 Welcome to module 20 of point set topology part one. So today we will discuss one small application of the Bares category theorem. As I have told you, all these three big theorems in metric spaces, they have many many applications in analysis. For example, the Benach contraction mapping principle is used in the existence and uniqueness of initial value problems, boundary value problems and so on in differential equations. So instead of calling theorem people will like to call them as principle because the result they want to use is the principle rather than the final result which Benach compract mapping theorem says. So today we will just give you just a small application in the elementary analysis. Since we cannot discuss the big applications of this in function analysis in this course, we start with a domain inside R, take a function, any function, real world function, take subsets of J which are open intervals, non-empty open intervals, those things I will denote by I, J could be any domain connected, open connected, need not be a some open subset of R will do that is all or even close subsets of a close interval or whatever. The function has to have a domain so J is that. For each fixed open non-empty open interval contained inside J, let us put this notation omega fi which is the difference of the supremum and the infimum of fx taken over the entire of i. So you take the supremum first, get the infimum and then take the derivative or you can just look at modulus of fx minus fy and take the supremum of these quantities as x and y freely range over i, i is fixed in here, for each i I have this number okay, maybe this could be infinity also, I do not mind okay plus infinity is also allowed alright. So only thing is I am assuming that i is non-empty so this will not be something like minus infinity plus infinity and so on which will not make sense, so this always makes sense. So this quantity is called oscillation of f in i, so you can say that how the difference between fx and fy how large it could be, so that is oscillation. Now for each fixed x you take all neighborhoods of i which are intervals, open intervals around i okay, put omega of fx equal to infimum of all this omega fi where i ranges now x is fixed there okay, so these are all subsets of the domain whatever. So A B should be J not here? A B could be J, so you know depends upon what you have taken for the domain of f, normally it does not matter because you will have to take i smaller and smaller, so it is around x that is what is important. So omega of fx is called the oscillation of f at x, for example I will just give you a very simple minded example, look at a step function, so here I am taking the domain as closed interval 0, 2 okay, it does not matter what you take, it does not matter, but I want to include a step function, so it should include some points wherein the breaking is over. I believe that you all know step function, so here is 1, fx is equal to 1 in the interval 0 to 1 open, it is equal to 2 from closed interval 1 to 2, so at 1 x equal to 1 there is a break here you can see okay, then take any sub interval around i around 1, this i is around 1 okay, look at or any sub interval open subset you have take which contains i, omega fi will be exactly 1 okay, now if you take the infimum it will be also 1 okay, so omega f1 will be 1, remember omega fx is the infimum of all omega fi okay, here is a picture what happens, this is a step function at 0 to 1 it is this 1 okay, I have deliberately changed here in the thing, here you could jump or this point you could be a put it here does not matter, the difference modulus okay, for any point here point here the value the difference is always 1, so the infimum will be also 1 okay, now you have definitions of continuous functions usually you take the domain to be an open set right, so I have taken open interval finally you can you discuss of continuity in at closed intervals also finally, but to begin with you take open intervals anyway, so take an open interval okay, take a point x inside that take a function it is continuous at that point if and only if the oscillation at that point is 0 okay, so this could have been right in the beginning you are a definition of continuity because if and only if but you have defined continuity in a different way, so let us just check it it is very straightforward okay, suppose omega fx is 0 then we want to prove that f is continuous in the epsilon delta definition, so given epsilon positive we must produce delta such that omega f okay, omega f y is less than or epsilon for all y such that modulus of y minus delta is less than delta y because look at this we want finally given a point x and y okay, y near x namely modulus of y minus x less than delta modulus of fx minus of y is less than epsilon if that is true for all x y inside that the supremum will be also less than that which is same thing as omega f, so I am using this part of the definition here these two are equal is obvious to verify only thing you have to know what is the meaning of supremum and infimum all right, so now omega fx is 0 first of all implies there exists an open interval i around x such that omega f y must be less than epsilon if the infimum is 0 this has to be less than epsilon right also you can see that you are taking infimum when you take a smaller set smaller set of values you will get you see larger set the infimum could be larger, so omega of i1 will be less than omega at i2 f is common here so that is f here choose delta positive such that i delta is x minus delta x plus delta contained inside i so this i delta is just in a short notation then omega of f i delta will be less than epsilon because I have already made omega of f i is less than epsilon, so now for any y in i delta what is the meaning of i delta that modulus of y minus x is less than delta okay look at all the infimum of fz such that z belong into i z that will be left i delta that will less than f y because I am taking infimum here y is one of the points here and that f y will be less than supremum of all the fz that belong this one because f y occurs here as well as here therefore when you take the difference f y minus f x that will be less than omega f i delta okay which is supremum of this minus infimum of that all right so therefore that will be less than epsilon okay omega f delta is already chosen to be less than delta less than epsilon because it is less than omega f i so this proves the continuity the other way around is even simpler so I will leave it to use an exercise this is just to warm up so that you might have forward continuity epsilon and the continuity definition and that is all so what I am interested in is take f from r to r any function and df be the set of all points at which f is discontinuous so d for discontinuous the set of points of discontinuity is an f sigma z okay so this is the proposition okay so here there is no continuity remember f is continuous some point it is not continuous at all the point you have applied this g delta f sigma for continuous functions now I am taking set of discontinuities is f sigma so how to prove this these are standard methods in real analysis so it is the method should be paid attention to you know they are educative that is why look at sn all points x in r such that omega fx is greater than or equal to one by n why I am looking at this I am looking at the points wherein omega fx is not zero omega fx is always non negative if it is zero corresponds to continuities of f so I am looking at points wherein it is positive it is positive it must be bigger than equal to one by n for some n so that is why look at this set okay then if you take df which is all points wherein it is positive will be the union of essence right because every point must be bigger than one by n for something at every point something this must be bigger than one by n so df will be union of essence we need to show that each s then is closed then this will be f sigma over okay so here is something I am not claiming omega fx is continuous or anything okay but greater than equal to one by n is a closed set is what I am claiming if it were continuous function this would have been obvious so let x be not in sn okay I should show that there is a open subset around x some delta whatever for that it is not in sn okay the whole open set is not in sn that means I am trying to prove that complement of sn is open okay suppose x is not in sn then omega fx by definition is less than one by n okay which in turn what is omega fx it is the infimum of all omega fi where i is some interval around x so that will this is less than f one by n means that there is an open set open interval i such that x is inside i and that oscillation inside i is less than one by n okay for other intervals it may be bigger but at least one of them must be there otherwise infimum will be bigger than equal to one by n so infimum is less means one at least one of the values must be less than one but then for all y in i okay i is an open interval omega f y will also less than one by n because now I have to take infimum over all these things one of the values is already one by n okay when I range y when I range i okay so where i is one of the things and y is ranging inside i so for every i fix i fix y around y one of the values is one by n so omega f y will be less than one by n okay so from x to y we are coming through this interval your i is a common interval here so once this this happens what happens the whole interval i is contained inside the complement of sn for each point you have what a interval so sn is open all right okay so coming to the proof of final theorem what you are going to do is combining this proposition that this tefsig mass heck okay along with what along with our example that what was the example let us have a look at it this example said that if you take any subset close subset of of irrational numbers then it is no inheritance so let us go here that is the example here okay the set of irrational numbers is not if f sigma right why if once it is contained inside irrational number it is no inheritance so that is that is the discussion here in fact what we have done is even if it contains a rational number it is it is f sigma because either rational numbers or irrational numbers they do not contain any interval but close subset you have taken all right so the conclusion here is there is no function from our power which is continuous at all rational numbers and discontinuous set all irrational numbers why df for such a thing would be exactly equal to r minus q then this proposition says that r minus q is what f sigma there is no problem about that but f sigma means what all these essence are close subsets of r minus q therefore it is no inheritance therefore I have written the rational numbers as what countable union of no inheritance sets that is not possible was the example okay the countable union of no inheritance sets is not possible for all the irrational numbers so there is no such function so this is not at all such a important result but what why I have put it here is just an illustration of the power of the BCT bias category theorem from very many things to very strong things it can control very important things can control just to complete the picture I want to remind you that if you interchange the role of rational and irrational here there are such functions there are continuous functions such that at all rational points they are discontinuous and at all irrational points they are continuous it is continuous so the so the df can be the set of rational numbers or any subset of it also no problem at a few points of rational irrational numbers also it is just the rational numbers it is not possible that is the conclusion of this theorem okay so here is that function which is which has many many names so it goes by idea goes back to Dirichlet Dirichlet add a other some slightly different put one here instead of zero here so Thome has slightly modified it and it becomes a quite interesting example so fx is zero if x is irrational and 1 by q if x is p by q with gcd of pq equal to 1 okay so just to remind you what will be f of zero in this f of zero by this definition see you have to write zero is a rational number how do you write zero zero as zero divided by one you have to write that is the only way gcd of p and q can be equal to 1 when p is zero therefore the value of this function at x equal to zero is 1 by 1 what is 1 okay I am just explaining the definition all this okay suppose we have 5 by 10 then you should write it as 1 by 2 and then it will be 1 by 2 suppose you have 15 by 25 then you will have to write it as 3 by 5 then this should be 1 by 5 and so on the continuity of this one is interesting the property of rational numbers or more more so of irrational number that I will not discuss this is very standard okay so let us stop here next time we will do one more serious result and that will be the end of this chapter okay so tomorrow we will do one more serious result about metric spaces namely completion of metric spaces any questions okay so this the theorem about discontinuity set of discontinuity is being f sigma set is there any analogous result for metric spaces metric spaces is can it be extended complete metric space is serious you see complete metric spaces you can formally say you know what you need to show first of all you can try to see that this omega f f x direct to 1 you should be able to show that it is closed so where how far you have used here what is the meaning of the oscillation etc you will have to be very carefully able to define okay so what is that you have to take neighborhoods instead of I you can you can just restrict yourselves to just close the open balls around that point no problem look at all x comma y it is ranging in the open ball okay modulus of f minus f y but the values have to be real value functions okay or not arbitrary then there is no infimum and supremum so domain can be arbitrary metric space but you have to put completeness okay then at least it makes sense so you have to check it it is a good exercise try it good question so try it and see how far all these things go okay countable union automatically comes just like this one so if you can show that these are close subsets which is not difficult any compliment is open is similar to proof is here okay now the problem problem here is why these essence okay are no evidence so you do not have irrational numbers and so on there in arbitrary metric okay so what is the conclusion it is f sigma up to there you are fine okay so you have to assume that the set of suppose I assume the set of discontent is is no evidence does not sorry has empty interior yeah only empty no content no that kind of things you have to show first assume that then that cannot be a set of discontinuities set of continuity sorry set of discontinuities okay so I have what I mean to say there is no concept of the set of irrational numbers or compliment irrational numbers and so on in arbitrary metric space so you have to replace the corresponding by correct hypothesis that's all okay all right let's stop here