 Welcome to module number 4 of point check topology course. So, today we shall take up one of the important concepts namely continuous functions on metric spaces. Starting with the idea or whatever we have learned our basic lessons in real analysis of functions, real value functions of real number, how you define continuity there using epsilon delta definition without any effort exactly same definition can be adopted here for continuity. So, and that will be called epsilon delta definition of continuity. You know it already. So, I am recalling it x 3d here, take a function f from x 1 d 1 to x 2 d 2, what of that mean? That means that x 1 is a set and d 1 is a metric on it, x 2 is a set, d 2 is a metric on it. The function is actually a function from x 1 to x 2, take a point x 1 in x 1, a function f will be called continuous at x 1 if the following happens. What is that? So, that is the epsilon delta part for every epsilon positive there exists delta positive such that d 1 of y 1 x 1 less than delta implies d 2 of f y 1 f x 1 is less than epsilon, x 1 and y 1 are inside capital X 1 obviously and I am applying d 1 to that, that distance is less than delta implies the corresponding distance between f of y 1 and f of x 1. Now, this time I have to take the d 2 distance from for x 2 that must be less than x 1. If this here happens for all points x 1 in x 1 namely f is continuous for all x 1 at all x 1, x 1 in x 1, then we say f is continuous on x 1 or just f is continuous. Let f from x 2 and x 2 continuous means it is continuous at all the points. We also use the word map which is a small word compared to all continuous functions. One of the important thing is what is called as using the sequences and that is again very much available to us in the case of metric spaces also exactly same kind of definition same kind of techniques and so on. So, let us make a first, let us make a one more definition what is the meaning of convergence of sequence in a metric space. Take a metric space x d, take a sequence x n in x. It is supposed to converge to a point x I am making a definition here so x n converges to a point x inside x if the following thing happens once again in terms of this epsilon n number k here. For every epsilon positive you must have a natural number k such that all n bigger than equal to k d x n comma x the distance must be less than epsilon. So in other language is after a certain stage all the points of the sequence are close to x what is this closeness the distance is less than epsilon. If you remove this d and replace it by modulus modulus of x n minus x what you get is the definition of continuity for k valued functions real or complex value function ok. So, we have just done the other way round you took the definition from complex analysis or real analysis whatever you have learnt the modulus you have replaced by the distance function here that is all ok. So, whenever a sequence converges we just write it in a very simple way x n arrow x. So, this I can read it as x n converges to x over then all these conditions are inside my mind ok. So, by just this symbol we are expressing all this idea. Now I am going to give you the equivalence of this epsilon delta definition with the what is called a sequential continuity ok. So, start its function f from x 1 to x 2 again take a point x inside x 1 f will be continuous at x if and only if for every sequence x n converging to x the sequence f x n must converge to f x ok. So, that is a condition f is continuous take a sequence it must converge take a sequence which is converging f of x n must converge if this happens for every sequence then f must be continuous x ok. So, these two things are equivalent is displayed here namely epsilon delta continuity is equivalent to sequential continuity. So, let us go through the proof here so that these things become solid in your mind. Suppose f is continuous x and x n converges to x then I should show that f of x n converges to f x right. So, to show that I take an epsilon positive then I must produce a number k such that for any gate current could k distance between f x to f x n and f x f x is less than epsilon this is what I actually produce. But what is the hypothesis f is a continuous at x and x n converges to x. So, given epsilon positive first I take using the continuity I take a delta positive such that distance between y and x is less than delta implies d 2 of f y and f x is less than epsilon. So, I use the continuity here now I will use the convergence also but this time I will replace epsilon by delta for this one. So, delta is also positive number for that I must get a k because k belong to some some natural number such that n bigger than equal to k implies distance between x n and x is less than delta. So, this is now playing the role of epsilon this delta is playing the role of epsilon for the convergence here. So, both the conditions are used now you just combine them moment s is bigger than a k this distance is less than delta. But now I can put this y equal to x n then what happens this distance delta so d 2 of f of x n comma f x will be less than epsilon that is this condition. So, one way we have proved namely continuity implies sequential convergence. So, now we shall prove the other way round. The other way round is not all that straight forward. Soon we can say these things are also straight forward straight forward itself is not all that straight forward. So, here you have to use the contra positive you want to prove the contra positive of the statement. Conversely assume that f is not continuous then produce one sequence which will violate the condition that is all produce one sequence x n converging to x but f x n does not convert to f x. So, what I am going to prove here assuming this is not true I am going to prove this is not true what do they mean this is not true this statement says that for every sequence something happens. So, what is the negation of that there is one sequence this does not happen. So, what is that that is one sequence converges but f x n does not converge. So, that is the final statement I have to prove. So, I have to produce such a thing now. So, this is where you have to learn what is the meaning of negation what is the meaning of contra positive and so on. So, I will give you one minute all of you to think about it and tell me what is the meaning of f is not continuous at x just write it down I cannot check it of course I will give you a minute and then I will proceed. So, if you have done it correctly congratulations in any case we in somewhere the answer that I am going to write down here I am going to prove that there is a sequence x n which converges to x and f x n does not convert to f x starting with the assumption that f is not continuous. So, what happens f is not continuous implies there is one epsilon positive for which blah blah blah those things are not true. So, what is not true for every delta you have to get a point such that something happens. So, that delta for every delta. So, I choose delta to be 1 by n. So, that is what I am doing for every n inside n there will be a some x n this depends upon the n I have chosen such that the distance between x n and x is less than 1 by n. So, this delta I am taking as no 1 by n I can take every any delta because for every delta it is supposed to be true there is for every delta there is something. So, for 1 by n there is something and that I am calling it x n d 1 of x 1 x n x 1 is less than 1 by n. But what is it the distance between f x n and f x is bigger than equal to epsilon if that is less than epsilon there is nothing. So, I have to say this is the whole condition is violated namely f is continuous it is not continuous at x 1 that is what I am doing. So, d 2 of f x n f x is bigger than epsilon this happens for 1 epsilon then I am then that is what it is f is not continuous. But what I have what is fixing 1 epsilon here I do not know what it is bigger than equal to 0 that is bigger than 0 that is all I have what a sequence now this sequence automatically converges to x f of x n on the other hand does not converge to f x because it is keeping away from f x a constant distance bigger than equal to epsilon over. The next theorem is one of the profoundest thing that we have to learn whereas in reality it is just very very simple you will see that this proof is very simple and if you use this one later on lot of things become very simple. So, this may be one of the things that you know why topology as a whole comes into picture after all. But right now there is no topology in that metric space is here f from x 1 to x 2 g from x 2 to x 3. So, I can talk about g composite f right suppose the function f is continuous at x 1 and g is continuous at f of x 1 I have read it in x 2 then g composite f is continuous at x x 1. So, this also theorem in real analysis as well as compression analysis proofs were identical there the proof I am going to give here is also exactly same the point is that it does not depend upon all that strong structures real and complex numbers. It is just the metric that we are going to use that is why it is there. So, look at the proof I want to show that g composite f is continuous at x 1. So, I start with an epsilon first I use the continuity of the g at x 2 that will give me a delta positive such that distance between x 2 y 2 is less than delta implies g x 2 and g y 2 the d 3 distance is less than epsilon. So, this is the continuity of g at the point x 2 x 2 is what f of x 1. Now, use the property that f is continuous at x 1 this time using this delta again in the place of epsilon. So, then I get a delta prime positive such that distance between x 1 and y 1 less than delta prime implies d 2 distance now between f x 1 and f x f y 1 is less than delta. Now, just like in the previous theorem you combine them start from here you get here. Now, play the role these things will play the role of x 2 and y 2 this is already x 2 put y 2 put f x 1 here and use this equation number 3 this is the implication. So, you all the way come here but what is g x 2 it is g composite f of x 1. Similarly, what is this one it is g composite f of y 1 because I have put y 2 equal to f of y 1 over. So, this implies that g composite f is continuous at the point x 1. So, see using combining these we have used twice in some sense. There it is it was for convergence of sequences and continuity here both of them are for continuity. So, let us now do some examples kind of thing but I will state them as theorems because they are very profound results. Once again we go back to real numbers or complex numbers take a sequence u n converging to u and another one converging to v. Then if you take the sum sequence u n plus v n is another sequence which is a sum of this sequence with that one. u n plus v n will convert q plus v and the product u n v n will convert to u. You remember this result or not in real analysis or complex analysis. Now here this is what I am not going to prove in real analysis you have this one. You can use this one to prove many things slowly. This you could have done in real analysis course also if you have not done it. Now I am going to do that because those things are crucial for me. Addition and scalar multiplication inside k right k cross k to k you take the addition or scalar multiplication they are continuous. This is the statement now. From convergence of these results about the convergence of the sums and product I am concluding this one. How do I do that? Because continuity I can convert it into sequential continuity epsilon delta continuity is equivalent to sequential continuity. So what happens I have to verify suppose you take I have to alpha alpha is continuous means what suppose you take a sequence here converging to u plus v whatever then alpha of that should convert to that is what I have to show right. So similar for example let us do this one here u n converging to u v n converging to v u n v n will convert it to u comma v. The product you know u n v n converges to u v therefore it is continuous. When you have a product that is f of that whatever here it is mu so here it is alpha. So use this theorem for both of them namely this theorem then you get these results that scalar multiplication and addition they are continuous. So I will explain this one once more. Look at how to how to show that alpha of u plus v alpha u is u plus v how it is continuous. I have to take continuity at a point u naught v naught if you want to take. Now how do I do I have to take two sequences u n converging to u naught v n converging to v naught. Then I have to apply alpha of these sequences what is this it is alpha of u n plus v n where does it converge it converges to u naught plus v naught. So I am using that convergence to show that so for every sequence converging to that I have shown that alpha of this sequence converges to alpha of the limits. Therefore alpha must be continuous similarly for mu mu is multiplication. Exactly same only thing is instead of plus you have to take the product here. Okay. So let us stop here next time we will do all these things we will use them and do a little more about examples of continuous functions. Thank you.