MA74 PRACTICE WITH THE REAL NUMBERS1. Basic Properties of RThe initial task in MA104 is to develop rigorously the real number system.In particular, it is necessary to go about a set theoretic construction (such as“Dedekind cuts”) to demonstrate the existence of a set having the properties thatwe “want” the real numbers to have. In other words, we have to show that a setsatisfying all of the properties stated below is not inherently contradictory.A naive example of something that would be inherently contradictory: supposethat we demanded a tax system in which everyone pays less and the governmentreceives more. We could prove all sorts of “If ... then ...” statements about thisnew tax system, all using this property. For example, “If everyone agrees to thissystem, then money would grow on trees!” But it is pointless, because no suchsystem exists. So when we say that R has all of the following properties, we aretaking for granted that there actually is a set out there that works.Field Properties: R is a (commutative) group under +, and R − {0} is a (com-mutative) group under ·, and the usual distributive and associative prope rties hold.In particular, 0 denotes the identity for addition, and 1 denotes the identity formultiplication, and 1 6= 0.Total Ordering: For every x, y ∈ R, one and only one of x > y, x = y, or x < yis true. The ordering on R agrees with the usual ordering on Q ⊆ R, and the usualproperties of equalities and inequalities hold.Least Upper Bound Property: If S ⊆ R is not empty, and if there existsM ∈ R such that for all x ∈ S we have x ≤ M , then there exists m ∈ R such thatm = sup(S). In other words, every nonempty subset of the real numbers that isbounded above has a least upper bound that is a real number. The book statesthis as an equivalent property for infima; you should know why the two statementsare equivalent.Archimedean Property: Given x ∈ R, there exists n ∈ N such that x < n.Equivalently, given ε > 0, there exists n ∈ N such that 0 < 1/n < ε. (You shouldknow how to show the equivalence.)Note: You should know the definitions of sequence, neighborhood of radius ε, andconvergence to a real number L from the text (page 201, defs. 34-36).122. Background to Cantor’s Diagonal ArgumentDefinition: A set X is countably infinite if there exists a bijection φ : N → X.(Generally, to say that X is countable is to say that X is finite or countably infinite.)Fact: The real numbers are not countably infinite. We will sketch Cantor’s “diag-onal argument” in lecture to informally verify this fact. Infinite sets that are notcountably infinite are said to be uncountable.Cantor’s argument requires us to formulate the following two definitions. Theformer should be familiar from calculus, and the latter is a fancy way to say whatyou learned in first grade (where you most likely assumed b = 10).Definition: Let a = (an) be a sequence. We say that the series determined by aconverges to S ∈ R, and we write∞Xn=0an= S ,if the sequence (sk), given by sk:=kXn=0an, converges to S, as a sequence of realnumbers. We call the sequence (sk) the sequence of partial sums determined by a.Definition: Let b ∈ N≥2, and let x ∈ R. Then a base b expansion for x is afunctiondig : Z → {0, 1, ..., b − 1}such that∞Xi=0dig(i)bi+∞Xi=1dig(−i)b−i= xExamples in base 10:x =13dig(i) = 0 for i ≥ 0, dig(i) = 3 for i ≤ − 1x = π dig(0) = 3, dig(−1) = 1, dig(−2) = 4, ...x = 1 dig(i) = 1 if i = 1, dig(i) = 0 if i 6= 1OR... dig(i) = 0 if i ≥ 0, dig(i) = 9 if i < 0Note: The last example settles the “debate,” that I’m sure you had in some mathclass at some point, about whether .999 = 1. We see that, for a given real number,the corresponding digits function dig may not be unique.(Hopefully, there is also a bit of embarrassment over debating something with-out even understanding the definitions...the symbol “point nine repeating infinitelymany times” is meaningless without the rigorous definition of convergent series.)33. Example ProofWe can now use our setup to prove a basic fact about the real numbers.Theorem: If (xn) is monotone increasing and bounded above, then there existsL ∈ R such that xn→ L.Proof: Let L = sup{xn| n ∈ N}. This supremum exists because the sequence isbounded. We will show that xn→ L.Let ε > 0. Then there exists N ∈ N such that:L − ε < xN≤ L(This was a homework problem.) Subtract L to get:−ε < xN− L ≤ 0Taking negatives gives:0 ≤ L − xN< εNow let n ≥ N. We must show that |xn−L| < ε. Since (xn) is monotone increasing,we know that −xn≤ −xN. Adding L, and noting that xn≤ L, we get:0 ≤ L − xn≤ L −xNSince L − xnis positive, we may replace it with its absolute value, and combiningthis maneuver and the above inequalities gives:L − xn= |L −xn| = |xn− L| ≤ L − xN< ε4. Homework due May 7thFor the final homework, you will prove the sequence of lemmas below, and thenuse them to prove the “geometric series formula.” This is arguably our most elegantexample of a concrete and useful formula that is derived from purely abstractprinciples. I will go over all of the problems in lecture, so your real task will b e towrite them up in a “proper” fashion.We start with the definition of a subsequence. If the formulation below seemsawkward, remember that we are striving for a rigorous definition, not a conceptualunderstanding.Definition: Let (xn) be a given sequence. Let ` : N → N be strictly monotoneincreasing. Then we say that ` determines a subsequence of (xn), and we write(xn`) to denote the sequence defined by x ◦` : N → R. (In other words, n`= `(n).)Example: Let (xn) = (−1)n= (1, −1, 1, −1, ...). Let ` : N → N be the mapk 7→ 2k + 1. Spelling out the notation:x ◦ `(0) = x(1) = x1= xn0= −1x ◦ `(1) = x(3) = x3= xn1= −1x ◦ `(2) = x(5) = x5= xn2= −1...4In other words, we have just rigorously described the “odd” subsequence. To obtainthe general construction of a subsequence, we replace the map k 7→ 2k + 1 witha map ` that prescribes which terms comprise the subsequence. We thus see thereason for the assumption that ` is increasing...this is exactly the rigorous conceptthat captures what we “want” a subsequence to be.HW Problems:Lemma: If xn→ L, and ` : N → N is strictly increasing, then x ◦ ` converges toL. In other words, every subsequence of a convergent sequence converges, and theyall share a common limit