 Hi, I'm Zor. Welcome to Unizor Education. Today we will continue talking about limits of the function. Primarily, we will discuss continuous functions and a particular property of the continuous functions on a segment. Actually, it's called boundedness, which means actually that continuous function is supposed to be bounded, and I will explain the details. This lecture is part of the course of advanced mathematics from teenagers presented on Unizor.com. I recommend you to watch this lecture from this website, because every lecture has a very detailed explanation. Also, the site has exams, so you can basically check yourself. The site is completely free. Alright, so back to boundedness. So, we're talking about continuous functions, which are defined on a segment, which is a continuous interval, including both ends. So, this is from A to B, including both ends, and the function is continuous. Well, maybe it's appropriate just to remind you what continuous function actually is. It's defined very simply. For any sequence of the arguments, which is converging to some point L, which is its limit, the corresponding function of these arguments is converging to f at L. This is f of L. So, whenever argument converges to a point, then the value of the functions at these arguments is converging to the value of the function at the limit point. You can also define continuous functions on epsilon delta language. Basically, the function is called continuous at point x0, if for any epsilon greater than 0, which is a distance from the f of it. That's probably a coincidence. Let's put another point. Let's say here. So, this is f of x0. So, the sense of the meaning of this definition of continuous means that if the function is close to this value, if the argument is close to this value, then the function will be close to the value of the function at this point. So, for any epsilon greater than 0, there is always some kind of a delta neighborhood of point x0, such that if x minus x0 is, if x is in the delta neighborhood of x0, then, follows from here, function of x would be in the epsilon neighborhood of the function value. So, if I would like actually to be within certain limit around function f of x0, which is epsilon, I can always find some neighborhood of x0 wherever argument is close to x0. My function will be closer than epsilon to the f of x0. So, that's the continuousness at the point. And obviously, function, if it's continuous at every point of the segment, it's continuous on the segment. So, these are just some considerations about continuousness. Now, my main point of today's lecture is that any continuous function defined on the segment AB with both ends included is always bounded. So, there is something on the top and something on the bottom, which bound the values of our function within this particular segment. So, the continuous function cannot have something cannot look something like this, where it's asymptotically goes to infinity, for instance, or in any other way. So, it's supposed to be bounded from the top and from the bottom. It's a very important theorem and it's kind of intuitively obvious. Well, you can say that from the intuitive standpoint what is continuousness, continuity. It means that you are pointing your pen at this particular point, which is f of a, and then without lifting the pen or pencil from the paper, you reach the point f of b, right? And obviously, you don't go to any kind of infinity. So, your movement is bounded from the top and from the bottom. There is always something which restricts your area, wherever you can draw this kind of things. But as always, things which are obvious is difficult to prove. It's kind of a typical in mathematics. Now, the purpose of this course is not as much to basically tell you that this is the fact. Okay, the function, if it's continuous in a segment, then it's bounded. Most likely, you wouldn't care about this. The purpose of the whole course is to teach you to logically think, to apply some creativity and analytical approach. So, in this particular case, your task or my task right now is to prove this obvious fact. So, prove means I have to really base my logical statement on some known facts. So, what's the known fact about this function? Well, first of all, the definition of the function. This one doesn't matter. But the function I said is continuous, which means the definition of the continuousness is definitely supposed to play a certain role. Another important role is the so-called theorem of Balcana-Beirstrass. Now, the theorem of Balcana-Beirstrass was proven for sequences in one of the previous lectures. And what it actually states is the following. And I will be definitely using this. If you have certain segment, let's put it this way, the same segment from A to B. And you have certain sequence of points here. Well, sequence means infinite number of xn. So, from the intuitive standpoint, if there is an infinite number of points between A and B, there might must be some concentration points, wherever they actually are concentrating in infinite numbers. Otherwise, there is no infinity here. If there is no such concentration points and the x of n are separated between each other, there are infinite number of members of our sequence. They cannot be separated by certain distance if you have only a segment wherever they are in between. Now, more precisely, the theorem of Balcana-Beirstrass is staging this from any sequence like xn. You can always extract a subsequence. Well, I should use a different index, I guess. So, yi is a subsequence of xn. And so, the theorem states that you can always extract, which will converge to certain number l in between A and B. Now, this actually is more mathematical state, more mathematically stating the fact that xn should have accumulations. So, wherever there is a point of accumulation, then I can always choose some kind of a subsequence which actually goes into this particular point. And by the way, if you remember, then there is also the axiom of completeness, which is applicable to real numbers, which says that for any set of numbers, whatever you have, if there is an upper bound, there is also the least upper bound, which in particular tells me that a segment from A to B is closed segment, which means that every sequence which has a limit inside, this limit also belongs to the segment from A to B. So, these are all preliminary facts. My proof of this boundedness will be actually made. So, how can I make this? Well, to prove that the continuous function is bounded on a segment from A to B, let's just assume that it's not. And we will come to some kind of a contradiction, right? Okay. So, what does it mean that the function is not bounded? Function is not bounded on this segment A, B, if no matter, well, we're talking about bounded from above, okay? So, from below it will be exactly the same. So, if we assume that the function is not bounded, now what actually it means? It means that no matter how large some number N I can choose here or here or here or here, there is always a point within this segment A, B where the function varies above that boundary, right? So, for any N there is some kind of X with an index N such that function value is greater than or equal to N. So, that's what unboundedness means, all right? Okay, unboundedness. So, the function is unbound. This is the statement. So, for any N however large, I can always find some kind of a point which belongs to segment A, B such that I didn't put index here, such that F at XN is greater than N. Okay, great. Now, these XN basically they constitute a sequence of certain points within A, B, right? So, this is a sequence of points. It's bounded from A to B. It's bounded on both sides, which means by the theorem of Bolzano-Baerstrass, I can always find a sub sequence. Let's use it YN. This is subsequence of this, which actually goes to a certain limit. A converging subsequence from any bounded sequence, bounded in both sides sequence, we can always extract a converging subsequence. And obviously this point A, this point L belongs to segment A, B. Now, function F is continuous, which means that its values at these points actually are supposed to converge to F of L, right? For a continuous function, but we have assumed, now F at L is some kind of a value, right? Because the function is defined from A to B on this segment. So, every point has some value. Now, we have actually assumed that the function is not bounded. So, these values are all greater than this one, greater than N. So, function value of XN is greater than N. Well, basically it means that as lowercase N goes to infinity, YN, it's a subsequence of XN, right? So, in theory, F of YN also should be greater and greater and greater as index is growing. And it should have absolutely no limit. It should be greater than any number, which contradicts the fact that it converges to F of L as F is a continuous function. So, this is the contradiction. We have assumed that the function is not bounded. We found a sequence of arguments which produce the sequence of functions going to infinity. From this, we took a subsequence which also exactly has the same property. So, the function of Y of N also goes to infinity. But here, the continuousness of the function tells us that function F of YN should actually converge to F of L, which is a concrete number, it's not infinity. That's the contradiction. Which basically proves that our initial assumption that the function is not bounded from above is wrong. So, it's bounded from above. Absolutely similarly, we can go to the lower bound. So, for any number N, however, small to the negative side, to the minus infinity, minus 1, minus 2, minus whatever, we can always find such a number N that function F of XN would be less than this negative number N. And then exactly the same thing. It will be just negative infinity instead of positive infinity. So, the function is bounded on both sides. And it's very, very important property which we will use for some other theorems. So, you better remember it. Again, it's obvious. And again, it would be very educational if any obvious statement you can actually either prove based on certain statements which have been proven before or take as an axiom. Well, in this case, it can be proven. So, everything is fine. I do recommend you to read the same material I was just talking about on the website. And, well, basically, that's it for today. Thank you very much and good luck.