 we were considering the real valued functions and the types of discontinuity. So, we shall continue with that discussion. Let us recall that we had tried following the type discontinuity. So, for this we had considered a function f defined on open interval a, b to r and x some element a, b and we had said that x is a discontinuity of let us say type 1 or another name is jump discontinuity or also called simple discontinuity. So, this means that the limit left hand limit and the right hand limit both exist and their values are different. So, this means f x minus and f x plus exist and the values are different f x minus not equal to this f x plus and the second type which is called type 2 discontinuity of type 2 this means that either f x minus or f x plus or both do not exist. So, we can simply say this means f x minus or f x plus does not exist. Before proceeding further let us see some examples of this kind of discontinuities. Again let us take the well known examples which we have considered earlier several times. For example, let us take f of x as integral part of x we can take any interval in fact we can take f defined from r to r. So, suppose I take let us say x is equal to 0 then we know that near 0 again let us remind remember that integral part of x is the greatest integer not greater than x. So, in this interval minus 1 to 0 its value will be 0 integral part of x will be 0 and in the interval 0 to 1 its value will be 1 and. So, it is clear that if you take f of let us say in this case f of 0 minus if you take any number here its value is value of f is going to be 0 and hence it is a constant function here. So, this limit will be 0 and on the other hand f of 0 plus for that we are considering the interval 0 to for whatever delta you take. So, integral 0 less than t less than delta whenever delta is less than 1 the function is going to be constant and its value is going to be 1. So, the right hand limit here is 1. So, this is a discontinuity of type 1 and you can now see why it is called jump discontinuity because the function takes a jump here. Of course, in this particular case it so happens that the value of the function at 0 it is same as this value of the but that may or may not be the case the value may lie somewhere in between also in in general. So, this is the discontinuity of type 1 or jump discontinuity alright. Let us take another example that is also something we have seen earlier let us say g of x is equal to sin 1 by x of course, this will not define the function when x is 0. So, let us give some while strictly speaking it is not necessary to define the value of the function x equal to 0 if I consider in the ring limit left hand limit or right hand limit. But it is since we are talking about the types of continuities let us take some value suppose I take the value 0 this value really does not matter. Now, in case of this you will see that neither g of x plus nor g of x minus exist that is the neither the right hand limit exist nor the left hand limit exist. Now, how does one show that I have said earlier also that the best way to show that the limit does not exist is that you construct two sequences both converging to 0 and limit of g let us say suppose you call those sequences x n and y n in such a way that x n and y n both converge to 0 in this case and g x n and g y n converge to different only thing is that if we want now we have to restrict this sequences x n and y n both either to the right of 0 if we consider the right hand limit or both to the left of 0 if we are considering the left hand limit. Let us take one of those things suppose we are considering right hand limit then in fact even this is also something I think we have seen earlier for example suppose you take x n as let us say 1 by n pi 1 by n pi and suppose I take n is equal to 1 2 3 etcetera then all x n are bigger than or equal to 0 all x n so this sequence lies on the right of 0 and what about g of x n g of x n is sin sin n pi so g of x n is sin n pi so that is 0 and in a similar way and of course x n tends to 0 x n tends to 0 and g of x n also tends to 0 g of x n also tends to 0 but what we can also do is that I can take let us say another sequence y n as 1 by let us say 2 n plus 1 let us pi by 2 then this also tends to 0 and g of pi n that will be sin 2 n plus 1 pi by 2 so that will be 1 that will be 1 so that is 1 for all so x n and y n both tend to 0 and g x n tends to 0 g y n tends to 1 and remember x n and y n both go to 0 from right so this shows that the right hand limit does not exist so this shows that g is 0 plus does not exist and so this is a discontinuity of type 2 of course we can also show that g of 0 minus also does not exist by similarly constructing two sequences but if the objective is to show that it is a discontinuity of type 2 it is not essential if you show one of these does not exist that is enough to conclude that the discontinuity is of type 2 in this particular case in fact both do not exist that is also recall another example which we had seen again in several context namely the so called Lebesgue function it is h x is equal to 1 if x is rational and 0 otherwise can be also described as the characteristic function of q in case of this function you will see that whatever x you take whatever x you take if you take any interval say x minus delta to x plus delta if you take any interval like this then any such interval will contain rationals as well as rational so what you can always do is that you can construct a sequence regardless of what whether x is rational or not you can construct a sequence of rationals converging to x in that case h of x n will converge suppose you take a sequence x n such that each x n is rational and x n converges to x then h of x n will converge to 1 so h of x n will converge to 1 in a similar way you can also construct a sequence let us say y n of irrational converging to x in which case h of y n will be 0 so that means this will have a discontinuity of type 2 for every x so h x is h x has a discontinuity of type 2 at all x for all x in case of this functions we have shown that the discontinuity of type 2 is at x equal to 0 for every other x it is continuous this function sin 1 by x for every x not equal to 0 it is a continuous function at x equal to 0 it is discontinuous and that is of type 2 whereas in case of this function at all integral values it has a discontinuity of type 1 at all other values it is continuous again whereas in case of this function it is discontinuous everywhere and has a type 2 discontinuity all right now we shall consider a class of functions again this class is something that you are familiar with what are called monotonic functions and we will see that as far as this types of discontinuities are concerned monotonic functions have some very interesting property namely that monotonic functions do not have this kind of discontinuity yeah for example we have seen this example integral part of x which I had shown just now that is how monotonically increasing function that is a monotonic function and we have seen that only discontinuity of that is discontinuity of type 1 what we shall show is that something like this is true for all monotonic functions so let us begin by recalling the definition of a monotonic function again it is convenient to define all these things in an open interval so let us start with so f from a b to r is said to be let us say monotonically increasing if suppose you take two numbers x and y in a b and x then x less than y this should imply f x less than or equal to f y we have defined what is meant by monotonically increasing sequence sequence is special case of functions so it is similar and we have seen that increasing monotonically increasing sequences have some special properties as far as convergence etc is concerned so similar thing is true for monotonically increasing functions when it comes to limits and continuity of course we can similarly define what is meant by monotonically decreasing function only thing is that this inequality will be reversed this will become f x bigger than or equal to f y so I can simply write it here decreasing so that will this will be f x bigger than or equal to f y and similarly we shall say it is strictly monotonically increasing if this inequality is strict that is f x less than f y and similarly strictly monotonically decreasing then finally what is meant by a monotonic function monotonic function means it is either monotonically increasing or monotonically decreasing so monotonic this means monotonically increasing or monotonically decreasing well you know several examples of monotonically increasing or decreasing functions or the functions which are not monotonically increasing or decreasing so we shall not go into that already we have seen like one example namely that integral part of x that is an example of a monotonically increasing function so let us go to the paint result what we want to prove is that if a function is monotonic that is the ultimate idea we will need some preparation for that what we need want to prove is if a function is monotonic then it cannot have the discontinuity of type 2 so let us start let us begin with a monotonically increasing function so let us say let f from a b to r be monotonically increasing then what we want to show is that it cannot have this discontinuity of type 2 and to show that it is enough to show that for every x both these limits exist suppose we show that that is enough right suppose we show that that is that is enough so then this is it for every x in a b f of x minus and f of x plus exist in fact we can say something more we can also say what those left and right hand limits will be what we can say is that this left let us say this is the interval a b and suppose this is x somewhere then if you take the set of all values of functions between a to x and take their supremum then that will be left hand limit of x that will be left hand limit of x that is in other words what we can say this l u b of this let us say f of t where t is between a less than t less than x we can say in fact this is what should happen this is f of x minus this f of x that is left hand limit will be supremum of all values of taken over these values of f t over this and similarly f of f of x this will be always less than or equal to f of x and f of x will be always less than or equal to the right hand limit f of x plus and what will be f of x plus for f of x plus you take this interval x to b and take all possible values of f on that interval and take their infimum take the infimum that will be the right hand limit. So, g l b that is infimum of f t where t varies between x and x less than t less than b and we can say something more if we take say two points x and y with x less than y then the right hand limit of x we are not at right hand limit as well as left hand limit exist right hand limit of x will be less than or equal to left hand limit at y. So, also we can say that also for a less than x less than y less than b f of x plus is less than or equal to f of y minus and I have stated this for broadly increasing function similar theorem for a broadly decreasing functions for decreasing functions what will happen is that this l u b etcetera that will be interchanged. So, you can suitably write the theorem for decreasing functions I will not write the complete statement. So, I will say similar statement for monotonically decreasing function. So, let us go to the proof of this. Let us look at this set first and why this is why the supremum of this should be same as your f of x minus. So, let me give some name to this set suppose I call this set a let a be equal to set of all these values f of t such that t varies over a less than t less than x. Obviously, a is non-empty you can always there is some t between a and x. So, that f of t lies here is it also bounded above because for every f t f t t is less than x. So, f t is less than or equal to f x. So, f x is an upper bound for this. So, a is non-empty and a is non-empty and bounded above by f x. So, l u b of a exists it is a non-empty set which is bounded above. So, by l u b x m l u b of a exists. So, therefore, l u b of a exists and not only that l u b will be less than or equal to f x f x is an upper bound. So, least upper bound will be less than or equal to f x. Let us now give some notation for that l u b also suppose let us say that let l be equal to l u b of a then this l is less than or equal to f x and we want to prove that l u b is same as this f of x minus that is we want to now put it l is same as f of x minus. We already shown that l is less than or equal to f x. Now, to show that is a limit let us use the usual epsilon delta definition. So, let us say that let epsilon be bigger than 0. Then we make an usual argument if you take any epsilon bigger than 0 then l minus epsilon is not an upper bound l minus epsilon is not the upper bound of this set because l is the least upper bound. So, there should exist some element here which is bigger than l minus epsilon. So, let us call let me call the element t naught. So, that epsilon bigger than there exist t naught such that a less than t naught less than x and l minus epsilon is less than f of t naught. Now, what is to be done after this is clear see f is a automatically increasing function. Remember what is what is the situation here that is we have this interval a to b x is let us say one of the points here and t naught is somewhere here between a and x naught. And f of t naught is bigger than l minus epsilon f of t naught is bigger than l minus epsilon. Now, is it clear that if you take any points say t naught to x any point lying between t naught and x then f of t will be bigger than or equal to f of t naught because f is automatically increasing and hence bigger than l minus epsilon. So, suppose I take this distance as delta suppose I take this distance as delta then for all then for that particular delta for all t lying between x minus delta to x we will have f of t bigger than or equal to f of t naught and hence bigger than l minus epsilon and obviously less than or equal to l. So, we can say that. So, let us just complete this. So, let delta be equal to x minus t naught then for all t such that x minus delta less than t less than x x minus delta is nothing but t naught x minus delta is nothing but t naught. So, what we have is that l minus epsilon is less than f of t naught and that is less than or equal to f of t l minus epsilon is less than f of t naught and see t is somewhere here between t naught to x. So, f of t is bigger than or equal to f of t naught f of t naught is bigger than l minus epsilon and f of t is anyway less than or equal to l, because l is the supreme of this. So, f of t is less than or equal to l, which will be anyway less than l plus epsilon. So, what did we show? That if t lies between x minus delta to x, if t lies between x minus delta to x, then f t lies between l minus epsilon to l plus epsilon. Now, this is same as saying that l is the left hand limit of f t s t goes to x. So, this proves that therefore, l is equal to f of x plus. So, what did we prove? We proved this equation that is this is what we had called l and now we have put it l is equal to f of x minus that is left hand limit. We have put l is equal to f of x minus and we have already shown that l is less than or equal to f x. So, all this we have proved. Now, what I would say is that similarly you can prove this other inequality. Just as here I have taken the set a as f t a less than t less than x, you can take the set b as say f t where x less than t less than b and take its infimum and show that that is nothing but the right hand limit. In the same way for example, if you take let us say m as the greatest lower bound of that, consider m plus epsilon that will not be the lower bound. So, there will be similarly some t naught and since the ideas are more or less very similar, I will not go into the details of that proof. So, prove that part yourself show that f of x plus is same as this number and then that number is bigger than or equal to f x. Now, what remains this last thing we need to show that if x is less than y then f of x plus is less than or equal to f of y minus. So, now consider this let a less than x less than y less than b. Of course, this proof is more or less trivial if we just look at the earlier it follows from what we have proved earlier. See if x is less than y that is let us say x is less than y here, I can consider some number t lying between those two. I can consider some number t lying between those two. So, consider x less than t less than y then remember one thing. Suppose, I look at f of y minus f of what is f of y minus by this result f of y minus is the supremum of f t lying between a less than t less than y if you replace x by y. So, is it clear that f of y minus should be bigger than or equal to this f t f of y minus should be bigger than or equal to this f t. So, f of y minus will be bigger than or equal to f what about f of x plus f of x plus is by this like it is greatest lower bound of this set f t x less than t less than b. So, is it clear from this that see t is bigger than x this t is bigger than x. So, f t should be bigger than or equal to f of x plus f t should be bigger than or equal to f of x plus is it clear. So, we have proved all the things required in this theorem. So, in particular we can say that what follows from this is the I will continue here is a corollary. Pornotonic functions cannot have discontinuity of type 2 that is the statement. Pornotonic functions cannot have discontinuity of type 2 which is same as saying that all the discontinuities must be of type 1 that is they must be jump discontinuities or symbol discontinuities. You can say that there is nothing new here it does not have discontinuity of type 2 means what the left hand limited right hand limit always exists and that is what we have proved that is what we have proved for every x left hand and right hand limit always exists. We can say something more we can say that if it is sub-rolonic function and if you look at this set of all discontinuities of f then that set cannot be very big that is that set has to be countable. Again you if you look at the function integral part of x what the discontinuity is the integral part of x it is a set of all integers that is a countable set that is a countable set. So, if you take any monotonic function if you and if you take the set of all discontinuities of that function then that set must be countable set which will also follow from what we have proved just now. So, let us write that as a theorem let f from a b to r be monotonic again here monotonic means monotonic increasing or decreasing then and let us say that let e be the set of all discontinuities of f. So, let us write the full form. So, it e is the set of all x in a b such that f is discontinuous at x then e is countable that is what we want to say. So, this shows that the set of all discontinuities of a monotonic function is a countable set usually in real analysis countable sets are considered small sets. So, this is explained by saying that the set of discontinuities of a monotonic function cannot be a very big set let us take the proof of this. For example, you cannot find a monotonic function whose set of discontinuities coincides with counter set that is clear or you similarly you cannot find a monotonic function which is discontinuous everywhere that is not possible. You cannot find a monotonic function which is discontinuous everywhere in some interval that is not possible because the set of discontinuities has to be a countable set. We want to prove something for a monotonic functions and as we know monotonic means monotonically increasing or monotonically decreasing. So, let us take one of the cases other will be similar. So, as usual we shall say that without loss of generality without loss of generality assume f is monotonically increasing and then let us say that e is the set defined like this. So, we want to show that e is countable let us take some x in e let us take some x in e as x in e means in particular it is also in this interval a b. Then we know that f of x minus and f of x plus exists that is there, but it is a point of discontinuity. So, they cannot be the same f of x minus and f of x plus cannot be the same if they are not the same then either one is less than or equal to one is strictly less than the other which is which should be less than which we know that f of x minus is always less than or equal to f of x plus. So, if the function is discontinuous at x then f of x minus plus b strictly less than f of x plus. That is clear then f of x minus must be strictly less than f of x plus. Now, we do the usual trick here. See these are two real numbers we do not know what kind of real numbers are there, but these are two distinct real numbers and we already shown that between any two distinct real numbers there exists a rational number. So, we choose one such rational number since it is going to there will be several choose any one of them and since it is going to depend on x I shall call that number r suffix x I shall just call that number r suffix x. So, there exists a rational number r suffix x such that f of x minus strictly less than this r suffix x and strictly less than f of x plus. Now, after that what is to be done is clear. So, suppose I collect all such rational numbers for each x I have one rational number for each x I have one rational number. So, suppose I take this set of all r suffix x the x belongs to e this is subset of rational numbers and hence it is countable. Hence it is countable this is countable remember for each x we have chosen one rational number and obviously that set is going to be countable. Now, does it say immediately that e is countable we are not shown that for example, it can happen that two different x may correspond to the same rational numbers. See this function x going to r x if we show that that is one one then that is fine if we show that x going to r x that is one one then it will be there will be a bisection between e and this set and then this is countable and hence e is also countable. But that we are not shown right now what we only know is that this map x going to r x is on to because we are collected only those numbers here. Now, let us give some name to this set suppose I call this set let us say f then there is a map which goes from e to f namely x going to r x that is x x going to r suffix x this is a map which takes e to f we know that f is countable. But we know that this map is on to but that does not say that e is countable that will not say that e is countable. But if we show that it is one one then it will mean that it is bisection and if f is countable that will also give that e is countable. Now, what is required to show that this map is one one we must show that if you take two different numbers x and y then the corresponding r x and r y are different. Suppose you take two different numbers x and y then corresponding r suffix x and r suffix y are different. So, for that we are going to use this result next part of this. So, suppose you take two different x and y so in particular we can assume that x is less than y then let us say that. So, to show that to show this part let us say that let x y be in e and x not equal to y obviously if x not equal to y either x is less than y or y is less than x it does not matter which one we call x or which one we call y. So, let us assume that x is less than y then what we know is the following we know that then f of x plus is less than or equal to f of y minus then f of x plus is f of x plus is less than or equal to f of y minus. What can we say about r x or r suffix x? R suffix x is must be less than f of x plus right r suffix x must be less than f of x plus so and what about r suffix y? r suffix y must be bigger than or equal to this r suffix y must be bigger than or equal to this not because strictly bigger than remember r suffix x is strictly bigger than f of x minus. So, r suffix y must be strictly bigger than f of y minus. So, now, can r suffix x and r suffix y be equal? So, what we have shown that if x is less than y r suffix x is strictly less than r suffix y. So, that shows that this map which takes x to r suffix x is 1 1 it is already on 2. So, f so there is a bijection between e and f and f is a countable set it is a subset of a rational numbers. So, e is a countable set. So, again to take the summary of this monotonic functions what we have shown is that monotonic functions cannot have discontinuity of type 2 all the discontinuities of r of type 1 and the set at which the function is discontinuous that set must be countable. Now, let us also discuss one more thing which something that we have discussed about the sequences, but we have not discussed about the functions. In case of functions we have seen that after defining the limits or convergence of a sequence and discussing various properties we extended those things to extended real number systems and we talked about when do we set a sequence goes to infinity or minus infinity etcetera. Now, similar things we will we also want to do for the functions that is see till now we have talked about this what is meant by saying that limit of f x as x tends to a is equal to l and of course we have done it in case of any matrix space and a function f goes from x to y also, but let us now say that f is a real valued function and it is defined on some interval on the real line and it is a real valued function. So, we have we have no problem in saying when a and l both are are both are real numbers. What we want to do next is that we want to consider whether when a or l or both can be extended real numbers that means whether l is plus infinity or minus infinity or similarly a is plus infinity or minus infinity and the way we are going to do it is as follows. Let us first recall what is the meaning of this let us let me let me recall what we have said is this that is for given any epsilon because we said there exists delta bigger than 0 such that such that for all y lying between x minus. So, for all x lying between a minus delta to a plus delta that is let us let me write this mod x minus a less than delta this implies mod f x minus l less than epsilon. We have seen another interpretation of this or different way of writing the same thing saying that mod x minus a is less than delta it is same as saying that x lies in the interval a minus delta to a plus delta that is let us say x lies in the interval a minus delta to a plus delta. Suppose, I call this interval as v then v is a open interval containing a or in other words it is an open neighborhood of a v is an open neighborhood of a. Similarly, saying that mod f x minus l less than epsilon that is same as saying that f x belongs to l minus epsilon to l plus epsilon again this l minus epsilon to l plus epsilon is an you can say open ball with center at l. So, it is an open set containing l. So, suppose I call that set as u then this u is a open neighborhood of l u is an open neighborhood of l what is the relationship between v and u if x belongs to v f x belongs to u is that same as saying that f of v is containing u because what is f of v? f of v is the set of all f x for x in v and so every such f x must be in u. So, saying that whenever x is in v f x is in u that is same as saying that f of v is in u. So, I can recast this whole definition in terms of this neighborhoods. So, what I can say is that for every epsilon bigger than 0 there exist delta etcetera what I will say is that for every neighborhood of l for every neighborhood open neighborhood let us say u of l there exist open neighborhood v of a such that f of v is containing u is that clear. So, I will say that instead of this definition I will say that this means for every open neighborhood u of l there exist an open neighborhood v of a such that f of v is containing u this is the meaning of this limit of f x. Basically, we have just written the same definition using the idea of open sets or open neighborhoods. Now, why did I do that? I want to extend this to the case when a and l can be in r star. What is r star? r star is r union these two symbols minus infinity and plus infinity with the usual operations which we have talked about earlier. Now, if I want to extend this definition what I will do is I will just use this definition. I should only say what is made by neighborhood of plus infinity and neighborhood of minus infinity. If I say that then the same definition will go and that is fairly easy. I will say that if you take any interval of this form m to infinity take any real number m take any interval this form m to infinity then that is called a neighborhood of plus infinity. That is what is this set? This is the set of all real numbers such that set of all x in r such that x is bigger than m. We say that this is a neighborhood of infinity. Similarly, if I take minus infinity to m that is considered neighborhood of minus infinity. Let me use some other register of m. So, suppose I take say minus infinity to l minus infinity to l. So, this is the what is the meaning of that this is the set of all real numbers such that x is less than l. This we regard as neighborhood of minus infinity. Now, once we fix that this is what we mean by neighborhood then we just use exactly the same definition. Now, a and l or both can be either it can be real number or the plus infinity of a and l both now can belong to the extended real like that means one of them can be plus or minus infinity or both can be plus or minus infinity. All that you do is that just use this definition and understand that neighborhood of plus infinity means intervals like this and neighborhood of minus infinity means intervals like this. Let us just take one illustration how this definition will translate into the usual definitions that you are familiar with. So, for example, let us say a is a real number but l is plus infinity that is. So, suppose let us say I want to say what is defined by this limit of f x as x tends to a it is infinity. So, a is a real number and l is plus infinity. So, what will be our definition for every neighborhood u of l for every neighborhood u of l there should exist neighborhood v of a such that f of u sorry f of v is contending u sorry not for u of l u of this infinity in this case it is infinity. Now, but what are the neighborhoods of infinity we have seen the neighborhoods of infinity of this state. So, saying that for every neighborhood u of infinity which same as saying that for every m for every m there should exist that is this is same as saying that for every m there should exist v is a neighborhood of a and if a is a real number neighborhoods are of the form a minus delta to a plus delta. So, we can say that for every m there exist delta bigger than 0 such that v is let us say v is a minus delta to a plus delta and u is m to infinity. So, what we want to say that f of v is contending u that means whenever x is in this neighborhood f x is in this. So, f of v is contending u that is same as saying that whenever mod x minus a is less than delta this implies f x is bigger than m. That means given any large number any number m large or small does not matter you can always find a delta such that whenever mod x minus a is less than delta f x is bigger than m this is the meaning of saying that limit of f x as x tends to a is infinity. Similarly, you can take all possible combinations either l is infinity x a is a real number or l is minus infinity and a is a real number or a is infinity and l is a real number etcetera or both are plus or minus infinity. You can write the definition just take this as a basic definition and use this that neighborhood of plus infinity is interval m to infinity neighborhood of minus infinity is interval minus infinity to l and then just translate all those definitions in this form you will get all the usual definitions in usual way in which this kind of things are written in analysis books. We will stop with that.