 So, welcome to module 5 of point set topology course. Last time we introduced the notion of epsilon delta continuity on metric spaces showed that it is equivalent to sequential continuity. That was a major result last time, so today we will use that and see a number of examples basically. So, here I have stated a theorem which is again using sequential continuity. So, first of all you must see what happens to sequences of functions, sequences inside k power n. So, that is the first theorem here. Consider k power n with the Euclidean metric which I have denoted by d 2 u e and I am just recalling it here instead of x and y I am writing u and v does not matter that is what you must be able to do. I range from 1 to n modules of u r minus v i square then take the square root. Suppose, u n now I cannot write this u n lower because u itself is a vector u and u 2 u n. So, I am putting a superscript here this is sequence in k power m. So, I have changed this n here that is why I have put m here now it is in k power m. So, this formula is two for all n after all right. So, I have to I cannot fix that one this n I am better I use for sequences. So, sequencing k power m where each u n is u n 1 u n m and so on. So, there are m of them m tuples of numbers here real number complex numbers whatever. So, let us say n tuple is a vector. So, it is a sequence of vector. So, this is a notation. Suppose this sequence converges to you under this metric and that u I can write as u and u 2 u n plain vector then and then only. So, this is if and only if each coordinate function here u i n converges to u i less than equal to m 1 for all n okay 1 2 3 up to m okay. So, each coordinate function converges to a point you take corresponding vector that will be the limit of the entire sequence here which is what it is. So, just like a sequence of complex numbers converges different only if each real part and imaginary part you get two sequences both of them converge. So, that has been generalized here for any k power n okay. So, let us just go through this one how this is just simple u n converges to u okay. Then what does it imply u n epsilon positive choose a k V wrong between such that p because any k implies distance between u p and u is less than epsilon. What is this distance? This is the square root of the sum of the squares okay. So, that immediately implies that each thing here must be less than epsilon because the square root of this whole thing is less than epsilon. The square will be less than epsilon square. So, each must be because these are all positive numbers each must be less than epsilon square. So, square root should be less than epsilon. So, each u i p minus u i must be less than epsilon okay. So, you have finished the same p we will do the job for all the sequences okay. So, what does it mean? u n converges to u means each u p n u i p converges to u i. So, that is the meaning. Now, conversely you have to work a little hard that is why I have put it here. Conversely suppose each of them converges the coordinate functions given an epsilon I should not produce a k such that this happens now right. So, what do I do? First of all what I get is for each sequence I get a k i belonging to n such that p is bigger than k i will imply modulus of u p i minus u i is less than something at something I am choosing not this epsilon but epsilon by square root of n. Apply it to the convergence for this number okay. So, this happens now you choose k to be the maximum of all these k i. So, I have reached for each i I have a k i take the maximum k i. If p is bigger than a maximum it is bigger than each k i. So, all of them will be true okay. You take the square that will be less than epsilon square by m. Take the summation there are m of them so it will be less than epsilon square when you take the square root you will get this one okay. So, I have I have not written down then I have asked you to check this one but I have told you how it works alright. So, as a corollary what we get is you get the projection maps pi i the coordinate projection k m to k okay. So, u is u i u i is a i th coordinate projection they are continuous. How do you get continuous? If u p tends to u pi i of u p is just u i u i p tends to u i. So, that is what we have proved already. So, this is a part of corollary here part of that theorem you do not have to use if and only if just one part. So, coordinate functions are continuous okay. Now, we will improve upon this fine. Take a metric space take a function from x to k power m any function in the product is the same thing as giving you m functions an m tuple of an m tuple of functions f1 f2 fm right. So, look at one function that is gives you m functions here what are they they are just fi f1 f2 fm which are just pi i composite f first apply f and then take the ith coordinate of whatever you get okay that is fi each fi. So, condition is if and only if each coordinate functions each of them is continuous okay all of them are continuous. So, how do you prove that same thing start with sequential continuity usually that is easier than finding epsilon delta and so on. Take x n tends to x in x now put u n equal to f of x n then you apply that theorem okay u n i will be what it will be fi of x n all right. So, you see sequential continuity helps you once that is why we converted this epsilon delta thing here into sequential continuity in this one okay. So, so that was the the theorem that you had proved last time. So, you can apply that a function is continuous if and only if each coordinate function is continued. As an additional corollary let us prove that the addition in K n as well as the scalar multiplication are continuous note that if you take two vectors u and v u plus v the ith coordinate of u plus v is nothing but u i plus v i the ith coordinates being added. Similarly, the scalar multiple alpha of u where u is a vector its ith coordinate is alpha times u i. Therefore, from our theorem to 1.20 the statement of this corollary reduces to the case when n equal to 1 by taking the coordinates okay. But once you put n equal to 1 this is nothing but the theorem 1.20 which you have stated last time since we have only stated and not proved it actually we said that you must have done it in your real analysis course and so on. In any case for completeness let us prove it now okay. Yeah consider the function first the addition which I denote by alpha right k cross k to k given by alpha of x plus x comma y is x plus y. To prove the continuity of alpha at any point say x naught y naught given epsilon choose delta equal to epsilon by 2. Now suppose this x y is in the ball b delta x naught y naught remember what is b delta of x naught y naught is all points x y is a set their euclidean distance from x naught y naught is less than delta okay. In particular if you look at the modulus of x minus x naught is always less than the euclidean distance of x y from x naught y naught because this is nothing but the square root of the sum of x minus x naught square and y minus y naught square. So it will be smaller than individual one square and then take the square root. So this is always true which we have used earlier also similarly y minus y naught is also less than delta right. Therefore what we get is alpha of x y minus alpha of x naught y naught will be less than what is this one this is nothing but the x plus y minus x naught plus y naught which is x minus x naught plus y minus x naught. The modulus will be less than equal to modulus of x minus x naught plus y minus y naught each of them is less than delta. Therefore the whole thing is less than twice delta which is epsilon okay. The proof of the multiplication continuity of multiplication is slightly more complicated okay. So let us go through that so I want to prove this multiplication scalar multiplication is continuous at x naught y naught right. So given epsilon I want to choose a delta appropriately. First of all look at x naught y naught take the maximum of the two okay add one call that number as 1 m m means maximum of mod x naught mod y naught plus 1. Then choose delta to be the minimum of epsilon by 2 m comma 1 okay. I would like I want it to be less than epsilon by 2 m but I also want it to be less than 1 that is why I am taking the minimum of the two okay. Now suppose x y is in the delta of x naught y naught as usual then I want to show that alpha of mu of x naught mu of x y minus mu of x naught y naught is less than epsilon that is what I have to show okay. So as before we have the moment x y is inside this ball x minus x naught modulus and y minus y naught modulus both of them will be less than delta. Moreover now x minus x naught less than delta implies more x itself is less than more x naught plus delta and this delta is less than 1 right so this should be less than m okay so this is less than m. So if you put all these things together what you get is now mu of x y minus mu of x naught y naught modulus of that is modulus of x y minus x naught y naught we rewrite it as x times y minus y naught plus x y naught minus x naught y naught add and subtracted. So this is modulus of x times y minus y naught plus modulus of x minus x naught times y naught okay and that is less than or equal to m times epsilon by 2m plus epsilon by 2m okay that m times comes out both for both of them here this is for y naught and this for x so so this completes the proof of theorem 1.20 that we are stated earlier and along with that the corollary 1.24 is also proved okay so we have the this very important theorem here I have just written corollary that does not matter okay. So as an example of application of this one any polynomial map p from k to k what is a polynomial map polynomial is something like c naught plus c1 x plus cn x power n where c1 c2 cn are scalars right so I want to say that any polynomial map is also continuous constants are continuous the first thing I wish to use that applies scalar the x going to x is continuous identity map so multiply the two of them that will be continuous which means x square is continuous multiply again that will x cube is continuous that means all monomial functions x going to x power n they are continuous so now we can add two of them at a time so c naught is continuous c naught c1 times x is continuous c naught plus c1 x is continuous and so on finitely many of them you have to add that will be also okay we can go one step further namely any polynomial map in any number of variables finitely many x1 x2 xn going to summation c alpha x power alpha mod alpha is less than or equal to d this is the notation if d is 0 this is just a constant that is no choice if d is 1 you can take x1 you can take x2 you can take x3 and all monomials of degree 1 you can take if you degree 2 then you have x square x1 square x2 square all of them not only that you have xi into yj sorry xi into xj so those things will be also there so how to write all of them it is a notation here take n ed which is a positive integer look at all multi indexes alpha which a1 a2 a n they are all integer non negative okay n two poles of integers because I have n variables here okay some total of these must be less than equal to d that is what you are putting the restriction because you want take only finite sum that is all okay then what is the meaning of this x power alpha it is one single notation for x1 raise to a1 x2 raise to a2 xn raise to an so that is the notation summation of all a1 a2 a n that is less than equal to d I write as mod alpha less than equal to d so mod alpha is this one this is like your taxi cab metric here okay and all these c alphas they are inside k they are coefficients they may be r or c more generally any field will do but we are concentrating only on real or complex numbers so such a polynomial is also continuous why because first of all all the xi's are continuous this is what we have proved so x1 into x2 x2 into x3 or x5 square and so all of them are continuous multiplying them by a constant they are continuous then you have to take the sum so instead of one variable you have one more namely here namely all you have to take various xi's and then take the product not just one x and its one power okay so that is all the differences so this is again the argument is similar that will give you all polynomials whether one variable two variable and variable they are all continuous functions okay to be sure you know that they are also differentiable functions right in fact they are more than that they are much more than that they are analytic functions and so on there is one more notion which you want to recall once again from function of one variable real variable or complex variable the definition is exactly the same again this time again you replace the modulus by the distance function that is all take a function from one metric space to another metric space it is called uniformly continuous then uniform continuity is defined on the entire set a there is no point in defining at a single point okay I can define it on some a a continuous x1 also that is possible but now here I am defining for over x1 so it will be true for any subspace is also okay so a uniform continuous function is defined for the entire domain here okay whatever domain you have chosen for the function so what is the definition for every epsilon positive you must have a delta positive such that whenever the points of a domain you know they are closed by delta d1 of y1 x1 is less than delta their images must be also satisfied that same relation but with this epsilon d2 of fx1 fy1 must be less than epsilon so you see there is no reference to given x given x and epsilon there exists delta was the definition for ordinary continuity at a point so there is no reference here this delta will just chip in upon epsilon it will serve the continuity for all points at all points in particular uniformly continuous functions are continuous also you know ordinary sense that is easy okay indeed I have not said anything other than replacing the modulus by delta on the d1 and d2 so you must be all familiar with these things okay so for instance what is the difference between ordinary continuity and uniform continuity in both the definitions this delta will depend upon epsilon okay but in the ordinary continuity it also depends because you start with x for each x and epsilon there is a delta so it may depend upon that x also okay in the uniform continuity there is no starting point at all you do not have to it starts with for every epsilon there exist okay and then the statement is for all points all pairs of points therefore uniform continuity implies continuity there is no problem okay but we will see some examples here that in general uniform ordinary continuity function may not be uniformly continuous so first example is the simplest example namely x going to x square okay you must have checked it if you haven't you should check it right now all right doesn't take much time not uniformly continuous means what again you have to do so given epsilon there existed delta that is the condition right for all t should happen so what is the negation of that there is some epsilon just that no matter what delta you take somewhere something goes wrong that point you can choose freely here that is the point because there is no for every given x x is not a fixed starting point or anything okay so you can take any number one for example and then show that there is no such delta which will give you the continuity for all the points that is what you have to show so having told that much I will leave it to you the same thing with t going to e power t is not uniformly continuous okay on the other hand closely related to that is the the the trigonometry function sin and cosine they are uniformly continuous okay I don't want to reveal there are tricks to see why they are uniformly continuous I don't want to reveal that to you right now you think about them all right if you have problems with this gun you should not hesitate to contact us on the platform namely the discussion form is there okay so there you can ask I tried this way I tried this way but I am not getting it okay please explain so our t s will explain that so let us look at another example here namely tangent and cotangent functions okay they are also not uniformly continuous so this time the domains are finite domains here but but the co-domains are the whole of minus infinity the entire of r okay so entire of r these these are actually homeomorphisms 1 1 mappings here but they are not uniformly continuous okay so uniform continuity is something somewhat funny thing it is not a topological quote end quote topological property at all so in topology you won't much see it at all in general topology there is no uniform continuity concept to bring it you have to do something like a metric space not exactly something like metric say what are they called they are called uniform spaces okay because they can they can have uniform continuous functions so you may you may have many functions which are continuous but not uniformly continuous but look at the look at the thing here it seems to have something to do with the domain and co-domain okay so indeed you have also studied in your real noise course namely every continuous function on a closed interval is always uniformly continuous okay so though I have given you these examples of not uniformly continuous functions if you restrict them to any closed interval they will be uniformly continuous the emphasis here is is not uniformly continuous whole of r similarly equal to equality is not continuous on the whole of r that is the point if you restrict it to close intervals they will be continuous that is one of the theorem that theorem namely a continuous function on a closed interval is uniformly continuous that one we can extend it in some sense and that is that that comes back in topology also definitely in metric spaces later on we will do that so here is one example which you might not have seen in your analysis course at all because this is about metric spaces take a metric space take any non-empty subset okay so the layman's language of distance I am going to use that so I am going to take arbitrary points and then talking about the distance of that point to the set okay so that is what d underscore a of x or d of x comma a whatever notation you want to use you can use so this is defined as infimum of all the distances between a and x where x vary x is fixed at a varies over a look at all these these are all positive or non-negative real numbers okay take the infimum why this is infimum makes sense because this is bounded by 0 below so infimum means at worst with the 0 or it may be some positive number it makes sense this infimum is called the distance of x from a okay so that is the notation d a of x or d of x a this function you can easily check that it is uniformly continuous you do not have to hunt for a delta it is there I will give you that a hint so write down the details yourself okay any doubts we will stop here until next time thank you