 We are trying to prove the theorem that we define for a function f on a domain d in R n to R to be uniformly continuous. If for every epsilon bigger than 0, there exists delta bigger than 0 such that whenever two points are closed by a distance delta that implies the distance between f x and f y is less than epsilon and we were trying to prove theorem namely f as before d in R n to R the following are equivalent saying that f is uniformly continuous is equivalent to saying whenever we have two sequences a n b n in the domain d such that they come close to each other. So, this goes to 0 then this implies the images also come closer goes to 0. So, we had proved one way but anyway let us recall suppose f is uniformly continuous. So, we are trying to prove one implies 2. So, let f be uniformly continuous. Let us take sequences a n and b n such that the distance between a n and b n goes to 0. So, we have to show that the image sequences also have the same property as of course n goes to infinity. So, this is what we want to show. Now, to show that this goes to 0 what one is to show that given epsilon bigger than 0, there is a stage after which everything comes less than epsilon. So, let be given we want to find a stage n naught such that after that stage this distance between f of a n and f of b n will be less than epsilon. But we know that will be less than epsilon whenever by uniform continuity if a n and b n are close and that happens because this goes to 0. So, let us choose some delta bigger than 0 such that no such that x minus y less than delta implies f x minus f y is less than epsilon. That is possible because the function is uniformly continuous that is given to us. So, whenever x and y are close by a distance delta that will imply that f x and f y are close by distance epsilon. So, if a n and b n come close to delta then corresponding will be less than epsilon and that is okay. So, find n naught such that mod of a n minus mod of b n is less than delta for every n bigger than n naught. That is possible because we are given that because mod of a n minus b n goes to 0. So, given any delta we can find a stage after which they are small. So, implies a n b n is less than delta so by the equation star. So, star implies f of a n minus f of b n is less than epsilon of b n bigger than n naught. So, that is what we wanted to show that if a n and b n are close then f of a n and b n are close so that is hence two holds. So, what we have shown is f is uniformly continuous then this implies this property 2. Conversely let us show two implies so conversely to show two implies one. That means whenever two sequences a n and b n are close we want to show that f of a n and f of b n are also close that is given to me. Second we want to show f is uniformly continuous. So, suppose one does not hold so what is the meaning of saying that f is not uniformly continuous that means uniform continuity meant for every epsilon something happens. Not uniformly continuous implies whole then there exist some epsilon bigger than 0 such that for existence we had there exist a delta. So, such that for every delta bigger than 0 we can find a pair of points there exists two points x delta y delta such that the distance between x delta and y delta is less than delta but the distance between f x delta and f y delta is bigger than epsilon. That is the meaning of saying f is not uniformly continuous something similar we had done when we were trying to prove that continuity implies a n converges to a implies f of a n. So, same proof basically negation is important. So, we want a sequence now. So, in particular for delta equal to 1 over n there exists x n y n in the domain such that x n minus y n is less than 1 over n we are specializing delta to be equal to 1 over n but f of x n f of y n is bigger than or equal to 1 but that contradicts the given hypothesis 2 which says whenever sequences are so this contradicts. So, that is what was given to us 2 implies 1 and what was 2 2 was whenever a sequence is closed images must be closed. So, we found a sequence x n y n they are closed. So, this goes to 0 but f of x n minus y n the distance always remains bigger than epsilon that is a contradiction. So, hence 2 implies 1. So, uniform continuity can also be expressed as in terms of sequences namely whenever 2 sequences are closed in the domain the image sequences must be also closed. So, that is in terms of sequences. So, let us look at some applications of this or some consequences of this. So, that is normally you prove so 2 statements are equivalent not just because we want to prove mathematically something nice but also in some situations one is useful, one criteria is useful in some other situations some other criteria is useful but both are equivalent. So, for example, saying that a sequence is convergent if and only if it is Cauchy the 2 statements are equivalent a sequence is convergent is equivalent to saying a sequence is Cauchy. Sometimes if you want the limit then you have to use the first one that sequence is convergent by finding the limit of it. But if you just want to prove that a sequence is convergent you are not really interested in knowing what is the limit then proving Cauchyness is good enough elements are coming close to each other. So, for example, here let us look at some example. So, here is let us look at f of x is equal to 1 over x whatever x not equal to 0. So, look at the function f of x is equal to 1 over x. So, if you look at the graph of this function geometrically as you come closer to 0 your graph blows up is increasing at is increasing much faster. So, it looks like this is so it will 1 the value is 1. So, here it goes like this is coming closer and closer to 0. But if you see that between 0 and 1 is becoming large very fast. So, same notion of closeness near 1 will not work near 0 you require much bigger kind of thing. But if you want to look at sequences. So, let us look at let x n be equal to 1 over n and y n be equal to 1 over n plus 1. What is the difference between the two? So, what is that? So, there is 1 over n into and that goes to 0. So, in the domain I have got two sequences one is 1 over n other is this and what happens to f of x n? What is the difference between these two? So, that is n absolute value that is equal to n plus 1 minus n absolute value that is equal to 1. So, I have got a pair of sequences in the domain x n y n such that the distance between them goes to 0, but the distance between the image does not go to 0. So, that is one way of saying hence f of x equal to 1 over x is not uniformly continuous. So, that is one example of how sequences are useful in proving these kind of things. Let us look at some more examples. We had already shown that f of x is equal to say for example, that x square we showed is not for x belonging to R is not uniformly continuous. But if I restrict the domain equal to say x square between say anything between say 1 to 3, x belonging to is uniformly continuous. That we proved as example we showed that if you restrict the domain then it becomes by using epsilon delta definition. In fact, so here is a theorem. Let f be be from D contained in R n to R be such that d is compact that means equivalently closed and bounded and f is continuous. Then, that is the kind of thing happening here in f of x is equal to x square that is continuous and the domain is a closed bounded interval 1 to 3. Whenever such a thing happens, f is uniformly continuous and f becomes uniformly continuous. So, let us see why is that. So, let us use the sequence criteria here. So, let x n in D such that in R n, I will just write absolute value meaning that in R n it is a norm. If you write, you can write norm of x n minus y n goes to 0. Everywhere, whenever we are looking R n instead of absolute value, you can think of absolute value of vector as the norm or whatever it is. Is it okay? It is a matter of notation. For a vector in R n, the norm, I sometimes call it as absolute value also of that vector. It is same as sigma i i square root that thing. So, that is the notion of distance in R n. So, this goes to 0 to show, so what we have to show? f of x n minus f of y n, we are in the real line. So, that goes to 0. So, that is to be shown. Now, look at x n. So, let us write and then probably explain why I am doing that. So, x n in D, D compact, what does that imply? It should have a convergent subsequence implies there exists x n k, a subsequence of x n such that x n k converges to something, let us call it as converges to x. Now, x n k converges. So, consider y n k, the corresponding choice of the subsequence y n k. So, what do we know? We know x n k and y n k are coming closer goes to 0, because x n, y n goes to 0. Subsequence is corresponding. We will also have the same property and x n k is converging to x. So, where does y n k converge? So, implies y n k converges to x. So, what does that imply? What is given to us? f is continuous. So, f continuous implies f of x n k converges to f of x and f of y n k also converges to f of x. So, that implies that both are converging to f of x, same limit. So, f of x n k minus f of y n k goes to 0. So, what we are showing? Are we showing what we wanted? We wanted that f of x n should and y n should converge, the difference should become. What we have shown is there is a subsequence for which that is happening. That is not good enough. That is not only shown for a subsequence. We want to show it for the original sequence. So, let us... So, this direct kind of argument does not work. So, let us go to... We want to avoid that going to a particular subsequence. So, let us assume that this statement is not true. So, let us assume that... So, not good enough. So, consider... So, we need to modify the proof. So, consider assumption, suppose minus f of y n does not go to 0. But we still have the fact that x and y n are going to 0. That is already given to us anyway. This does not go to 0. So, implies what? Something, a sequence of numbers. This sequence of numbers that does not go to 0. That means what? Negation of the statement implies there exists some epsilon bigger than 0 and subsequence x n k, y n k such that mod of x n k minus f of y n k remains bigger than or equal to epsilon for every k. Negation of the statement that does not go to 0 should remain away from 0 for at least a subsequence. Now, I can work with the subsequence. So, now consider x n k in D which is compact. D compact will imply what? Will we learn to pin the same track? x n k will get a subsequence which is convergent. So, that will be convergent. We are lending up at the same blockade in our proof, thinking a sequence proof should be good enough. But the problem will be reaching at the same place because we will get a subsequence for which this contradiction will be there. But for every y n k, that should be okay probably. For every k, I think may be okay. Let us just write and see whether that is good enough. If not, we will change the argument. D compact implies there exists, for this there is a subsequence. So, same argument as before, but for the subsequence there exists a matter of writing. There exists a subsequence x k n is already a subsequence. What would we write to write? So, let us write x k n l such that x n x n k l, not k n, but I just, this subsequence of a subsequence x n k l, x n k l converges to some x. So, let us write. So, implies, that is good enough, I think. It is greater than, that is okay. So, implies what? Implies, if I look at the corresponding subsequence y n k l, that also converges to x because x n k l minus x y n k l, that distance goes to 0. So, that subsequence is subsequence. I probably will explain again what I am saying. So, let us negation means this thing, that is okay. Saying that this does not converge to 0, that means there is at least one subsequence, where the distance remains bigger. Now, look at this sequence x n k, which is actually a subsequence of the given x n, but look at that as a sequence. That should have a convergent subsequence because d is compact as before. So, it is a matter of writing x n k l, that means it is a subsequence of x n k. So, in turn also a subsequence of x n, subsequence of a subsequence, that is a subsequence of the original also. So, let us, so that converges to x by compactness, but the corresponding y subsequence because x n's and y n's are coming closer. So, y n k also will converge to x because they are coming closer. So, that implies what? So, that implies by continuity f continuous, that f of x n k l converges to f x, also f of y n k l also converges to f of x. So, that should imply what? There are two sequences, which are converging to same limit f of x. The sequences must come close to each other now. So, implies mod of x n k l minus f of y n k l, that goes to 0. The same proof as before, but only for subsequences. But now, look at this statement, call it 2 and call that earlier statement as 1, where is that? This statement as 1. So, look at, for the original subsequence x n k, the distance remains bigger than or equal to epsilon for every element. So, for the subsequence also of this sequence, the distance will remain bigger than epsilon. So, x n k and y n k are subsequences whose distance should remain bigger than or equal to epsilon by 1. But by second, it says that should go to 0. So, that is a contradiction. So, this contradicts 1. And what was that 1? So, just remind you again, because of our assumption, the distance between the corresponding elements of the subsequence must always remain bigger than epsilon. But here, we have gotten two subsequences of the given same sequence. So, either the distance goes to 0. So, that is a contradiction. So, hence, the assumption must be wrong. And that means, hence, mod x n minus y n going to, so that, so what we have proved? We assumed it is not, so suppose it is not true, then that must happen. So, that proves, so we are proving 2 implies 1. Hence, 2 implies 1 is ok. No, what is 2 implies 1? I am sorry, we are not proving 2 implies 1. We are proving, if a function is continuous on a compact domain, then it is uniformly continuous. So, hence, that was earlier one, hence f is uniformly. Let me just, because I change things too much, so let me revise this. You are given f is a function on D, r into r, D is compact, f is continuous. You want to prove that f is uniformly continuous. So, we have to prove, whenever 2 sequences x and y in a such that the distance goes to 0, the corresponding distance between the image must go to 0. So, the basic idea is, if this does not happen, I will have a sequence x n k, y n k in the domain where the distance of x n k and y n k goes to 0, but this distance does not go to 0, it remains bigger than epsilon. That is a negation. So, that is a negation we wrote to show. So, this was not the correct one here. So, suppose this does not go to 0, so given epsilon bigger than 0, there is, so this is what is the assumption. So, if this does not go to 0, then for any, there is an epsilon, so there is a subsequence, so the distance remains bigger. If it does not go to 0, there is at least one subsequence for which the distance should remain bigger than 0. That is bigger than epsilon. Now, look at this x n k in the domain. x n k and y n k are two sequences in the domain. Look at x n k. It should have a convergent subsequence. So, x n k converges to some x for some subsequence. Corresponding subsequence of y n k should also converge to x, because x n k and y n k are close to each other. That is given to me. So, I have got a subsequence of x n k. I have got a subsequence of y n k, both converging to same point x. By continuity, the image of this subsequence will also converge to same point f of x. But that means what? If two sequences are converging to same point, then they should come close to each other. So, that implies for the given sequence, I have got a subsequence y n k and so on, such that their distance goes to 0. So, that is what we got here. But that cannot happen, because corresponding terms of the original sequence always remain bigger than epsilon. So, this is a contradiction. So, that proves that every continuous function on a compact set is uniformly continuous. So, as a consequence, one could have said that in that example, this is uniformly continuous. But we proved it by using definition alone.