the first one or one of them is finite. Okay. This is, of course, true, because we are within the interval. Okay. So this is equal to the limit as n goes to plus infinity of the measure of k n, which is two over three to the n, which goes to zero. So, by now we can say that the counter set has measure zero, but at the same time, has the same cardinality of zero, one. So this is somehow something new. Okay. Okay. Now we will see that another property of this set that the counter set has empty interior. So the counter set will denote with this dot at the top of k to the node interior is equal. Okay. So before proving this, I just recall you. So recall the following property. So let u be an open set. Okay. Then, if u is not the empty set, then the measure of u must be positive because this is true because any open set in the real line must contain a non-degenerate interval. Okay. Again we argue by contradiction we assume that this is not empty. Okay. So we argue by contradiction and we assume that k, the interior of k is not empty. Okay. So the interior, of course, is open. Then, there must exist an open set u. Okay. Such that is contained in k and such that the measure of u is positive. And this is absurd because we just prove that the measure of the counter set is zero. Okay. So, indeed, we have that k is the empty set. Okay. Then, so we are investigating some topological property of the counter set. Okay. Okay. Now we focus on the complement of the counter set and we see that the complement is dense in zero one. Okay. So another fact is that kc, which is the complement is dense in zero one. Okay. This comes from the fact that this is empty. Okay. So we have that since is empty. Okay. We have that the closure of the complement is all the interval zero one. Okay. Then from this it comes also the fact that the measure of kc is equal to the one minus the measure of k, which is one. Okay. Because this is zero. Okay. Okay. So we have somehow an example of a set which is dense in zero one and as the same measure of zero one. But the fact that the set is dense in another doesn't mean that the measure is big. Okay. So why? A first example. A set that has which is dense. For instance, we focus on zero one. A set which is dense in zero one is a whole range. Yeah. Exact. This is a set they are dense but the measure is zero. So somehow you can also construct okay, the set duration of course is measure zero. But you can construct also some set which has t-positive measure but as small as you want. So now we will see some construction. So I mean the aim of all this argument of this lecture is somehow to relate some topological property of the set with the notion of measure. Okay. So you can construct a set which is dense in zero one but whose measure is less than one and as small as you want. Okay. Okay. So consider again an enumeration of the rational number where we start from that of rational number in zero one. So okay. Call them so this is q intersected zero one call them for instance and fix an epsilon so quantity which might be small as much as you want between zero and one over two. And consider this opening interval this is actually accountable sequence of opening interval and consider open intervals of the types qn minus epsilon over two n qn plus epsilon over two n. Okay. Okay. We complete the measure of this. Okay. The measure of qn minus epsilon to n qn to n is equal to epsilon over two n. Okay. And again consider the union over n of this interval. Okay. So consider their union so call it this italic u if you want this is the union over n of qn minus epsilon to n qn plus epsilon. Okay. And take the intersection with to be with zero one. So we have that all of this is contained in zero one. Okay. So we have that u is open and u contains q intersected in zero one. Okay. Now we can see that u the closure of u contains q intersected zero one which is zero one but the same time u is contained in zero one so it makes it must coincide so u is dense in zero one. Okay. But what about the measure of this set italic u? We have that the measure of u is less or equal by monotonicity to the sum of the measure of set our interval which is the sum of okay, you have two epsilon two to the minus n which should be equal to two epsilon which is less than one. So you can make it as much as you want. Okay. And so the measure of u is less than one. Okay. Now we constructed this set italic u. So what about the boundary of u? How it looks like? Which property it has? What about is the boundary of u? Okay. We observe that since u is dense in zero one just prove it. Okay. We have that we have that delta u is equal to zero one minus u. And the measure of delta u is larger than one minus two epsilon. This comes from subordinitivity. Okay. Then we have that delta u is a closet set in the complement of an open set. And we have that the interior of delta u is empty since the complement since the complement of delta u u which is dense in zero one. Okay. So all this construction so what can we learn by this? Just to somehow to warn you that the measure of a set does not coincide in general with the measure of its interior or with its closure. This happens for interval or for elementary set. But in general this is a kind of warning that this example you have that as positive measure but the interior as zero measure. So the measure does not coincide. So this is the warning that the measure of a set a set in general is not the measure of the closure the set or the measure of the interior. So think. Okay. Because if you think elementary set interval you cannot find account or example. So you have to to use this somehow some construction. Yeah, yeah, yeah. But this is sure. This is also to an example for the interior of the set. I mean it's true. The closure of the set does not coincide with the measure for the closure for the interior. Okay, now we have that. We will prove that the counter set is uncountable. Okay. To prove this we will need to study the ternary expansion of the point in zero one. Okay, so I just recall you some very easy fact concerning decimal numbers. Okay, so for instance we know that the representation of decimal numbers is not unique. But at least we know that there are at most two types of non uniqueness. Just give you some example. Okay. A remark. Okay, the decimal representation of no, okay. Okay, decimal representation of number, no. The representation of decimal number is not unique. So we will recall you this by an example. For instance, if you have the number nine, nine infinity many nine this is what this is what this is one, no? I mean, if you have infinity many you can express one either as one or as zero with infinity many nine after the comma. Okay. Is not okay? I mean this is a... Okay, believe me. No, no. You have infinity many. Think of the series if you want. Okay. Is it okay for you? Okay. This is a somehow element. Okay. In general, it's true. If you express this as a series, okay, no? You can easily check that it is indeed leads you to one. So here the aim is to say express for instance one in two ways. Just easily as one or in this way. Okay. And what is interesting that somehow it's good for us that there are just these two kind of non-uniqueness. You cannot express in other way one, either as one or in this case. So in general, you have the following situation. So focus on number in zero one. So you have that. Okay. If you have if you have two representation if you have two representation of the same number in zero one so you can... You will end up just with two situation at most two. So for instance you have zero comma a1 okay, a2 a3 so on. And here we have a zero dot b1 b3 Okay. Assume that the number is the same. So what about their digits? So you have two possibility. So so you have two representation of the same number. So the number they are equal they are equal, they are the same number. Okay. Either the digits are they coincide. So either ai is equal to bi or what? Or you have that or there exist some index k such that bi is equal to 9 for any i larger than k and ai is equal to zero for any i larger than k and moreover you have that bk is equal to ak minus one and aj is equal to bj for any j less than k. If you have two representation, you are sure that this is the same number. So either this is the trivial case they coincide for any index or you have this situation which maybe with an example is more clear which is the same situation of here zero, two you can write it as zero, two or zero one and infinitely many nine. Okay. This is the trick. Okay. This is in this is just to write but just focus on this. These are only two. This is the second case. I mean the first one is trivial. Okay. But now, I mean, we are interested in in the ternary expansion. Okay. You can, this is we are used to deal with the decimal one but we are interested in the ternary expansion because in the counter set we started to construct it by removing this the interval for instance was one over three, two over three, so of length which is multiple one over three. But you can say the same things. No? Okay. Also for the ternary expansion. Okay. Okay. For the ternary expansion of course you are allowed to use just three digits which are zero, so the ternary expansion of numbers in zero, one you have three digits, zero, one and two. Okay. So in analogy with you have three digits. These are digits. These are the digits that you use when you deal with the ternary expansion. I mean this is in analogy for instance with the binary representation of numbers is the same thing, so but you use Bayes tree. Okay. So for instance if you have just to recall the number when you the number zero, zero two, two one, one what is this number in the ternary representation is two over three to the power two because this is place two this is place one this is place three this is place four and this is place five plus two three over three three over three plus one, three over four plus one over three over five. This is an example. So we need this because what we want to do is to characterize the points in the contour set by means of the ternary representation. We need to do this in order to study the cardinality of the contour set. It is very easy. So this is a question for you. So what are the numbers whose representation in Bayes tree has the first digit after the comma equal to zero. For instance of this type. I just recall you the first approximation of the contour set. So where they are located in this interval. I recall you that zero one is what? Is one. This is the limit. One over three. They are here. These kind of numbers are here. In analogy you can see that the numbers whose representation Bayes tree has digit equal to two. In the right will be located here. This is a way to pass to characterize the number of the contour set. So we start by characterizing them with the approximation of the contour set. Here we have this is the first digit. This is the first digit. This is two. This is two. This is two over zero dot two. This is just a warning because we can characterize with no problem the point which are in one over three in that way. But when dealing with the contour set we want to take the close of the interval. We are dealing with the close of the interval. So we should include also this number one over three which is in the ternary expansion is zero dot one. So it doesn't fall here because it has one after the comma. But we recall that we can express it in another way. You can express it as for instance zero zero with many two. So we use this. In that way we know that we can exactly represent all the numbers in the closet with this way and the same argument is valid for this. So the role of nine is replaced by two here. The role of nine, the decimal representation is replaced by two in the ternary expansion. So what is K nine? We want to characterize the end approximation of the contour set. K nine is the set of point in the close of the interval zero one such that kx has just to be precise at least because we know that it can have another representation at least at ternary expansion expansion whose first digits whose first end digits belongs to the set zero two. Ok. So this is the generalization of what we observe for the set K one. If you think you can characterize KN. Ok, we know that K is the intersection of KN. So what can we say using this representation? So x, the contour set will be the set of point x in zero one such that kx has at least at ternary expansion whose digits so all of them belong whose belong to zero two. Ok. This is zero. In this way we characterize the element of the contour set. Ok, so what we learn by this. So by all this construction we can deduce the following things that K, the contour set is in one to one correspondence so one two one correspondence of the set of sequence with the set of sequence with values in zero two. Ok, because of this. Ok, this is, you can denote these set of sequences this is a way to denote it to the riser to the N. Ok. Ok, so, but you have that zero two this set of sequence with values in zero two that I denoted with this symbol what are our function which is a function which whose domain is N, the domain of natural numbers with values in zero two. Ok. And this is in one to one correspondence with the set of parts of PN. Ok, this a way to to see this you can see this if F is a set is an element in the set of parts of PN you can define a sequence which is associated to F in this way so you can define AN which is equal whose general term AN is equal for instance to two if N belongs to F and is equal to zero otherwise Ok, this is a way to put this to establish this one to one correspondence Ok, but if we put all together we see that K is in one to one correspondence in one to one correspondence with P of N Ok, and this is by let me continue this chain of correspondence this is in one to one correspondence with zero one Ok, this is a result of set theory ok one to one Ok, so if you gather all these correspondence together what you get so combining result we have that K our counter set is in one to one correspondence with zero one so ok, you have that K is not countable has the same cardinality of zero one even if it is measure zero ok because it's in one to one correspondence Ok Ok, now Ok, ten minut five minutes we know how many we will define the counter function ok, you can also ok, if you want you can also say that if we saw that as a remark if you want this ok, we saw it from the beginning that if A is countable K is zero K but the converse is not true ok, because of the counters because of K for instance Ok, now thanks to the counter set we can define the counter function which has also very nice properties so in step with what we did for constructing the counter set we will also construct the counter function by means of approximation of it so this will be somehow some properties of the function of this counter function that we are going to construct so it will take it is defining zero one takes value here and we have that is continuous counter function f ok, it is continuous it is non decreasing is surjective and constant on each connected component zero one minus K ok, this is given as a union of open intervals and constant constant constant on each connected component so, what is nice for instance one of the things that should be observed that the counter function is as zero derivative out of the counter set because it is constant there but it is not but it is not it is an increasing function ok, because the counter set somehow it is small has zero measure but somehow not that small the function to increase ok so we call a n set zero one minus k n, k n is the end approximation of the counter set ok, then we have that a n consists of two to the power n minus one open intervals he is made of two and minus one ok, intervals call them i j ok, ordered for instance from the left to the right these are the intervals which are removed on each step ok, of the construction ok now we want to define the counter function as an approximation now as a limit of some approximation so let fn be the continuous function zero one which such that we have that fn of zero is equal to zero fn of one is equal to one and in the middle you have that fn of x is equal to j to the minus n on this interval and then it is defined so j one two two n minus one and which is linear on each interval ok so for instance just to visualize you have something like this ok, assume that you you consider f one and f two so you have something a situation of this type ok, here is zero here is one and then you have one over three and two over three so the approximation f one is like this two is something like this this goes to ok, this is f one ok, what about f two f two is you have to consider this two over nine and here seven over nine eight over nine so the other one one over four and three over four so you have here is made like this here is like this here you have that goes from here to here then it's like this and like this this is ok, maybe I can dash it you can represent it so f one is denoted with the straight line and f two is the one with the dashed line I don't have a color so I don't know how to distinguish then ok, now we observe what we have that by construction we have that fn plus one is equal to fn on nj in this case they coincide here in the middle and you have that fn minus fn plus one is less than two minus n so what you get finally is that the sum of fn minus fn plus one converts uniformly in zero one so you can deduce that fn is a Cauchy sequence in in zero one so it admits limit and so we set f, so the counter function has the limit in uniform norm of course of the fn f is the uniform uniform limit of fn and then you can see that of course it preserves some properties of the fn so you have that f of one is one is non decreasing and it satisfies the property that I mentioned before sorry f1 you have a problem here ok f1 is increasing is increasing but you make that or f1 is not correct ok, ok, then sorry, then reverse ok, no, I did yes, I know what you mean ok, so here ok, then it should be like this ok now it ok, here they go inside ok, so I think that for today we can stop here