 Hi, I'm Zor. Welcome to Unisor education. I would like to add a few very important pieces of information about sequences. In particular, we are talking today about bounded sequences and very important theorem which has again a name, Bolzano-Weierstrass theorem. It's a very important theorem for many other aspects of using derivatives and other parts of this higher level of mathematics, I would say. So this is one of the fundamental theorems and the fundamental theorems are always very close to axioms and that's why you really have to be very careful not to jump to any kind of conclusion without really thinking about how that actually jump is founded upon, based upon. I mean it should be either axiom or a previous theorem which you have already proven. So I will try to be careful about this. So now this lecture is part of the Unisor.com advanced course of mathematics and the purpose of the whole course is just to develop your logic, your creativity, your analytical thinking and that's why I'm actually devoting a lot of attention to proofs and solving the problems. So in this particular case, it's one of the fundamental theorems, which I'm going to basically present to you. So we are talking about sequences. Now first of all, we are not talking about any sequence. We are talking about sequences of real numbers. Now, that's the first very interesting requirement. Second requirement for our sequence is that it is bounded. So if you have a sequence, we are talking about certain left and right bounds, where all the members of this sequence are located in between of. Now, sequence has an infinite number of members. So it is kind of intuitively obvious that there must be some concentration points between A and B. You see if the sequence is not bounded, we can actually just go forward and forward to infinity like one, two, three, four and there is no concentration point at each point around each point. You don't have anything else, but this particular point, right? But in this case, when the sequence is bounded from both sides, all these infinite number of members of this sequence must fit in between A and B. So they must concentrate somewhere. And it can be expressed in the following statement that the bounded sequence always has a convergent subsequence, which means I can actually take certain particular members of this sequence. Let's say x2, x25, x100, etc, etc, which also contains infinite number of members. And it's convergent to some point in between A and B. So being obvious obviously is great, but as usually the challenge is to properly prove it. And here is, I suggest, I would say, a rather elegant way of proving this particular theorem. Now it's not based on nothing, it's obviously based on some previous material, which we have already learned about sequences and real numbers. Now what's important about real numbers, and we have already addressed this before in one of the lectures, is their completeness. Now completeness basically means the following. If you have certain number of points, a set of points if you wish, and it's bounded, let's say, from above, so if this certain number of points, a set of points, is bounded from above, then there is so-called lowest or least upper bound. So this is an axiom of completeness, which is very particular for real numbers, because if it's, for instance, we are talking about a set of all rational numbers, this is not true, because this set of numbers, et cetera, obviously has a limit of zero, and zero is rational number. But this set of numbers, which is all the numbers which are less than, well, not less than square root, but xn square is less than 2. So these are all rational numbers, and I'm basically getting the approximation of the square root of 2 here, right? So these are all rational, but their limit, which is square root of 2, is not rational. So sometimes bounded, in this case, from below, a sequence of rational numbers does have a rational limit, and it belongs to the set of rational numbers, but in some other cases, that's not true. With real numbers, this is always true. So whenever you have some kind of a sequence of monotonic sequence, for instance, or even not monotonic, that doesn't really matter, any set of numbers here. And it's bounded from above, let's say, then there is a least bound, and it's also real number. Now, same thing from below. If you have a certain number of bounds from below, there is an upper, the most upper lower bound. So this is called an axiom of completeness, and we will be using this particular axiom. Then another, and this is already a theorem, by the way, which we have proven before, was a theorem with a funny name. It's a theorem about two policemen and a drunk man, or there are many other names which are kind of similar. So if you have a sequence of three different sequences, and these two converged one particular limit, and for each particular m this one in between these guys, so these are two policemen which are actually leading the drunk man between them on both hands to exactly the same limit. So then this also has a limit, and it's exactly the same as this common limit of a and b. So this is also a theorem which we have proven before, and obviously you're welcome to go to previous lectures. You see my lectures about limits, sequence limits, first were introduced as part of algebra because I needed it for some other purposes. So these relatively simple theorems are proven over there, and in the description of this particular lecture I do put exact reference to the lecture which provides all this information. So let's consider we do have an axiom of completeness, and we do have a theorem of two policemen and a drunk man, and that should be sufficient to prove that from any bounded sequence we can always extract a convergent subsequence. Now the whole sequence doesn't really have to be convergent, for instance, have for instance this particular one minus one, one minus one, etc. Now it's obviously bounded because it's only two numbers minus one and one, so it's bounded from left by minus one and by right by one. But it does not converge to anything, however we can always pick a sub sequence which is for instance this one, this one, this one, etc. which are one, one, one, one. So one, one, one, one, obviously converges to one, or minus one, whatever, minus one, minus one, minus one converges to minus one. So not necessarily we can talk about convergent of the entire sequence, but we definitely know that there is some kind of a subsequence which converges. And here is an interesting proof which again I would consider rather elegant. Let's do it graphically. So this is your a, this is your b, and somewhere there are points of our sequence xn, somewhere here. Now xn is an infinite set. Now even if you have the same point, let's say minus one, one, minus one, one, even if you have the repeating numbers in this sequence, it's still different numbers. I mean considering their quantity it's still infinite. So it's one and then another one and then another one doesn't really matter that you repeat it. It's still multiple points, well namely infinite number of points. So this particular sequence has infinite number of points and they are in between a and b. Let's divide this particular segment from a to b in half by point m0. Well we do have some infinite number in between a and b so it might be infinite number either here or there or in both cases right? Because if there is a finite number here and finite there then together it will be finite number which is not the case. Okay so let's pick the half which contains infinite number of points. Well let's say it's this half. Let's divide it by two, m1. Now there are infinite number of points between m0 and b. Now m1 is a midpoint so there might be infinite number either here or there or in both. We pick any one of them. Let's say it's from this to this. We divide it again. Now what do we see here? We see the nested intervals right? The interval m0, b is inside ab. The interval m0, m1 is inside m0, mb. So these intervals let's call ab is let's say it's i0. Then m0, b is interval i1. What next? m0, m1 is i2. So what I have is I have a sequence of nested intervals. Each of them contains infinite number of points of my or members if you wish of my sequence. Almost there. Now what's obvious is that we are dividing the length in half which means that the length is obviously converges to zero. The length of these intervals is infinitesimal value. It's one half of the original length, then one quarter, one eighths, etc. And we know that this sequence converges to zero. Okay. Now how can I choose a sequence, sub-sequence, which is converging? Well, very simply. I will take one point from this one, one point from this one, one point from this one, etc. They are inserted into each other. They are nested. And these are obviously getting closer and closer to something. And that something actually must be a limit. So all we have to do is to rigorously prove that there is a limit and these points which I have chosen without any restrictions, except that the corresponding point belongs to the corresponding interval. So we have to just prove it. Idea, this is idea. Idea is obvious. I divide the interval by two and then another by two and then another by two. And in each consecutive interval I have still infinite number of points. And I take one point from each interval. And it must converge. So now we have to prove the convergence. Okay. Let's consider left ends of these intervals. Well, it's A, M0, again M0. Next maybe it would be, let's say M3. And for instance, this is where infinite number. So it will be M3, M2. Right? So in any case, if I consider just left boundaries, this one, this one, again this one, and then this one, they are, since intervals are nested, so the left boundaries are always monotonically increasing. So the left boundaries, let's call them L0, L1, L2, they are monotonically increasing. Now they obviously form certain set of numbers. It's bounded from above, by B for instance, right? Which means that for using the axiom of completeness, there must be least upper bound. So there is a least upper bound. And they are infinitely closing towards monotonically closing to this one. Now let's talk about right boundaries. Right boundaries are B, then B again, then M1, then M2, etc. So obviously the right bounds, which I can call R0, they are monotonically decreasing. And they are also bounded on the left, monotonically decreasing, bounded on the left by A, which means by axiom, sorry, of completeness, must be the upper bound, which is R. But now let's think about the difference between L0 and R0, L1 and R1, etc., etc. That's the length of the interval, which goes to 0. Which means that these two points cannot be different points, because the difference between them, in this case, would be greater than 0. And since these guys are infinitely close to this, and these guys are infinitely close to this, then eventually the difference between them would not be infinitesimal, right? So if you have, this is my left, this is my right, so the left goes to here, and the right goes to here. So the difference between them obviously would be greater than, let's say, one-third of this distance eventually, because one-third would be here, and one-third would be there. So after certain number of iterations, my difference will definitely be not infinitesimal. So L and R must exist, must be equal to each other. And not only that, what I also here right now have is that left-eyes is less than my freely chosen member of the interval I, and it's obviously less than R. Well, which basically proves that this kind of picture is completely impossible. What is possible is this picture. So right boundary must always be greater than left. But it doesn't really matter. Matter is that they must be equal to each other. And since they are equal to each other, so this is monotonically increasing towards L, this is monotonically decreasing towards R, which is equal to L. So by the theorem about two policemen, our drunk man, Y-eyes, is supposed to go to converge to the same number. And that is the end of the proof. Our sequence of one particular point from each corresponding nested interval is exactly the subsequence from X, which is converging. That's it. I suggest you to read the proof of this theorem on Unisor.com. It's a very, very detailed explanation of the same thing, basically, just to make you a little bit more comfortable with this theorem. And again, it has a very fancy name, Bolzano-Weierstrass, because these are two mathematicians who really were the first one to prove this theorem and use it in subsequent research, which they have. They have a lot of contributions to calculus. Okay, that's it for today. Thank you very much and good luck.