Okay well, this whole issue of of least upper bounds is is one of these things that's, when you're learning it is notoriously difficult to get your mind around. once you've mastered it, it's really pretty straightforward. eh, but there's something about the way the human mind works that makes it difficult for most of us if not all of us. to master this, to get our minds around it when we first encounter it. I remember when I first encountered this as an undergraduate, I just found it completely baffling, I just really didn't know what was going on. now I understand it. I realize that there's actually nothing deep here. but I shouldn't have to be careful, because you have to use language in a very careful, precise and somewhat intricate way. and it's very easy to sort of go off track. So, I don't think this ever gets easy in a, in a sense of being able to just rough it off without thinking about it. But, once you've mastered it, providing you think about it and you keep your clear head and drink enough coffee in advance. Then it's usually fairly easy to, to do these things, so they're not terrible difficult, you just have to be a little bit careful. But as I said, and let me state it again, if you're meeting this for the first time you're going to have trouble with this, we all do. Okay, let's see what we got. This one is definitely True, okay, and here's the reason. Let's say, if a and b are two least upper bounds then to see this a as a least upper bound means that a is less than or equal to any other upper bound. Okay, that's another way of saying a has to be it cannot be bigger than b. If a is the least upper bound, then it cannot be bigger than other local bound. That's what least means, least means it's not, but you can't find a smaller one. Well, if b was less than a, b would be a smaller one. So if a is the least upper bound, you cannot find an upper bound b which is less than a. Which means that if there's another upper bound, b, it's greater than or equal to b. And similarly, because b is the least upper bound, b is going to be less than or equal to a. If b is the least upper bound, a cannot be less than b. Because if a was less than b, b would not be the least one, okay? it's, [LAUGH] I realize as I speak this, that when you meet this for the first time, this, what I've just said is not going to make any sense. it sounds almost as if I'm saying the sky is blue, because the sky is blue because the sky is blue. I'm not, there's actually more substance to this, but you have to think very carefully. If these are both least upper bounds, you cannot find an upper bound less than a, so b cannot be less than a. And if b is the least upper bound, you cannot find an upper bound less than b, so a cannot be less than b. So they're equal. Okay, well, as I said, this makes sense to me, and you may well have to work pretty hard to follow that. That's just the way it is, you're not being stupid. You just got a human mind, and the human mind just does not like this. Okay, this one's also True. This one's actually less difficult, because if a is a lower bound, any lower bound, of the set a, then so is any b less than a. If a, if, if little a is to the left of everything in this set, then so is everything to the left of a. So, providing there is a lower bound, then anything to the left of that lower bound is still a lower bound. We're not talking about least upper bound or greater lower bounds, we talking about just plain old lower bounds. And once you've got a lower bound, anything to the left of the lower bound is a lower bound. Okay, this ones False, this is third one, if a set reals has a lower bound and an upper bound, its finite. And I'm going to do that by giving an example. And the most obvious one, I think, it to take the unit interval, zero, one. Then it has a lower bound, namely zero. It has an upper bound, namely one. In fact, the lower bound is a minimum element. And the upper bound is, not the least upper bound. At least the one we've mentioned, is one. So this element, this is, set actually has a least element and a maximum element. So it's saying that it has a lower bound and an upper bound. Anything to the left of zero is also a lower bound. Anything to the right of zero. Anything to the right of one is an upper bound. So this thing is bounded below and bounded above. But this is an infinite set, it's got all of the real numbers between zero and one. So having though in different bounds does not mean that the, the set is, is infinite, is that the set is finite. Indeed, if it's an interval it could be infinite. Incidentally, this is not the same, there is a phrase, finite interval. that's got a different meaning. That basically means that the interval does not stretch to minus infinity or plus infinity. That it is an interval between two points. Well, this is a special use of the word finite. It doesn't mean it's a finite set. It just means it's got finite lower and upper bounds. It's got endpoints if you like. so, having said that, I'm going to get rid of that, because I don't want to confuse the issue. As a set, this is an infinite set. Okay, and finally, let's look at the last one, this is also false. let's just look at what the set is, let's write it out. The set is, starts down here. A long way down here, actually. Then we've got minus four, minus three, minus two, minus one, or negative four, negative three, negative two, and negative one, if you're pedantic on those issues. And then, this one actually has a maximum element. That's also the least upper bound. Okay, if there is a maximum element, it has to be the least upper bound. So, there is a least upper bound, it's the maximum element, and it's negative one, it's not zero. Zero is well to the right of this thing. So, the least upper bound is negative one. And zero is not the least upper bound. Okay, well how did you do on that one? you know, if it's, if I'm reading it soon now, you throw your for, you throw your, your fist to your forehead. And say oh, golly, how could I be so stupid. Well, you weren't being stupid, you were being human. this is just tricky when you first meet it. you know, and if it, if it looks simple now, then good. And if it still looks complicated, then you're going to have to work for a bit more, but you're not you're not at all unique, if having seen this you're still confused. we all are when we first meet this. Okay, well let's go to number two now. Well as with the previous question the approach is to move slowly and painstakingly and be careful about what things say. Just do some positive parts, so that we know exactly what it says. So let's do that one step at a time. B less than or equal to a for all a in a, that says b is a lower bound of the set a. If c is, so that says, c is a lower bound of the set a. So b is given as a lower bound of the set a, and if c is another lower bound of the set a, then b is greater than or equal to c. In other words, b is the biggest one, b is bigger than any other one, or at least it's greater than or equal to any other one. So b is a lower bound and moreover, b is at least as big as any other lower bound. So he answer is yes, that one does say it's the greatest lower bound. What about the second one? That says b less than, so this says again it's the same thing, b is a lower bound. And this is, if c is any other lower bound, then b is bigger than c. there's a problem here, because b itself, satisfies that, that would mean, b is bigger than, b. This statement, would imply, b, is bigger than b. Which is impossible of course, right? No number can be strictly bigger than itself. So this thing, it's just, it's not only is this, does it not say that. So this is, in fact, a False statement. It doesn't say anything sensible. What it implies is that b is bigger than itself as some b. So certainly, this does not say as the greatest lower bound. It doesn't say anything meaningful at all. Let me say a little bit more. What's going on in here and here? It means we've got implicit quantification over c, implicit quantification over c. This is really saying for all c, if c less than or equal to with... This is a quantification, in the, in this situation, it's implicit quantification over c. And whenever you've got universal quantification, implicit universal quantification, whenever you've got universal quantification, you can hit every possible target. So, whenever there's something like this going on with universal quantifier, among the possible c's, is the b itself. I mean the b isn't really been quantified within this. B is just some number that we're trying to classify. So the b is just a number, it's a fixed number if you like. The a's are being quantified over, for all a's, for all a's, and the c here is being quantified over. Now, we haven't used the traditional language of for alls here. But it is a universal quantifier, and there, it, because it, it just says any c less than or equal to a, and so the c can be anything. And in particular, the c can be the original b. In the first case, it was no problem. Because b is greater than or equal to b. This case its a big problem it makes it false in fact, because the number can't be bigger then itself. Okay, lets look at the next one, well lets see lets why don't we take A to be the closed unit interval zero, one, okay? The greatest lower bound of A is 0, the minimum element, right? Why did I do that? Well, because zero is not less, strictly less than everything in A. In the case of this set, you would have to have zero strictly less than zero, which is impossible, okay? So, it cannot be the case, I mean, this can't [LAUGH], it just can't be, I mean, it's just not true. This, this thing is not true because to be a, to be a greatest lower bound you would have to, this tells you you'd have to be strictly less than everything in the set. Then in the case of a closed set, it doesn't have to be the unit interval zero, one. Any set with a minimal element will do this. All we need is a minimium element. And if the set is a minimum element. Then the greatest lower bound cannot be strictly less, than everything in, in the set, because if there's a minimum element, it is the greatest lower bound. So that rules out this one. Okay, what about this one? well, essentially the same reasoning. You can't classify your greatest lower bound, bas, by including the requirement that is stricter that's in everything in the set. If there's a minimum element, it is the greatest lower bound, and it's not strict or less than this. So, now it doesn't capture it. Okay, let's look at this one. This says, b is a lower bound of the set A. And what this one says, by the way, there's a reason why it's expressed in terms of these epsilons. A lot of the theory of the real numbers that was developed in the late 19th or the early 20th century. it turned out that doing this stuff with, with epsilons was very powerful. But if you look at what it says, it's going to turn out to actually say the same as the one right at the top. That says, if you combine that with that. This says, forget that bit for a minute. This says, any number slightly bigger than b, bit bigger than a, any number bigger than b. Good grief, okay. You're going to add an epsilon to b, so you gotta go slightly to the right of b. And what it says is, if you go slightly to the right of b, you can find an a to the left of that. Let me just draw a little picture, okay? We've got this set, a, here, so okay. So a is somewhere in here, okay? And I've got a b, just to the left of everything in A, or at least not to the right of everything in A when it, it's less than or equal to everything in A. And then what I'm saying is, if I go slightly to the right, to a number b plus epsilon, which we'll put in here. Then already I've gone to the right of some element A, of the, the, the, of A of the set, here. So, b is certainly to the left of everything in A, but if I go slightly to the right by adding an epsilon onto b, then I will be able to find an element in the set to the left of it. So, this can't be a lower bound. So b's are low bound, but if I go, if I pick anything to the right of b, and this is saying, I really shouldn't of wanted that, maybe. This is saying everything to the right, anything to the right of b, anything to the right of b, already dominates something in the set A. So anything in the right of b will fail to be a A lower bound. So b is the greatest lower bound. Okay, so you just have to parse apart that combined with that. For any epsilon greater than zero, b plus epsilon has a property. So these two together is basically saying, for any number bigger than b. That's really what I was saying, for any number bigger than b, there is an a such that. In other words, it's saying these thing, this is exactly the same as this, expressed differently. Okay, well, you probably have to think a little bit about this. As, as I've, mentioned a couple of times already now, the human mind just seems to find its way difficult when it first meets it. An as I, demonstrated by a couple of misspoke, misspoken remarks I've made. Even when you're familiar with material it's tricky to say these things precisely unless you really sort of rehearsed it out. And it's just the way the mind works. Okay, let's move ahead to the, to the next question. Well, question three, deals with the delightfully named Sandwich Theorem. incidentally, this is not to be confused with the Ham Sandwich Theorem, which is a very different theorem altogether. I'm not kidding you, there is a thing called the Ham Sandwich Theorem. You can look it up on the web and find out what it is. This something quite different. and what it says is that if you have two sequences. one always less than another one, and you've got another sequence in between them. Then when one sequence and another sequence when they're sent to the same limit. the thing in between them gets forced to the same limit. it's called the Sandwich Theorem, because the a sequence and the c sequence are slices of bread, and the b sequence is in between the slices of bread. And what this says is, if you make a, a sandwich with two slices of bread and some meat in between them, or some, some filling, and you squeeze the two slices of bread together. Then the filling will get squeezed between them. actually, of course, what happens is is that mayonnaise runs out all over your lap, or ketchup and things, so it's kind of messy in real life. But in principle, what we're saying is if we squeeze the two slices of bread together, then the then the fillings get squeezed in between them to the same thing. Okay, now in, in, in question four, we're going to have to prove the Sandwich Theorem. Here we're just going to apply it. And it's a very useful theorem. It's a simple theorem. It's, it's intuitively obvious. And we're going to give a very simple example here. And here's the theory we're going to prove, using the ha, using the Sandwich Theorem. That the limit of sin, goes to sine squared of n divided by 3 to the n equal zero. Okay, and the proof is rather straight forward. what we do is we observe that for any n. Zero is less than or equal to this term because we got the square which is always positive, of course, and we got the positive number on the bottom. So this is always, designate non negative. and it's clear that this is less than equal to 1 over 3 to the n because the sine, the maximum, value of the sin of any, any number is one. So this is less than one over n, and so we've got a sandwich. Incidentally there doesn't, there's nothing to stop the sequence, an being a constant sequence. So in this case, a zero equals a one equals a two. I mean all of the a's are zero. Okay, actually we start at 1 usually so a one, a two, a three, etcetera zero. So we've got the an's are the zero's. The cn's, the top slice of the sandwich if you'd like, this top piece of bread is one over three to the n, and this is sandwiched in between. Well clearly well we know that this sequence is a constant sequence of zero, so that a in sequence ten is to zero. So the cn sequence ten is to zero as well, okay? Because, it's, it's quite clear that the denominator [INAUDIBLE], it grows without bound. And so this gets bigger and bigger and bigger. and, it tends to infinity if you like, and so the quotient ten to zero. So, these terms are squeezed between zero, the constant sequence, and 1 over 3 to the n, the sequence which turns to zero. So, by the Sandwich Theorem, this turns to the limit zero as required. That's all there is to it, it's logically correct. Everything's there, it's clear. There's an opening, very, you know, it's not much of an opening to do in a sense. You know, arguably one should have said, one could have said at the beginning we're going to use a Sandwich Theorem. But that was actually given to us in, in the question. And so as long as we cite it when we use it that's fine. Okay, so the opening in a sense was, was provided to us by the context of this one. there was certainly a conclusion. There's the conclusion, clearly stated. The reason, well, there, there's there's a couple of reasons. As this is part of the reasoning. But this is the key one that we're going to cite the Sandwich Theorem. That's the main reason, and overall, absolutely correct, 24. it's, it's not, I mean the, the complexity, the depth is actually in this statement here, in the Sandwich Theorem. The Sandwich Theorem does all the work for us. This is really just a corollary of the Sandwich Theorem. but, you know, proofs don't have to be long. So long as everything's there, so long as it's clear, it's proof. Now, I, I, I just said that I didn't prove this. I said it was, it was clear because in the context of applying the Sandwich Theorem, you're entitled to say that this is clear. Clearly one can give the epsilon proof of this, you know, given that an epsilon is an n and. You could use the definition of, of, of limits, but it's not necessary. You know, you don't normally prove that seven plus five is, is 12, when you're doing some arithmetic. You just assume that. I mean, it could be proved, you could prove it from first principles. you can do a various things if you want to. But in the context of applying the Sandwich Theorem, you're entitled to assume something like this. it is important to say if it's clear. and if you'd said because of that, we can apply the Sandwich Theorem, eh, then it's a, it's a little more mysterious. But saying something is clear is a valid reason, so that's an okay method of saying that's something's a reason. Say it's clear, if it's clear, simply saying it's clear is adequate to give a reason. you don't need to have length and depth in, in this proofs. So long as, well, you need to have the depth but it doesn't need to be unnecessary long. This is a good one, 24 max. Let's go on, let's go on now and and prove the Sandwich Theorem itself. So, here we are, let's again, what it says. It says you've got three sequences. And that from some point onwards, the a sequence is less than everything in the b sequence. The corresponding member of the b sequence is less than the corresponding member of the c sequence. This happens on a term by term basis, from some point onwards. the nth term of this is less than the nth term of that is less than the nth term of that. So from some point onwards, you've got a sandwich. Where the b is the, is the filling in between the two slices of bread, okay? And then, the the, the outer sequences, the lower sequence and the upper sequence tend to the same limits. They converge in terms of the same limits. Then, the conclusion is that the sequence that's in between is convergent, and moreover, converges to that self same limit. Now, intuitively it's obvious and the, the obviousness is actually captured by the name of the theorem. That's why people call it the Sandwich Theorem. Also known as the Squeeze Theorem, which I, I think was actually mentioned in the original statement of the problem. Yeah, obviously when I'm doing these abbreviated forms for the for the, for the demonstrations. I cut things down because I want to get everything on a single, on a single slide. Okay, am I going to prove it by going to the definition of convergence, which is a one of these epsilon arguments. And oh, dear, already I'm very unhappy. Because this guy needs to begin with a statement that epsilon greater than zero be given. Without that, we do not have a proof. We simply don't have a proof. this epsilon appears here somewhere, suddenly comes in But what is the epsilon? It's not in the statement. It's something we introduce in the proof. And what makes this a proof is the fact that the argument we carry out here, can be carried out whatever epsilon we're given in the beginning. It's the arbitrariness of the epsilon, which makes this a proof. This is a big deal and it's going to affect the, the grade I give. Okay, come back to that. Let's just follow the steps. Since the limit of that is L, we can find an integer such that at some point onwards, an minus L is, is, is within a distance. So an is within a distance epsilon of L. Ditto, since the cn sequence tends to L, we can find an integer n2, so from that point onwards, the cn numbers are within a distance x not of L. And now I can take M to be the maximum of n zero. The point from which we get a sandwich. We get a sandwich from n zero onwards. And n1, the point from which we get this happening. And then to the point from which we get that happening. So beyond the point M, all of these three conditions are going to be satisfied. So for any M beyond there, I'm going to have this, which is the is the first condition except I've gotten rid of the absolute value terms. And I've written with a minus and a plus. Okay, ditto here. I've got it between minus and plus epsilon for the cn's, so that implies, and what I'm going to do now is I'm going to combine these together with the fact. That we've got a sandwich, this is good. I'm going to give credit, credit for this because this reminds us that we've got an extra piece of information. So what I'm doing is combining this. I'm saying I've got a minus epsilon is less than this guy. This guy is less than this guy, because a n is less than b n. So a of minus l is less than b of minus l. So that gives me that one. This inequality is also going to come from there, because bn is less than or equal to cn. And so I could subtract the l from both sides of there, and then finally this one is going to give me c n minus l is less than epsilon. So what I'm doing, is for the first inequality here. I'm taking this, and that feeds into that inequality. Then, this here is going to feed into the second inequality. This here is going to feed into the third inequality, and this here, is going to feed into the last inequality. So, that inequality here, gives me the first one. This gives me the second, that gives me the third, and that one gives me the fourth. So it's all concatenated and sort of pressed in, but it's, but the reasons are all given. This is there, that's there, and that's there, those are the three reasons. They all combine to give me that. Then I can say that minus epsilon is less than bn minus L is in the middle and plus epsilon is on the outside. And I now I can put absolute values back in and say that means bn minus a, that absolute value is s in epsilon. So by the definition of limits, this proves that this thing is convergent and moreover it's convergent too well. Well it does that providing this argument is carried out for an average epsilon. So now let's give some, some, some numerical grades, logical correctness. I'm going to say four for that, the, the logic was absolutely brilliantly laid down. I mean, this, this is very succinct, nicely done, very impressive. clarity, absolutely clear, absolutely clear. Opening, there wasn't an opening. The opening has to be that epsilon greater than zero be given. That's how the story starts, this is the equivalent of, once upon a time, the, the, in a fairy story. Yeah, I mean, you have to start somehow, somehow. You have to open the story. This is the opening for this kind of argument. It's equivalent to saying, we've gone to use induction for an induction proof. and that plum proof has to being by stating that the epsilon's epsilon, that the epsilon is an arbitrary positive number. This is crucial, first of all, it, it tells you it's positive. You've got to make, make that, that explicit. And you've gotta say that it's arbitrary. So there's no opening here. state the conclusion. Oh yes, the conclusion is beautifully stated. Four marks for the conclusion, what about reasons? I'm going to give four marks for reasons in particular. I love this one here, this is really nice that they, that this has been written here. Let's put a little smiley stick here, that this has been done here. That's nice to put that in here and rather than expect people to sort of pull it back out of the hypothesis. yeah, in fact, without that, I, I think I would have deducted my actual reason. this is the the, that's there and that's there. You know, arguably, one could have put a bit more detail in, the way I explained it when I went through it. But in terms of expressing what needs to expressed, this does it very well. Overall evaluation, in all conscience as a mathematician, I can't go any higher than zero for overall. Without that, the whole thing falls apart. This is not a triviality. This is what, this is what makes the theorem work. It hinges on the fact that it works where epsilon greater than zero be given, okay, and you can't simply say, well, you know, we all know that. Well, we don't know that. This is a proof, and then, this is part of a proof. It's not just part of a proof. It's an absolutely critical part of a proof. So, can't go beyond beyond that. Not only did it have a critical , not only is it an incredibly important part of the proof, but this approach, this way of dealing with things. Came after the result of, of hundreds of years research in mathematics. It was only late in the 19th century, when people figured out how to do this kind of thing and make sense of limits. So, this not just important within this proof, this was a, a major advance in mathematics in the late 19th century. incredibly powerful. So I've got four lots of four, I've got 16 for this one. and I don't think that's unduly harsh. even though I feel sorry for the person because this is very clever. You know, manipulating inequalities and getting them right and that, that's tricky, and people have difficulty with that. but I've given credit for that. Okay, that was in the correctness and the clarity and given the reasons. Where this falls down is it's not a proof, it is simply not a proof and without that there's no way it's going to be a proof. So that, that's a big deal and the fact that we can, we can make this into a routine, little opener, is by the by. guess it may look routine when we write it down, but it's a huge part of the logic. Okay, one more question to go on this problem sheet. question five, we have to prove this theorem. and that means proving it from the, the definition of a limit. Okay, so we're proving that this particular sequence, n plus one over two n plus one, tends to the limit one half. Okay, I, I will tell you that this is actually my proof. And the reason I'm doing that rather than taking the proof that I have from students or composite from students. Is that these proofs tend to be sort of longish because you're manipulating inequalities and fractions and things and they wouldn't fit neatly onto a single slide. So I know that they fit onto a slide, I actually give my own proof, and so this is a very succinct proof, okay? and I'll tell you up front, that I'm going to give myself four marks, which I'm likely to do, aren't I? But, in fact, this is a, is a correct proof. So let's take away all of the all of the suspense, are saying I'm going to give myself four marks on this one. instructors can do that, right? but in fact it, it, this is fine, but I, this is not the kind of proof I would expect to see from a student. and certainly not with a beginning student. I'd expect to see quite a bit more. But let's just, see what's going on. Okay, so, we begin by saying that epsilon is greater than zero be given. Okay, so I start off well. Okay, I give the, the, the opener. So that's the, that's the opener right here, taken care of. Choose N large enough so that N greater than or equal to two epsilon. Well, if epsilon then the idea, the intuition, of course, is that. Is what makes the proof where. Is that the epsilon can be arbitrarily small. Which means one over two epsilon will be arbitrarily large. But of course I can always pick an n bigger than that. that actually is a, is a mathematical principle known as the Archimedean Principle. But we won't go into that now. Okay, intuitively it's clear that you can always pick an integer bigger than any, any real number. Okay, so why am I picking that? Well, because I'm looking ahead to what I want to the end, and after doing this kind of thing for years I can look ahead and see what I've got. In practice what you've probably got to do, is just work through with, with, without saying what the N is. And then see what you have to do at the end, you know, so I could break this down in the beginning because I could see ahead. But typically you might have to go through the logic first before you can decide on this steps. On this sentence for most people, comes at the end of the manipulation. Okay, then for N great or equal to, and I'll come back to that point in a minute. For N greater or equal to N we've got the following, okay? we need to show that this is within epsilon, okay? So were going to show that this guy is less the epsilon. Well let's just work it out. Write it as a single fraction, common denominator is two into two n plus one, so then I've got two, two n plus one, minus two n plus one. do the algebra, that just becomes one over two, two n plus one. everything's positive now, so I don't need the absolute value symbol. so and I can get rid of the half now, by saying it's less than that over that. Just simplifying things and I can say that that's less than one over two n because the denominator is bigger here so it makes it smaller. And now, you should be able to see why I picked N that way because what I wanted to say was that one over two n is less than epsilon and I can do that. By picking that N, because if the little n is bigger than the big N, then this one is less than than one and that one is less than epsilon. So this last step is where that came from. Why did I do that? Because having manipulated this through at the end, I realized in order to get this guy, the epsilon. This whole point is to show that this is less than epsilon. That's what the, that's what were going to show. Show that that's less than epsilon. How do we do that? We do that by going, by picking n's big enough, so that one over two n is less than epsilon because this is actually bounded by one over two n. So it was doing this manipulation that gives me the, the choice of the big N. As I say, I've been doing this for so long that I can see ahead to see what's going on. But even if you can't, you can just do. Just start with what you're trying to prove, you're trying to prove that this is less than epsilon. See what it is you have to prove, and make simplifications. Get rid of the two, you don't need it. Because that makes it bigger. Get rid of the, the plus one, you don't need it and then you get to something simple, and you can pick your N that way. Okay, and then by definition of a limit. So let's see why I gave this, the logical correctness, well. It has a start in there, the opener. the, [INAUDIBLE], let's take them one by one. The logical correctness. These are just manipulations. This is just inequalities. Okay, it's pretty slick, I did this in a few steps. You might make a few more steps. but it's logically correct. It's absolutely clear. There is an opening, there is a strong conclusion. the definition of a limit, reasons. Yeah, well there's a reason there. and there's sort of reasons, embedded in here, in what I'm doing. I didn't write reasons down here in my explanation, because this is just straightforward manipulation. You know, you could say, you could say what I've just articulated. You could give explanations and reasons but you don't need to. This is, this is, this is transparently clear which is basic algebraic manipulation of inequalities. so reasons are okay I mean, this is an important reason, I think, it's definitely worth saying that. and if you go back to previous question, question four, I didn't make a big deal about it. But if you look at question four, it also ended by, by citing the fact that we're using the definition for a limit. And then overall yeah, you, you give four for this, because there's absolutely nothing missing. It's, it's a clear, concise, vigorous, logical complete proof, okay? They, these take a while to get used to, though. Because you are working backwards. you have to sort of play with the thing until you get it into a form where you know how to pick the N. But it comes down to picking an N so big that everything happens that you want to happen. Okay, well that's the end of problem set eight which is the last problem set of this course. How about that.