# limsup stuff

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<p>NOTES, MAT 472, INTERMEDIATE ANALYSIS, FALL 2010JACK SPIELBERGContents1. Axioms for the real numbers 22. Cardinality (briey) 83. Decimal representation of real numbers 94. Metric spaces 115. The topology of metric spaces 146. The Cantor set 177. Sequences 198. Continuous functions 219. Limits of functions 2310. Sequences in R 2411. Limsup and liminf 2612. Innite limits and limits at innity 2813. Cauchy sequences and complete metric spaces 2914. Compactness 3115. Continuity and compactness 3616. Connectedness 3717. Continuity and connectedness 4018. Uniform continuity 4219. Convergence of functions 4320. Dierentiation 4521. Higher order derivatives and Taylors theorem 5122. The Riemann integral 5323. The Darboux approach 5624. Measure zero and integration 5925. The fundamental theorem of calculus 6326. The Weierstrass approximation theorem 6527. Uniform convergence and the interchange of limits 6828. Innite series 7129. Series of functions 7430. Power series 7531. Compactness in function space 7632. Conditional convergence 7812 JACK SPIELBERG1. Axioms for the real numbersIn this course we will more-or-less follow an axiomatic approach. Namely, we will giveaxioms for the real numbers, and prove everything in the course from these axioms. Well,this is not strictly true some things will be stated without proof. These may be assimple as ordinary high school algebra, which we will assume is well-understood already(and that you have had some experience in deriving from the axioms). We will also makeuse of various functions familiar from calculus, such as the trigonometric, exponential andlogarithmic functions, even if we havent yet proved their existence and properties fromthe axioms. However, in this case we will eventually at least sketch how this can be donerigorously, and we promise that even if we never talk about these proofs, the use we makeof these functions isnt needed for the proofs (thus we avoid any circularity in the logicalstructure of the material). There is one theorem at the foundation of the course that we willneither prove nor sketch. That one we will just take on faith. (You can read a proof inthe rst chapter of Rudins book, and if you do, you will understand why we wont use classtime for it.)So now we begin. The axioms that dene the real numbers come in three parts: the eldaxioms, the order axioms, and the completeness axiom.Denition 1.1. A eld is a set F with two binary operations, addition (denoted +) and mul-tiplication (denoted ), that satisfy the following axioms (we assume that these are familiar,so we only give them briey).(1) Addition and multiplication are associative and commutative.(2) There exist identity elements for addition and multiplication, denoted 0 and 1, re-spectively.(3) 0 = 1.(4) Every element of F has an additive inverse.(5) Every non-zero element of F has a multiplicative inverse.(6) Multiplication distributes over addition.All of the usual algebraic rules of arithmetic follow from these axioms. For example: Additive and multiplicative identities and inverses are unique. (1)2= 1. xy = 0 if and only if x = 0 or y = 0. (a +b)n=nj=0</p>
<p>nj</p>
<p>anjbj, where the binomial coecients are dened by</p>
<p>nj</p>
<p>= n!j!(n j)!.We will assume familiarity with this stu. It is interesting, though, to consider what isactually included in the phrase this stu. What facts from high school algebra are coveredby the eld axioms? Here is an example of something that is not covered.Example 1.2. Let F be a eld. Are the elements 1, 1 + 1, 1 + 1 + 1, 1 + 1 + 1 + 1, . . .all distinct? In fact, if we just have the eld axioms, we can neither prove nor disprovethat these are all distinct elements. Notice that these are what we normally refer to as thenatural numbers (denoted N). So it isnt clear that the natural numbers even make sense inan arbitrary eld.Exercise 1.3. Explain why the fact stated in the previous example is true.NOTES, MAT 472, INTERMEDIATE ANALYSIS, FALL 2010 3As another example, it is impossible to dene the absolute value function in an arbitraryeld in an intelligent way. To resolve these problems (i.e. to make sure that our axiomsreally do pick out the real numbers) we have to give more axioms.Denition 1.4. Let F be a eld. Then F is an ordered eld if there is a distinguishedsubset F+of F (called the positive elements of F) satisfying the following properties.(1) For each x F, exactly one of the three statements x F+, x F+, x = 0 is true.(2) If x, y F+then x + y F+.(3) If x, y F+then xy F+.Now we dene the usual symbols to express order. For any elements x, y F, we writex > y to mean x y F+, x y to mean x > y or x = y, etc.All of the usual rules of inequalities follow from the order axioms and the eld axioms.For example: If x y and y x, then x = y. If x = 0 then x2> 0. If x < y and z > 0 then xz < yz. If 0 < x < y and n N then xn< yn. xy > 0 if and only if x > 0 and y > 0, or x < 0 and y < 0. A nite subset of F has a minimum. (Of course this is not true for innite subsets.)As a further example, lets note that the order axioms resolve the ambiguity mentionedabove. If F is an ordered eld, then 1 + 1 + + 1 > 0. It follows easily (exercise!) that 1,1 + 1, 1 + 1 + 1, . . . are all distinct positive elements of F. Thus N is contained in everyordered eld. But then the integers, Z, are also contained in every ordered eld. But then therational numbers, Q, are also contained in every ordered eld. In fact, the rational numbersthemselves are an ordered eld (this is obvious, isnt it?). Thus the rational numbers can bedescribed as the smallest ordered eld.We will now dip back into high school algebra to review the absolute value function. Eventhough we assume familiarity with this stu, absolute value is so important that its worthstating some of the details.Denition 1.5. Let F be an ordered eld. The absolute value function [ [ : F F isdened by[x[ =</p>
<p>x, if x 0,x, if x < 0.When we think of the real numbers as a number line, we can view [x[ as the distancebetween x and 0. The number line is great for intuition, but must not be used for proofs.Here are the basic properties of absolute value.(1) [ x[ = [x[.(2) [x[ 0.(3) x [x[.(4) [x[ < a if and only if a < x and x < a (written a < x < a).(5) [x[ > a if and only if x < a or x > a (Note that this cannot be written withoutusing the word or).(6) [x +y[ [x[ +[y[ (the triangle inequality).(7) [x +y[ </p>
<p>[x[ [y[</p>
<p>.4 JACK SPIELBERG(8) [x a[ < r if and only if a r < x < a + r (draw a picture on the number line).(9) If a < x < b and a < y < b then [x y[ < b a.(10) Let x F. Suppose that [x[ < for every positive element F. Then x = 0.Property 10 above can be strengthened a bit, in a way that can be very useful. (Dontcite property 10 when proving this.)Exercise 1.6. Let F be an ordered eld, and let x F. Suppose that p, q F+are suchthat for every F with 0 < < p, we have [x[ < q. Then x = 0.Remark 1.7. Here is another consequence of the ordered eld axioms. Let b > 0. Then(1 + b)n= 1 +nb + > nb.Now let 0 < a < 1. Then 1a > 1, so1 aa = 1a 1 > 0.Let b = 1a 1. Then a = 11+b, andan= 1(1 + b)n < 1nb =</p>
<p> a1 a</p>
<p>1n.Now we ask the following question (assuming some familiarity with the concept of limit,but only for the sake of the discussion): if 0 < a < 1 does antend towards 0, as n ?Another way to put this is to ask: if c is any xed positive element, does there exists n0 Nsuch that an< c for all n n0? Using the above computations, we see that we can answerthis question armatively if we could show that for any xed positive element c, there existsn0 N such that a(1a)n < c for all n n0. Now observe that we could do this if we couldnd n0 N such that a(1a)n0< c. In other words, we could prove that an 0 if we couldnd n0 N such that n0 > a(1a)c. But since c is an arbitrary positive element, then sois a(1a)c. So this all comes down to trying to prove that for any positive element x, thereis a natural number n0 such that n0 > x. An ordered eld in which this is true is calledArchimedean.Denition 1.8. Let F be an ordered eld. F is called Archimedean if for every x F thereexists a natural number n such that x < n.It is evident that Q is an Archimedean ordered eld, and we know that R is one too.But we cant prove it yet, because not all ordered elds are Archimedean!! In other words,we dont yet have enough axioms for the real numbers, since we cant prove the most basicfact from advanced calculus. Along with the eld and order axioms, there is one more axiomthat is necessary to characterize the real numbers. We need some denitions before we canpresent it.Denition 1.9. Let F be an ordered eld, let S F, and let x F.(1) x is an upper bound of S if y x for every y S.(2) x is a lower bound of S if y x for every y S.(3) S is bounded above if there exists an upper bound for S.(4) S is bounded below if there exists a lower bound for S.(5) S is bounded if it is bounded above and below.Exercise 1.10. Is the empty set bounded?NOTES, MAT 472, INTERMEDIATE ANALYSIS, FALL 2010 5Denition 1.11. Let S be a subset of an ordered eld F, and let x F. x is a supremum(or sup, or least upper bound, or lub) of S if(1) x is an upper bound of S.(2) For every upper bound z of S, x z.The condition (2) can be expressed in the equivalent forms:(2</p>
<p>) For any z F, if z < x then z is not an upper bound of S.(2</p>
<p>) For any z F, if z < x then there exists y S with z < y.In a completely analogous manner we dene inmum (or inf, greatest lower bound, glb).The details of the precise formulation are left as an exercise.Remark 1.12. It follows immediately from condition (2) of Denition 1.11 that if S has asupremum then it is unique (and similarly for inmum).Exercise 1.13. Let S be a subset of an ordered eld F, and let x F. Let S = y :y S(1) S is bounded above (respectively, below) if and only if S is bounded below (respec-tively, above).(2) x is an upper (respectively, lower) bound for S if and only if x is a lower (respec-tively, upper) bound for S.(3) x is a supremum (respectively, inmum) of S if and only if x is an inmum (respec-tively, supremum) of S.Now we are ready to state the last axiom of the real numbers, the completeness axiom.Denition 1.14. Let F be an ordered eld. F is complete if every non-empty subset of Fthat is bounded above has a supremum.The following is an easy consequence of Exercise 1.13.Corollary 1.15. Let F be a complete ordered eld. Then every non-empty subset of F thatis bounded below has an inmum.The next theorem is the foundation of the course, but is the one result that we wontattempt to prove. As mentioned earlier, you can read a proof in Rudins book.Theorem 1.16. There exists a unique complete ordered eld.The one and only complete ordered eld is called the eld of real numbers, and we willwrite R as an abbreviation. This is the same number line that we (think we) know and love.But even though we have lots of intuition about it, we will insist on proving EVERYTHINGabout it. For example, since R is an ordered eld, R contains the rational numbers Q as asubeld. Are there any other elements of R besides Q? Well, we think we know that thereare but how do we prove that there are? The usual way is to bring up the classical proofthat2 is irrational. But this is sophistry! That proof merely shows that no rationalnumber has square equal to 2. Its possible that there is an element of R having square equalto 2. If there is such an element, then it cant belong to Q, so it would be an element ofR` Q. But we dont know yet that there is a real number having square equal to 2. In fact,there is, but this fact must be proved.Lets return to an even more basic point, the Archimedean property. We mentioned earlierthat R is Archimedean, and of course this is a fundamental property of the number line: the6 JACK SPIELBERGnatural numbers march o arbitrarily far to the right. Our rst theorem about the realnumbers is this fact. As we pointed out before, the proof must rely on the completenessaxiom, since not all ordered elds are Archimedean.Theorem 1.17. R is Archimedean: for every x R there exists n N such that x < n.Proof. We suppose that R is not Archimedean, and derive a contradiction. So let x Rbe such that x n for all n N. This just means that x is an upper bound for N. Thusthe (non-empty) subset N of R is bounded above. By the completeness axiom, N has asupremum. Let z = sup(N). Now z 1 < z. By Denition 1.11 (2</p>
<p>), there is an elementn N with n > z 1. But then n +1 > z. Since n +1 N, this contradicts Denition 1.11(1). Therefore R is Archimedean. We now present some corollaries of the Archimedean property.Corollary 1.18. If x R with x > 0, then there exists n N with 1n < x.Proof. By the Archimedean property there is n N with n > 1x. Then 1n < x. Before stating the next corollary, we recall the well-ordering principle (WOP) and one ofits variations. The WOP states that a non-empty subset of N...</p>