A guide I made to help others study for the first exam of Real Analysis at UNC.

read moreAxiom of completeness:Every nonempty set of real numbers that is bounded above has a least upper bound. We do not have to prove that this is true. It is accepted as being true (hence “Axiom”). For an upper bound to be a least upper bound, the following must be true: i) it must be an upper bound ii) for any upper bound s, s must be less than or equal to the least upper bound candidate.Nested Interval Property:The intersection of infinitely many inclusive subintervals is nonempty. This property relies on theAxiom of completenessto work. We break apart each interval and put the left hand endpoints in one set and the right hand endpoints in another. We then state that any element in the right hand set of endpoints is an upper bound for the set of left hand endpoints. By the breaking the left and right hand endpoints into sets, we can use the Axiom of Completeness to say that the left hand set has a supremum, which we can call x, and by the nature of the setup, the right hand set is always an upper bound for the left hand set and so we can say that x is always less than or equal to any element in the right hand set and by the properties of the supremum we can say that x is greater than or equal to any element in the left hand set. From this, we can say that no matter how nested you get, because this supremum exists, there will always be something inside the intersection. The intersection is nonempty.Archimedean Property:i) Given any real number, there exists a natural number that is larger than it. ii) Given any real number greater than zero, there exists a natural number whose reciprocal is less than that real number. This is a property that applies to the relation between two sets. The Archimedean property can therefore be true for certain pairs of sets and false for others. The proof also relies on the Axiom of Completeness. For part i), we prove by contradiction. The proof seems only sort of direct to me because we’re proving that something that would imply that this is false is false itself. We know that this would be true if the natural numbers were unbounded, so we’re actually just proving that the natural numbers are unbounded. We start by assuming that the natural numbers are bounded, and show that this leads to something that is ridiculous and therefore our assumptions must be false. Assuming the natural numbers are bounded (above, specifically), means by the Axiom of Completeness that they have a least upper bound. We will call this supremum ‘A’. If A is a supremum, then A-1 must not be an upper bound and therefore there must exist a natural number n such that A-1<n. But if this is true, then A<n+1, and assuming that the natural numbers are closed under addition means that n+1 was also in the set of natural numbers, and we therefore did not have a least upper bound to start with and because we made the least upper bound general, we have shown that it is generally impossible to set the supremum to any number because a larger natural number can always be found. Therefore, we have shown that the supremum cannot exist which implies that the set is not bounded above (once again by the Axiom of Completeness which states that every set that is bounded above has a supremum) which implies that we can always find a natural number larger than any arbitrary real number. The proof of ii) follows pretty simply from i). ii) is also only contingent on the natural numbers being unbounded. We can morph the statement of ii) into one that has already been proven in i) by taking the reciprocal of both sides which is equivalent to the original statement. Therefore, if we prove this equivalent statement then we prove the originally intended statement. In words the equivalent statement is: for any small real number, there exists a natural number that is larger than this small number’s reciprocal . It’s quite clear now why this is also only contingent on the natural numbers being unbounded.Density of Q in R:For any two real numbers such that one is strictly greater than the other, there exists a rational number that lies between them. Let’s let the smaller real number be a and the larger one b. We must find two integers m and n such that a<m/n<b. Imagining a and b on a number line lets us realize that we don’t want 1/n to be so large that we skip over it when we increment m a little bit. So 1/n<b-a will ensure this doesn’t happen (we can say this is possible by ii) of the Archimedean Property). From the first inequality we know that na<m<nb must hold for our rational number to be between these two reals. So let’s choose m such that m-1<na<m. From this, we know then that a<m/n which is half of what we need (we also want m/n<b). We also know from that that m<na+1 and we can substitute a with b-1/n and we get that m<nb which implies the second half that m/n<b. We are done. Now, this is a kind of proof that we haven’t seen before. We use the properties of real numbers (this includes the Archimedean Property) to show that an m and an n can be chosen for any two arbitrary reals. We use the A.P. first, then we use the fact that any real is between two integers, and then we may use the fact that the reals are ordered, but I’m not entirely sure.The Existence of Square Roots:There exists a real number a such that t^2=2. We prove this with the Axiom of Completeness (as surprising of an axiom that may be to use in such an endeavor). We actually take the approach of proof by cases, and show that the two undesired results are impossible (each of which is done by contradiction). We start by defining a set T such that for all t in the real numbers, if t^2<2, then t is in T. We know by the Axiom of Completeness that this set has a supremum, and we will call this supremum a. We want to show that a^2=2. We do so by disproving all the other possibilities (a^2<2 and a^2>2). We will only disprove the case that a^2<2. We start by assuming this is true and then show that this assumption leads to a contradiction and therefore the assumption must be false. Here’s something to keep in mind: we are proving something that we already believe/have some intuition for, so it’s wise to try to employ that intuition to try to arrive at a contradiction. In this instance, before starting the proof, we believe that the supremum of this set is in fact the square root of 2. So, knowing the cases we have to disprove, we should use this belief to guide us in arriving at a contradiction. Because we believe that the supremum is the square root of 2, when trying to disprove that a^2<2, we would want to show that this assumed supremum is not actually an upper bound (we know we should disprove this point because we know/believe that the least upper bound is the square root of two which is larger than an a^2 that is <2). With this approach, we can figure out what conclusion we want to reach before even starting the proof. Now, some preliminary discussion before we start the proof. We know we want to show that an a^2<2 is not actually an upper bound and we’re assuming that a^2<2. We can show that it is not an upper bound if we show that something slightly larger than it is in the set. This shows that it is not an upper bound because an upper bound would be greater than or equal to everything in the set, and finding something greater than it that was in the set would mean that it was not an upper bound and therefore not the supremum. This something that we find that is greater than it that is in the set may or may not be the maximum of the set, but that actually doesn’t matter. Showing the existence of something greater that’s in the set disproves that the a chosen such that a^2<2 is an upper bound and therefore also disproves that it is the supremum. Now for the proof. Let’s take something slightly larger than a (our supposed supremum) and try to show that this is also in the set. Let’s take a+1/n. This is in the set if its square is less than two. So, let’s take (a+1/n)^2 which is a^2 + 2a/n + 1/n^2. We can simplify this a bit by replacing the 1/n^2 with a 1/n. This is possible because 1/n>=1/n^2 for any input of n, and we are just trying to show the existence of some arbitrary element of the set that is slightly greater than our supposed supremum, a. And if we show that a^2 + 2a/n + 1/n is in the set, then surely its smaller counterpart (the same polynomial but with 1/n^2) which is also larger than just ‘a’ should also be in the set. So we proceed with just 1/n. We can combine like terms to get a^2 + (2a+1)/n. The next part is a little tricky. Remember our objective: we want to show that a^2 + (2a+1)/n is in the set and because we’ve assumed that a^2<2, we now just want to make sure that the (2a+1)/n term fits in whatever room before 2 that the a^2 term left us. Imagine a number line with a^2 on it which is to the left of 2. We know this is true by our assumptions, but we want to make sure that adding the (2a+1)/n term doesn’t push us past 2. So, we want to make sure that (2a+1)/n is less than the distance between 2 and a^2. Now, because the n is an arbitrarily small number, we can employ the Archimedean Property to ensure that (2a+1) is in fact less than the distance between 2 and a^2. We want to choose an n such that (2a+1)/n < 2 - a^2. Because of the way the Archimedean Property is presented to us in the book, this is rewritten as 1/n < (2-a^2)/(2a+1) so that we can better see the employment of the AP But, this somewhat hides the intuition and at least for me, made this step hard to follow. One follows from the other, but not necessarily easily in a certain order. Now that we have shown using the AP that it is possible to fit this term in the gap left between a^2 and 2, we have shown that this a+1/n is also in the set which contradicts our original assumption that an a^2<2 was the supremum of the. Therefore, we conclude that an a^2<2 could can not be the supremum of the set. I leave the alternate case (a^2>2) for someone else to answer.Convergence of a SequenceNote that e is meant to denote epsilon in this section. Definition: A sequence a_n converges to a real number a if, for every positive number e, there exists a natural number, N, such that whenever n>=N, it follows that |a_n - a| < e. This is saying that a sequence converges to a given limit, a, if no matter what positive value you choose, the distance between all the terms in that sequence and the limit after some term is less than the positive value you chose. That’s a little hard to follow so we can try to spread it out a bit more. A sequence converges to some limit if the following is true: you can tell me however close you’d for the terms in the sequence to be to the limit (by giving me some e), and I can assure you that there exists some point (some term) in that sequence, such that all the terms after that term are as close as you asked them to be. I believe the notation N(e) is used to highlight the dependence of the term I spit back at you (N) on the closeness you want me to get (e). It’s a function in a sense in that it has an input and output. It seems to me like the neighborhood function takes two inputs and is a variation of N(e). We denote the neighborhood function V_e(a). a is the value that I have to find terms around and I have to find terms that are within +e and -e around a. I don’t suppose a has to be the limit of the sequence, but I believe it often is (at least in our examples). I think we can cycle through different values of a (imagine V_e(a) sliding up and down the number line) and for each of these values of a, we vary e. If we have a value of ‘a’ such that no matter how e>0 is chosen, the neighborhood of ‘a’, V_e(a), is non-empty, then the sequence in question must converge to a. I think it’s important to be able to differentiate between the “topological view” of a sequence and its graphical view (on the standard cartesian coordinate plane). To me, it seems like the topological view is better suited for imagining neighborhoods while the graphical view is better suited for visualizing N(e). At the heart of all of the convergence problems we do is the relationship between N and e. Page 40 of the book gives a good explanation of an explicit relationship between N and e. In many of the problems we do, we do not try to explicitly define N in terms of e, but rather, we are given that a relationship exists between some N and an arbitrary e and we are told to derive, based on this fact, that a relation exists between this given N(e) and some N1(e1). There should be more on this later. yo Review the template for a proof that x_n converges to x on page 41.The Algebraic and Order Limit Theoremsi) the limit of any constant times a sequence is equal to the constant times the limit of the sequence. The proof of this theorem is an example of using an implicit definition of N for one sequence to show that an N1 can be defined for another sequence. This is pretty tricky. We start with something we are given, which is that the limit of the sequence a_n exists and that this limit is ‘a’. Whenever we’re given a convergent sequence in our assumptions, we should make note that we can rewrite it in the for all e>0 there exists some N(e)... notation. This is useful (and in most cases necessary) for making the deductions in the proof. So once again, we are given that a_n converges to a. Which means that no matter what e is chosen, |a_n - a| < e for all terms after some single term a_N, which is determined by N(e). It’s important to differentiate between the cases when N(e) is given explicitly, and it is implied that some N(e) exists. In most of the problems we’ve done, I think it has just been implied that some N(e) exists. All we need to know is that an N(e) exists for any input of e>0 to know that a sequence converges. For the proof of this theorem, we know that some N(e) exists, but it is not explicitly given to us. However, we can use the fact that one does exist to show that a similar function, which we’ll call N1(e1) as it will be derived from N(e), exists for some other sequence. It is this derivation of some N1(e1) (or in some cases N2(e2) as well) that we have done over and over again in class. Let c be the constant that the sequence is being scaled (multiplied) by. What we want to show is that no matter what value of e we choose, we can find some N such that the following holds for every term after N: |ca_n - ca| < e, where n is any term after N. Okay, so we know we can do this for the sequence a_n, that is, the statement above but instead we insert |a_n - a| < e. Well, because we know this is true (it is our assumption so in our current little proof universe we say that it is true), we can try to algebraically manipulate our assumption so that it matches our goal. This would be a forward way of doing things. But because it’s commonly algebraically easier to manipulate the goal statement to match the assumption, we do the algebra backward and make sure that the reverse direction still holds. So, we want to manipulate our goal statement, so that both of the left hand sides of our goal statement (|ca_n - ca| <e) and our assumptions statement match. That’s something that’s very important to remember. The reason we don’t care about the right hand side (on which there should only exist epsilons and constants), is because part of our assumptions is that epsilon can be any arbitrary positive number. Now, don’t get too caught up with epsilons because the concept just wants the right hand side to be able to be arbitrarily small. So just because you need to multiply epsilon by a positive integer doesn’t mean that the the statement would no longer hold. This follows from epsilon (better put, the right hand side) being arbitrarily small. If you scaled up epsilon by a constant, or even by epsilon itself, the value of epsilon could just shrink to take care of that growth. So now we can try to manipulate our goal statement to match our assumption. We can take |c| out of both terms within the absolute value of our goal statement which leaves us with |c| * |a_n - a|, which is great because we can now see our assumption statement easily and standalone. Now let’s recall our original assumption statement: |a_n - a| < e. Because of the properties of e in this instance, we can do whatever we wish to it and still be sure that this inequality still holds (the reason for this is something I don’t think we have gone over in class but it has to do with the cardinality of sets and I think having a bijection between two sets). So in this case, we divide it by |c| to get |a_n - a| < (e)/|c|. Because it was true that we could take e to be any arbitrary number larger >0, we can simply define a new e1 in terms of the old e such that e1 = e/|c| and by the properties of epsilon, we can conclude that |a_n - a| < e1 also holds. So essentially all you’re doing is taking any scaling or otherwise obfuscating factors out from your underlying convergent sequence and you’re moving them over to the epsilon side. ii) the limit of the sum of two sequences is equal to the sum of the individual limits of the two sequences. This proof is trickier, but it employs the same technique as before (and so will the subsequent one as well). Once again, because the algebra is easier to do backwards, we start with our goal statement and try to find our assumption statements within. The concepts here are the same, and the reason we can make these algebraic changes are the same, the only tricky part here is that we have to know to apply the triangle inequality to get something manageable out of this guy. After using some properties of the absolute value to switch some signs and applying the triangle inequality, we find that our goal statement is less than or equal to |a_n - a| + |b_n - b|. Because of our assumptions, we know that we can make each of these terms arbitrarily small. Let’s say that we originally said we could make |a_n - a| < e1 and |b_n - b| < e2 for any arbitrary e1,e2 > 0. Well, this holding true we could take the sum |a_n - a| + |b_n - b| and be sure that this value is less than e1+e2. The important thing to note is that both e1 and e2 are arbitrary, and so we can define an arbitrary e such that e=e1+e2. Now how do we make sure that an N(e) also exists? Well, we know that an N(e) exists for both of our two assumption statements. If you’re taking the max of the N(e)’s of two convergent sequences, it’s essentially like whatever’s returned as the max is less or equal to both N(e)’s because the N(e) that’s returned should be the one that converges slower.Specific reminders for exam 1:Triangle inequality and its derivatives: |a+b| <= |a| + |b| |a-b| <= |a| + |b| ||a| - |b|| <= |a - b| NOTE: [ ] includes [min, max] ( ) excludes (inf, sup) Review the template for a proof that x_n converges to x on page 41.Questions:Do we need a blue book?? What algebraic manipulations (if any) can you apply to epsilon so that the resulting equality no longer holds? What does it mean to find the max of two N’s? Can anyone explain that? Page 19, why is it that from b_n being an upper bound for A means that we are justified in setting x = sup A? What exactly is the logic there? Ans: They just want to verify there is a x in the union. So, the union is not empty. See pg.6, there is a similar example but tricky part, the intersection is empty. Page 19, likewise could we have let x = inf B? Ans: if x = inf B, that also means x is an upper bound of A. Thus, a_n < x < b_n also could be equal here. Questions: Does our exam cover the section 1.5? I don’t believe so. I think it’s only the stuff between 1.1 and 2.3 (inclusive) that we went over in class. We didn’t go over Cantor’s Diagonalization method in class, did we? *If interested in why Cantor’s Diagonal Argument is interesting -> see Godel’s Incompleteness Theorem or “Godel, Escher, Bach.” I can say you will not be sad. Gotcha, Thx! Ummm, I think we didn’t and we also do not have any homework from section 1.5. Question: on pg. 22, here is countable and uncountable sets, didn’t we go through this part, did we? Ans: I don’t believe we went over this in class so it shouldn’t be on the test. Cardinality can help with understanding some of the concepts that we have been introduced to, though.