CSIR NET MATHEMATICS DEC 2017 | Complete Solution of Real Analysis | Section B

Hello, Dor, I'm Mahesh, and you're watching the Next Gen Math YouTube channel. If you haven't subscribed to the channel yet, please subscribe. And press the Bill icon named Kule. Inside this video, we discuss the Real Analysis problem from CSI NIT, December 2017. For example, let Z denote the set of integers and Z greater than or equal to 0. Denote the set {0, 1, 2, 3}. Consider the map F, Z greater than or equal to 0, cross Z to Z Given F(m, n) equals 2 to the power m times (2n plus 1) Then the map F is surjective but not injective, i.e., not one-to-one. Second option: 1, 1, but not onto. Third option: both 1-to-1 and onto. And the fourth option, neither one-to-one nor onto. If you want notes on this solution, please visit our website www.nexgenmaths.com In the description, a link is available. This is the problem: 1, 1, and onto function. First, let us recall the definition of the 1, 1 function. A map F from X to Y is said to be one-to-one or an injective map. If F of A equals F of B, then A equals B I mean, if the images of two elements are the same, then the elements themselves are also the same. And the equivalent definition: if A is not equal to B, then F of A is not equal to F of B. I mean, if two elements are not the same, their images will also not be the same. For the given map F, Z greater than or equal to 0, cross Z to Z The map is F(m, n) equal to 2 raised to the power of m multiplied by (2n plus 1). We will use the first definition here. F(m1, n1) equals F(m2, n2) We have taken two elements, m1, n1, m2, n2, from the domain That is, here m1 and m2 are greater than or equal to 0. and n₁, n₂ belong to ℤ By definition, F of m1, n1 means Replace m with m1 and n with n1 And F of m2, n2 Replace m with m² and n with n² Here we can observe one thing 2 to the power of m1 and 2 to the power of m2 are even numbers. It's just even numbers. 2n₁ + 1 and 2n₂ + 1 Both are odd numbers There will be no common factor between these two. Therefore, we can directly compare 2 to the power of m1 equals 2 to the power of m2 and 2n₁ plus 1 equals 2n₂ plus 1 After simplification, we have m1 equal to m2 And n1 equal to n2 That means m1, n1 equal m2, n2 I mean the images of the two elements are the same. Then those two elements will be the same. So by the first definition We can say that this function is one-one. That means it is an injective function Second part is an onto function Then we will also recall its definition. A map f, x, y is said to be onto or a surjective map. if y belongs to y That means from the codomain There exists at least one element x belonging to s such that f of x equals y Take any element from this codomain Then I will get at least one element from the domain. The one whose image will be y itself I mean, the element of the codomain will be itself. So in our case, inside Here m belongs to We set z equal to 0 That means non-negative integer and n belongs to z I mean that n is an integer If m equals 0 We are fixing the first dimension. That means m equals 0 So f of 0, n Put 2 to the power of 0 here in the definition. 2n + 1 So 2 to the power of 0 will be 1 So it will be 2n plus 1 but 2n plus 1 all will represent an odd integer If we take m equal to 0 So from the codomain All odd integers represent become If we take m equals 1 So what will happen in the definition when m equals 1? Keep it, it is 2 to the power 1 So all even integers will represent So here the codomain is the set z All odd integers from z have now appeared. All even integers have come, but what's left behind? 0 That means f(m, n) equals 0. But for 0, we can have any We cannot distinguish between m and n. Therefore, we call 0 cannot represent So from the codomain 0 is an element for which I don't have any element in the domain Coming my way This function will not work on I mean, it is not a surjective map. So the given map is 11 but not surjective and which is the option number 2 Then that is the correct answer will be, for example an, n greater than or equal to 1 a sequence of real numbers satisfying a1 greater than or equal to 1 and a plus 1 greater than or equal to a plus 1 for all greater than or equal to 1 then which of the following is necessary We have 4 options. the series sigma n equals 1 to infinity 1 upon a square diverges second option sequence an, n greater than or equal to 1 is bounded third option series sigma n equals from 1 to infinity of 1 over n squared converges and the fourth option series sum from n equals 1 to infinity 1 upon an diverges if you want If you are aware of this solution, please visit our website. www.nextgenmath.com The description link is available This is a problem. sequence bounded or unbounded or series diverges or converges So first we are given a1 greater than or equal to 1 and a plus 1 greater than or equal to a plus 1 So we will use this. if you set a equal to 1 So will you come then? a1 plus 1 means a2 greater than or equal to a1 plus 1 But a1 is greater than or equal to 1 therefore this is greater than or equal to 1 plus 1 So a² is greater than or equal to 2. Now we will take a equal to 2 So a squared plus 1 that means a3 greater than or equal to a squared plus 1 But a2 is greater than or equal to 2 so this is greater than or equal to 2 plus 1 will be 3, so a3 greater than or equal to 3 if we continue this process Then we will have a4 greater than or equal to 4 and a5 greater than or equal to 5 Finally, we can say this such that a is greater than or equal to an, on top of what was given to us And thus we arrived at this result. a_n greater than or equal to a_n Now we will use the comparison test. So, let's recall the comparison test first. for positive term series sigma a_n and sigma b_n If a is less than or equal to b and sigma bn is convergent then Σaₙ is also convergent I mean that If a large sequence The series is convergent. Then the smaller sequence's series will also be convergent. if bn is less than or equal to an and sigma bn is divergent then Sigma a_n is also divergent that means Small sequence series divergent This is a simple comparison test If the large series converges So the small series is convergent. If the small series is divergent So, a highly divergent series This can be applied only to positive term series. We will discuss option numbers 1 and 3. Who sigma 1 over n It is about the convergence and divergence of the square. We have just proved it a greater than or equal to n Therefore, 1 upon n is less than or equal to 1 upon n. About both squaring positive values will become Also about applying sigma to inequality It will remain the same. since The series sigma 1 over n squared is convergent. series If the larger series is convergent, then by the comparison test The small series will also be convergent. therefore sigma n equals 1 to infinity Which is our option number 3 So option number 3 is the correct answer. And option number 1 is the wrong answer. Now we will discuss option numbers 2 and 4 Regarding the bounded sequence and sigma 1 upon an It is about convergence. We have just proved that a_n is greater than or equal to n. So this An unbounded sequence has occurred Because for any number we A very large number will be obtained. So therefore it is a unbounded sequence, so option number 2 It's wrong to say that this is a bounded sequence. next 1 upon n less than or equal to 1 upon n if an equal to 2n greater than or equal to n Here it will tell us about an greater than or equal to n So we took an example: 2n greater than or equal to n then sigma 1 over equal to sigma 1 over 2n that It is a divergent series, and another one option equals n cubed greater than or equal to n then sigma 1 over n cube is convergence series means We have Two such sequences came up that one in the series convergent is and one divergent which is this satisfying the condition that a is greater than equal to n therefore this condition For the series that satisfies us, we have some Nor can they even say 1 upon a convergence will occur because divergence is therefore Option 4 is not always Correct, so therefore it is wrong. Will go, so option number 2 and 4 are the incorrect answers Let D be a subset of the real line consider the assertion every infinite sequence in D has subsequent which converges In D, this assertion is true if We have been given four options; the first option D equals the closed open set from 0 to infinity D equals closed interval 0 to 1 closed interval [3, 4] third option D equals open interval minus close One-to-one union open closed interval from 1 to 2 and the fourth option D equals to open close interval minus 1 to 1 if you want For notes on this solution, please visit our website www nextgenmaths.com Link in the description is available This problem is very easy. You just need to use the result. If you know the result, you can solve it. immediately then first us We will remember the result. So the first result is D is compact if and only if every infinite sequence In D has a subsequence which converges in D You can see, this is the same statement as provided What happened to this D? Compact Set But what is D a subset of? Real Line of We use the second result. This subset D on R is compact. if and only if it is closed and bounded So D is a subset of the real line. When will it be compact? What do you mean, which D is given? Closed End Bounded We have the option. First Option No, it's not closed, and no, it's not bounded. I mean, this is not compact. So this is not compact, so it will not satisfy this statement. Then the first option will be rounded. Both are closed sets. The union of a closed set is also a closed set. So this is also a closed set as well as bounded. So this will be compact. And this is our correct answer So Option Number 2 is Correct It is bounded but not closed. Therefore, this will not be compact. Option number 3 will also be canceled. It is bounded but not closed. Therefore, this will not be compact. Option number 4 will also be canceled tomorrow. So we have arrived at the Correct Option Option number 2 We have four options. Fela Option We have Option This is the definition of a Uniformly Continuous function. So first, we will recall its definition Sunte Chanti Armath Floं Agan Sayilbeṇ Dāgdaṃ Dā Jab whose AAIPLON is that if a Mordils x is Mary In Elze, Delta Mardilane is an exceptional case. Fupu Washin Prachau, yihaka Nice Saky, Sanumen Dance mod x - y less than delta implies |f(x) - f(y)|, where f(x) is equal to x and f of y is equal to y so x minus y, which is already less than delta if we take delta as epsilon then it satisfies this condition so f of x equals x is a uniformly continuous function but limit x approaches infinity of f of x is equal to limit as x approaches infinity of x, which is infinity So this limit does not exist. for f of x is equal to So option number one which states that the limit of f of x as x approaches infinity exists and option number 3, the limit of f(x) as x approaches infinity exists Both of them will be wrong. Next, now we have to check limit x tends to 0 plus f of x exists, so first let us remember the result, every uniformly continuous function is always continuous, therefore the given function f is continuous on the open interval from 0 to infinity result 2 a function is said to be continuous on an open interval if and only if It is continuous at every point in this interval. So when will the function be continuous? when it is continuous at every point of that interval means therefore the given function f is continuous on every point of the open interval from 0 to infinity that means the limit of x as it approaches a f of x exists for a belongs to open interval (0, it is continuous means the limit exists So its limit should exist at every point. so limit as x approaches 0 from the right of f of x exists because where will 0 plus come So a larger number than 0 will be in the range from 0 to infinity. interval, so where will this a be from 0 to infinity, therefore the limit as x tends to 0 plus f of x exists, so that is Option number 2 Option 2 will be correct, and option 4 will be wrong. example Then we have 4 options as If you want notes on this solution, please visit our website www.nextgenmath.com The description link is available We have been given there exist so They will be uniformly continuous functions. Or else those will be bounded functions. Or they will be constant functions We are constructing a function f of x is equal to 0.5 for x belonging to q and 0 for x does not belong to q so the modulus of f of x minus f of y less than or equal to 0.5 Its maximum value will be 0.5 So we can write such an article, it is less than 0.6 So we Let ε = 0.6 So for any value of delta modulus of minus y less than delta We have a value epsilon of Available 0.6 So this It will satisfy the condition of our S. therefore f belongs to f but f discontinuous function, it is discontinuous everywhere on the real line So, the option is that f is continuous It went wrong. It is discontinuous, so not uniform. Not only that, but he/she couldn't even do it continuously. Then the second option will also become wrong. This is bounded because What is its maximum value? So it is a bounded function, so option number 3 It will be corrected. And it is not constant; it is not a constant function either. Option number 4 will also become incorrect. problem which of the following is natural true for a function defined from x to y, if f is injective then there exists g from y to x such that f of g of y equals y for all y belonging to y, second option if f is surjective then there exist g y to x such that f of g of y equals for all y belongs to third option f is injective, y is countable then x is finite fourth option f is surjective and x is uncountable then y is countable finite If you want notes on this solution Please visit our website www.nightchainmath.com In the description link is available This is a problem regarding injective and surjective functions. Therefore, we first have injective and Let us recall the definition of a surjective function. a map f from x to y is said injective map or one-to-one mapping if f of a equals f of b then a equals or equivalent definition if a is not equal to b then f of a is not equal to f of b What does it mean when an F1-1 function occurs? when this condition is satisfied If f of a equals f of b, then a equals b means if the image is the same, then the element is the same or a not equal to b means if the element is not the same then The image is the same. onto map or surjective map a map from x to y is called an onto map or a surjective map if y belongs to y There exists at least one element. x belongs to x such that f of x equals y I mean that for every element in the co-domain for here at least A value of x will be obtained such that f of x equals y For each element, an x will be associated with this It is not 1-1, so here More than one element will be associated Which we will use Option 1 if f is injective then there exists g y to x such that f of g y equals y us are constructing Set x equal to 1 and the codomain set y equal to 1 as well defined from x to y f of 1 equals 1 There is only one element in this domain. And it is associated with one element. So it has become a 1-1 function. now we are going to Define G Y to X So, how many elements are in y, 2 elements, 1, 2 And how many in the co-domain? element is, x in 1 And we are interested in f of g y equal to y So f of g 2 should be 2 But where should this 2 be? the codomain of g, the codomain of g x is, but in x there are 2 elements Not only that An air-conditioned function won't be possible at all. So none of us has any g No, it won't be available for the image of Because f of g 2 will be the 2, then this Not possible, option number 1 And option 2, the second option, is If f is surjective, then there exists g from y to x such that f of g y equals y for all y belongs to y, then what is given to us f is surjective means that the co-domain is y At least for each element One x will be obtained, in which f of x equal to y, at least one x For y, more than one x could have been, we had a set Will create, i y equal to x such that f of x equals y I mean the image of all these x's What will happen, y, but More than one element is possible here because we need It would not have given a one-one function. We are defining g in this manner g y to x, g of y equal to x, and for only If x belongs to i y, then we say so defining the function that g of y equals x, and Where is x coming from, from among y? Li is an element, and there is more than one of it, so we We will fix it, where the same element is appearing. So this is g y What will you do, f of g y, and What is the image of y under g, the image of y under g is x. So f of x equals y What we have assumed So f of g of y equals and, then this is ours satisfying the condition Therefore, option number 2 will be correct, option number 3, f is injective and y is countable then x is finite We are entering into a function contract. f of n to n, f of x equal to x, it is the identity function Then there will always be one one function. Okay, so that's an injective map. So the first part is satisfied. y equals n So it is also countable. And x is also countable, x will not be finite Therefore, the third option will be wrong, option number 4, if f is surjective, x is uncountable then y is countably finite, we we will define f(r) to r, f of x equal to r, identity map So it has become surjective. The first part of this condition satisfying, x I mean, domain, domain r is there So r has become uncountable. So the second statement has also been corrected. now, f should be countably finite, but what is y here y equals r But that's uncountable. This is countable is not finite, then option Number 4 will also be wrong for example, let k be a positive integer and let s equals k x belongs to the closed interval [0, 1] a decimal expansion of x has a prime digit at its place, then the Lebesgue measure of S. K. is We have four options; the first option is 0. second option 4 by 10 third option 4 by 10 to the power k and fourth option 1 If you want notes on this solution Please visit our website www.nextgenmath.com in the description The link is available This problem is about Lebesgue measure. If set S K is, then first Lebesgue measure some properties Recall it, any open interval a, b and closed interval a, b closed interval [a, b] closed open interval a, b of real numbers is Lebesgue measurable and this Lebesgue measure is its length b minus a Any type of open interval is, closed interval is, open closed interval is, closed open interval is So that one Lebesgue measurable, and second Her measure will be. b minus a, meaning its length upper end minus lower end, second if a is a disjoint union of countably many disjoint Lebesgue measurable set, then a is itself Lebesgue measurable and Lebesgue measure Ma a is equal to the sum of the measures of Lebesgue measurable set, meaning a countable many disjoint Lebesgue measurable The union of set Then this will itself be Lebesgue measurable. And its Lebesgue measure will be All of these sets measures Sum, what has been given to us set x belongs to closed interval [0, 1], a decimal expansion of x has a prime digit at its kth position k equals 1 Then what will s1 become? x belongs to the closed interval 0.1, a decimal expansion of x has a prime digit at its first place Us Elements will be found in s1 in which the decimal expansion of You will get a prime number in the first place. So, which numbers will be prime numbers? So which elements will they be? So first we were assuming that first place with 2 prime numbers So which elements will come? 0, 2, 0, 2, 1, 0, 2, 1, 2, 3 By doing this, 0, 2, 9, 9, 9, 9 So there will be infinitely many elements We will write this in interval form. So closed open interval 0, 2, 2, 0, 3 What will be its labbing measure? b-a 0.3–0.2 Just like that in first place Prime numbers that are 3 How many elements will be formed? 0, 3, 0, 3, 1, 0, 3, 1, 2, 3 0, 3, 9, 9 open interval 0.3 to 0.4 So, what will be its labbing measure? 0,1,0,4-0,3 Similarly for 5 And likewise for 7 Closed open interval for both Will be 0.5 to 0.6 and 0.7 to 0.8 And both of their labbing measures 0, 1, 0, 5 So S1 Of these four intervals It will be a union and four sides measurable set And four disjoint Also is S1 of these four intervals It will be a union, and that one disjoint, then S1 What would be the measure of all these measures? sum μ of S1 means measure of S1 equal to μ₁ plus μ₂ plus mu3 plus mu4 will go 0.1 plus 0.1 plus 0.1 plus 0.1 So 0.4 you can write down in the 4 by 10 Similarly, k equal to Then what will S2 be? x belongs to the closed interval 0.1 is a decimal expansion of x has a prime digit in second place So the second decimal expansion The digit at the place will be prime. So prime numbers 2, 3, 5, 7 So which elements are in this? They will come, then support will be there. Zero is in first place So 0.0, then in second place The first prime number is 2. So 0.021 0.02123 0.029999 So this will become a closed-open interval 0.02 to 0.03 And its It will become 0.03 -0.02 0.01 Then which other numbers will come? in second place The prime number 2 is at the first place. It could be number 1. 0.12 0.121 in the same way as 0.12999 Then this will become a closed-open interval 0.12 to 0.13 And its measure will become 0.01. Then first place to How far will it go in first place? Will go up to 9 0.92 0.921 0.92999 So this will become a closed-open interval 0.92 to 0.93 And it will become measurable. 0.02 So how many have we got now? These are the total. 0 to 9 means that 10 intervals have passed. In the same way Give me 3 prime numbers whose sum is 10. Similarly for 5 prime numbers I'll get 10, in the same way You will get 10 for 7 prime numbers And the labeg measure for all will be 0.01. So it's all totaled up 10 plus 10 plus 10 The union of a 40-interval And 40 intervals are disjoint Therefore, labeg measure of S2 will be the sum of all these. 40 times 0.01 It will become 0.4 So it will become 4 by 10 We will continue the process in the same manner. Similarly for S3 as well It will be measured there at the lab. 0.001 Then after S3 We will search for S4 in the same way Anyone will find for S3 Then our labeg measure will become 4 by 10 So S3's It will become 4 by 10 It is option number 2. Option number 2 is the correct answer. example let S equal to x belongs to closed interval from -1 to 4 such that sin x > 0 Which of the following is true? We have been given four options. first option infimum of S less than 0 The supremum of S does not exist supremum of S equals pi and the fourth option The infimum of S equals pi over 2 If you want notes on this solution Please visit our website www.nextgenmath.com The description link is available This is the problem of supremum and infimum First, we will recall its definition. supremum of S is its least upper bound that means S is a subset of R if M belongs to R is an upper bound of S such that M is less than or equal to M' for every upper bound M' of S then M is called the supremum of S denoted M equals supremum That means M. The tightest upper bound is the smallest upper bound. So we set its list upper bound Will do and to him we We will take the supremum of S. infimum of S is its greatest lower bound I mean that If N is its lower bound Then the rest of them Greater than the lower bound That means N is greater than or equal to M' for every lower bound M' of S If the greatest lower bound is So we take its greatest lower bound. I mean that infimum of S We are given that S is equal to x belongs to the closed interval -1 to 4 sin x greater than 0 strictly greater than 0 So first, let's draw the graph of sin x. sin x is greater than 0 Between 0 and Pi Zero for zero Zero is also for Pi Then they won't come into the set x belongs to 0 to pi open interval sin x > 0 So what has our S become? Open interval from 0 to Pi and for us Supremum and infimum are to be found Thus, the supremum of S equals π. Because this set's will be the upper bound of pi and A smaller upper bound is possible. No. and in the same way, infimum of S 0 because This set's will be the lower bound, and from this Any lower bound possible No. Then the infimum of S equals 0. So, what are our options? infimum of S less than 0 So that one turned out to be wrong. The supremum of S does not exist — that is also wrong. The supremum of S equals π So, it had become the correct answer. Option 3 is the correct answer, and option 4 infimum of S equals Pi divided by 2 So the answer to this equation is supremum of S equals π Thank you for watching for more videos Like, share, subscribe press the bell icon of next 10 minutes