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Handout from Today's Lecture MA532 Lecture Timothy Kohl Boston University March 26, 2020 Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 1 / 26 The Real Numbers The constructions of Z from N and Q from Z were born out of a desire to extend the arithmetic operations by incorporating in additive (Z) and multiplicative (Q) inverses. However, the construction of the real numbers R is born out of the desire for another kind of ’completeness’ but this more tied to order rather than arithmetic. Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 2 / 26 For a poset (A, ) we have defined the concept of an upper bound for S A namely an element M such that x M for all x S. ⊆ ∈ We observe that this upper bound is neither (a) unique, nor (b) a member of S. And we also defined what a maximal element of S A is, namely an ⊂ M S such that x M for all x S. ∈ ∈ We observe/recall that if S A has a maximal element then this element ⊆ is unique. Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 3 / 26 An ’in between’ concept is the notion of a least upper bound which appears regularly in calculus and real analysis. Recall that ℓ is a least upper bound (denoted ℓ = lub(S)) if ℓ is an upper bound for S but no element less than ℓ is an upper bound. Note, lub(S) must also be unique, if it exists, just like a maximal element, but it need not be a member of S. n Example: Let S = n N = 1 , 2 ,... Q. { n+1 | ∈ } { 2 3 }⊆ We can show that ℓ =1= lub(S). Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 4 / 26 n How? Well clearly n+1 < 1 for all n, so consider a number a bit less than 1 such as 0.9 and observe n 0.9 n + 1 ≤ ↓ n 0.9n + 0.9 ≤ ↓ 0.1n 0.9 ≤ ↓ n 10 ≤ So 0.9 is not an upper bound, and similarly 0.99 isn’t an upper bound, nor is any number less than 1. And observe that indeed, lub(S) = 1 S. 6∈ Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 5 / 26 A somewhat different example, and one which highlights the key difference between Q and R is the following. Let S = x Q x < 0 or x2 < 2 Q. { ∈ | }⊆ Observe that the conditions x < 0 or x2 < 2 imply that S contains every negative rational number, as well as every non-negative rational number whose square is less than 2. Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 6 / 26 The equation x2 = 2 has no solution in Q, i.e. √2 Q. 6∈ a Why? Well if √2= b where a and b are integers with no common factor then if we square both sides we get a2 2= a2 = 2b2 b2 → and so a2 is even (which implies that a is even), so that a = 2m and thus 4m2 = 2b2 which means 2m2 = b2 which implies that b is also even, which contradicts the fact that a and b have no common factors. (This proof is due to Euclid.) Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 7 / 26 So back to S = x Q x < 0 or x2 < 2 what we can do (by cheating a { ∈ | } bit and assuming the usual stuff we know about the reals) is to observe that S = ( , √2) Q. −∞ ∩ The reason for this is that the condition x2 < 2 is equivalent to √2 < x < √2. − S already contains all the negative rationals, and it also includes all non-negative rationals up to, but of course, not including, √2 since √2 Q. 6∈ Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 8 / 26 And indeed this is the point of the discussion since if we defined S in R, namely S′ = x R x < 0 or x2 < 2 { ∈ | } then S′ = ( , √2) and therefore lub(S′)= √2 R. −∞ ∈ However, S as a subset of Q has no least upper bound in Q, even though it is bounded above, namely by any rational number greater than √2. Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 9 / 26 This motivates the following. Definition A poset (A, ) is complete if S A and S is bounded above then S has a ⊆ least upper bound in A. So the above example demonstrates that (Q, Q) is not complete. ≤ And this is what motivates our construction of the reals as the ’completion’ of the rationals. Our construction will be a bit different since it won’t consist of equivalence classes, but rather will consist of a particular subset of (Q), the power P set of Q. Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 10 / 26 The example of S above as a kind of ’infinitely long interval’ that extends to the left, is what we have in mind in the following definition. Definition A partition of Q into sets L, R is a Dedekind cut, if (a) L = and L = Q 6 ∅ 6 (b) If x, y Q where x < y and y L then x L. ∈ ∈ ∈ (c) For every x L, there exists y L such that x < y. ∈ ∈ Condition (b) is described by saying the L is closed downward, in a sense analagous to an infinite interval like ( , 3) for example. −∞ Condition (c) is basically the statement that L (think left set!) has no maximal element. Note, the definition really only depends on the choice of L since R is then Q L, so indeed it suffices to consider just the sets L which are closed − downward. Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 11 / 26 Examples: If we set L = S = x Q x < 0 or x2 < 2 from earlier we { ∈ | } have a Dedekind cut. Why? Condition (a) is obvious and if y L then either y < 0 and so if ∈ x < y then x < 0 too (hence x L), and if 0 y 2 < 2 and x < y then ∈ ≤ again, either x < 0 or 0 x2 < y 2 and if x > y then x2 > y 2 so again we ≤ conclude x < y. As to condition (c) if x L then either x < 0 in which case x < y = 0 and ∈ x if 0 x2 < 2 then consider y = 2 +2 where ≤ x+2 2x + 2 x < x + 2 l x2 + 2x < 2x + 2 l x2 < 2 and so x < y and x2 < y 2 < 2 and thus L is a cut. Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 12 / 26 As we just saw L is a cut, but the point is, it corresponds to a number, namely √2 which would be the least upper bound of L except for the fact that √2 is not in the ambient set Q that contains L. The intuition is that L correpsonds to L = ( , √2) and −∞ R = Q L = (√2, ) where we use the ’interval’ notation to convey − ∞ intervals of rational numbers, where, of course √2 Q. 6∈ But this is actually the key to ’build’ the real numbers in terms of cuts. Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 13 / 26 Before we get to this, let’s point out a few things about cuts that are quite interesting, the first of which is that one can put an order on the set of all cuts. Lemma Given two Dedekind cuts L and L′ either L = L′, L L′, or L′ L. ⊆ ⊆ Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 14 / 26 Proof. First, we observe that L L′ cannot be empty since if a L and b L′ ∩ ∈ ∈ then, being rational numbers, either a < b or b < a. So if a < b then since b L′ then by downward closure a L′ as well ∈ ∈ which is a contradiction, and similarly if b < a. If neither L L′ nor L′ L then L L′ is a proper (non-empty) subset of ⊆ ⊆ ∩ both. So suppose L (L L′) = and L′ (L L′) = then there exists −′ ∩ 6 ′∅ −′ ∩ 6 ∅ x L (L L ) and y L (L L ). ∈ − ∩ ∈ − ∩ As x, y Q then either x < y or y < x. ∈ If x < y then as y L′ then by downward closure x L′ which means ′ ∈ ∈ ′ x L L which contradicts the fact that x L (L L ). ∈ ∩ ∈ − ∩ A similar contradiction arises if y < x. As such either L (L L′)= which implies L′ L or L′ (L L′)= − ′ ∩ ∅ ⊆ − ∩ ∅ whichTimothy Kohl implies (BostonL University)L . MA532 Lecture March 26, 2020 15 / 26 ⊆ If we define R to be the set of Dedekind cuts of Q then what we’ve just shown is that provides a total ordering on R. ⊆ (i.e. all cuts are comparable so (R, ) is not just a poset) ⊆ So instead of the subset symbol, we can write (R, R) although where ≤ needed for certain arguments, we will use the subset formulation. What this also shows is that R satisfies the law of trichotomy in that for a given pair of reals x, y R exactly one of the three possibilities holds, ∈ x < y, x > y or x = y. Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 16 / 26 We take a brief moment to also point out that the collection of Dedekind cuts is a subset of the power set (Q). And since Q is infinite (since, after P all it contains Z and, in turn, N) it turns out that R will turn out to be a larger order of infinite, in particular we shall demonstrate later that there is no bijective correspondence between Q and R.
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