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.

Indeed, we shall show that there is a bijective correspondence between N Z and Z Q, but that R > Q . ↔ ↔ | | | | (More on this later...)

Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 17 / 26 We also note that if (L, R) where R = Q L is a cut, then it represents a − rational number if R has a least element r with respect to the usual ordering of rational numbers.

Think L = ( , 1/3), R = [1/3, ) represents r = 1/3. −∞ ∞ And if R has no least element then (L, R) is irrational so that it represents a number not in Q.

Also, we can use this conceptualize ’approximating irrational numbers by rational numbers’.

Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 18 / 26 Suppose α = (L, R) is a Dedekind cut (real number) then if r = (L˜, R˜) is rational then r < α iff L˜ L. ⊆ For example suppose √2 = (L, R) where L = x Q x < 0 or x2 < 2 . { ∈ | } If L˜ = y Q y < 1.4 and R˜ = z Q z 1.4 then we can write { ∈ | } { ∈ | ≥ } 1.4 = (L˜, R˜).

Moreover, L˜ L since 1.42 = 1.96 < 2 and we can say that 1.4 = (L˜, R˜) is ⊆ an (under) approximation of √2 = (L, R).

Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 19 / 26 Furthermore, we can quantify what it means to have a better approximation.

If Lˆ = y Q y < 1.41 then Rˆ = Q Lˆ = z Q z 1.41 and so { ∈ | } − { ∈ | ≥ } 1.41 = (Lˆ, Rˆ) where now L˜ Lˆ L ⊆ ⊆ so that 1.41 is closer to √2 and therefore a better approximation than 1.4.

Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 20 / 26 As far as the arithmetic of these real numbers (Dedekind cuts) one can define an addition operation.

Definition If α = (L, R) and β = (L′, R′) are real numbers represented by Dedekind cuts of Q then α + β is defined to be the Dedekind cut (A, B = Q A) − where A = z Q z = p + q where p L and q L′ { ∈ | ∈ ∈ } which is roughly the sum of the numbers in the sets L and L′.

The intuition behind this is that if p L and q L′ then p < α and q < β ∈ ∈ so that p + q < α + β.

Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 21 / 26 In a similar way one may define subtraction as follows: Definition If α = (L, R) is a Dedekind cut, then let α = (L′, R′) where ′ − L = y y R where y is not the least element {− | ∈ } For example √2 = (( , √2), (√2, )) implies −∞ ∞ √2 = (( , √2), ( √2, )). − −∞ − − ∞ A perhaps easier example is 1/3 = (( , 1/3), [1/3, )) so −∞ ∞ 1/3 = (( , 1/3), [ 1/3, )). − −∞ − − ∞ (i.e. the rational number 1/3 is in the ’right’ set, just as 1/3 was in the − right set for the cut defining 1/3)

And from this, we therefore define subtraction as α β = α + ( β). − −

Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 22 / 26 Similarly one can define notions of multiplication, division, and absolute value, of real numbers using Dedekind cuts.

However, we should circle back to the primary motivation for the development of the reals, namely: Theorem

The real numbers (R, R) = (R, ) (i.e. the set of Dedekind cuts of Q) is ≤ ⊆ complete, that is, every subset of R that is bounded above has a least upper bound.

Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 23 / 26 Proof.

Let be a non-empty subset of (R, R) = (R, ) that is bounded above S ≤ ⊆ and define T to be the set T = [ L L∈S which is the union of all (left sets) of the Dedekind cuts.

The set T is clearly non-empty since is non-empty. S Moreover, each L in T is a subset of Q which is closed downward.

The first point to show is that (T , Q T ) is a Dedekind cut as well. − To see this, say t T then t L for some L . ∈ ∈ ∈ S As such x L for all x < t but since t was typical element of T then T is ∈ closed downward.

Moreover, T has no largest element since if y T is such that t < y for ∈ all t T then x < y for all x in any given L which is impossible since ∈ ∈ S each L in is a left set of a Dedekind cut. Timothy Kohl (BostonS University) MA532 Lecture March 26, 2020 24 / 26 Beyond real numbers...

The real numbers, as it turns out are also a field, just as Q is a field.

Moreover, we can, also starting from this number system, R construct a still larger one, namely the complex numbers C, whose construction is also motivated by the need to fix a ’deficiency’ (so to speak) in the reals, namely that not all polynomial equations (with real coefficients) have solutions in the reals, in particular x2 + 1 = 0.

Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 25 / 26 The addition of the principal solution of this equation i = √ 1 allows us − to define the system C = a + bi a, b R { | ∈ } with addition and multiplication defined by (a + bi) + (c + di) = (a + c) + (b + d)i and (a + bi)(c + di) = (ac bd) + (ad + bc)i which makes C into a field as − well.

What one has is that all polynomials with coefficients in C have solutions in C which is the so called ’Fundamental Theorem of Algebra’.

What’s also kind of interesting is the following. Theorem If F is a field such that R F C then either F = R or F = C. ⊆ ⊆ i.e. There is no field in between R and C.

Timothy Kohl (Boston University) MA532 Lecture March 26, 2020 26 / 26