Math 410 Homework 3 Due Friday, September 25 1. A (Dedekind) cut is a subset α of Q such that (c1) α 6= ? and α 6= Q, (c2) If x 2 α and y < x, the y 2 α and (c3) α doesn't have a largest element (more formally, for every p 2 α, there exists q 2 α with p < q). Which of the following are cuts? Show your work (as in give why they pass/fail specific conditions). ∗ a) 2 = fx 2 Q j x < 2g This is a cut. ∗ ∗ ∗ ∗ (c1) 1 2 2 so 2 6= ? and 3 2= 2 so 2 6= Q. (c2) If x 2 2∗, then x < 2 by definition. Therefore, if y < x, then y < x < 2 so y 2 2∗ ∗ p+p p+2 2+2 p+2 ∗ (c3) If p 2 2 , then p < 2, so p = 2 < 2 < 2 = 2, so there exists q = 2 2 2 larger than p. b) α = fx 2 Q j x ≤ 2g This is not a cut since it has a largest element, namely 2. 2 c) β = fx 2 Q j x < 2 or x < 0g This is a cut. (c1) −1 2 β so β 6= ? and 4 2= β so β 6= Q. (c2) Let x 2 β and y < x. If y < 0, then y 2 β. If 0 ≤ y < x, then since both sides of the inequality are positive, y < x is equivalent to y2 < x2 < 2, so y 2 β. (c3) We proved something very similar to (c3) in class. Just like before, let p2−2 2 q = p − p+2 , but this time, p < 2 so we are subtracting a negative and p < q. Just 2 (p2−2) 2 like in class, q − 2 = 2 (p+2)2 which this time is negative so q < 2 and q 2 β. 2 d) γ = fx 2 Q j x ≤ 2 or x < 0g This is a cut. In factp it is the same exactp cutp as inp part c) since the only difference between the set (−∞; 2) and (−∞; 2] is 2 and 2 2= Q. 2. Prove that if α is a cut and p 2 α and q2 = α, then p < q. This Lemma will be useful for subsequent problems. Proof by contradiction: Assume that α is a cut, p 2 α; q2 = α but q ≤ p. If q = p, then q 2 α and we have a contradiction, so assume q < p. However, since α is a cut, it satisfies condition (c2), and since p 2 α and q < p, q 2 α as well which is again a contradiction. Therefore q ≤ p can't hold and we can conclude that q > p. 1 ∗ ∗ 3. Let R be the set of all cuts and let α < β in R if α ( β as subsets of Q. If q 2 Q let ∗ ∗ q = fx 2 Q j x < qg 2 R . ∗ ∗ ∗ a) Show that 2 < 3 as elements in R . 2∗ = (−∞; 2) and 3∗ = (−∞; 3). Let x 2 2∗. Therefore x < 2 < 3 so x 2 3∗ so ∗ ∗ ∗ ∗ ∗ ∗ 2 ⊂ 2 as subsets of Q. In addition, 2:5 2 3 and 2:5 2= 2 so 2 ( 3 as subsets of Q. ∗ ∗ ∗ ∗ Therefore, by definition of the order < on R , 2 < 3 in R . b) Show that q∗ < p∗ if and only if q < p. ( This proof is almost identical to part a), just replace 2 with q, 3 with p and 2:5 p+q with 2 . ) This is the lemma from problem 2. ∗ c) Show that < describes an ordering of R . (That is, show triality for the set of all cuts and transitivity of containment.) Hint: to show triality, you need to show ∗ that given any two cuts α; β 2 R , one of α < β; β < α; α = β must hold. Start with the assumption that there exists x 2 α with x2 = β. What does this imply about all y 2 β? An ordered set is S such that 1. for all s; t 2 S, one of s < t; t < s; or s = t holds and 2. for all s; t; u 2 S, if s < t and t < u, then s < u. (These two properties are called triality and transitivity respectively.) Let α; β 2 R and assume that there exists x 2 α with x2 = β. Then for all y 2 β, y < x by the lemma in problem 2. By property (c2) for α, since y < x and x 2 α, y 2 α and therefore β ⊂ α as subsets of Q (we just picked an arbitrary element of β and showed it is in α). Since x 2 α and x2 = β, α 6= β and therefore β ( α as subsets of Q ∗ and therefore β < α in R . Similarily, if there exists z 2 β with z2 = α, then α < β. If there doesn't exist either such an x or such a z, then α − β = ? and β − α = ?, so either α \ β = ? or α = β. If α \ β = ?, for all x 2 α, x2 = β so by the lemma for problem 2, for all y 2 β; y < x for all x 2 α. However, by condition (c2), this im- plies y 2 α for all y 2 β which is a contradiction. Therefore α = β which proves triality. ∗ Assume that α; β; γ 2 R with α < β and β < γ. Therefore, as subsets of Q, α ( β and β ( γ. Let x 2 α. Then x 2 β by the first inclusion and x 2 γ by the second inclusion and therefore α ⊆ γ. Since α ( β there exists y 2 β with y2 = α. By the second inclusion, y 2 γ, so there exists y 2 γ with y2 = α and therefore α 6= γ and ∗ α ( γ as subsets of Q and hense α < γ in R which proves transitivity. 4. Define addition of two cuts α and β to be α + β = fx + y j x 2 α and y 2 βg. a) Show how this definition works for 2∗ + 3∗ and demonstrate that 2∗ + 3∗ = 5∗. By definition, 2∗ + 3∗ = fx + y j x < 2 and y < 3g Since x < 2 and y < 3 implies ∗ ∗ ∗ ∗ x + y < 5 (previous homework), 2 + 3 ⊆ 5 as subsets in Q. Additionally, if z 2 5 , i.e. z < 5, then let = 5 − z > 0. Then z = 5 − = (2 − /2) + (3 − /2). Since > 0, 2 − /2 < 2 so is in 2∗ and 3 − /2 < 3 so is in 3∗ and we have written z = x + y where x 2 2∗ and y 2 3∗. Namely, we just showed that 5∗ ⊆ 2∗ +3∗ and therefore 5∗ = 2∗ +3∗. b) Show that α + β is indeed a cut. (c1) follows directly from the facts that both α and β are neither empty nor all of Q: Since ?( α (alpha is a cut so satisfies condition (c1)), there exists x 2 α and since ?( β, there exists y 2 β. Therefore, x + y 2 α + β, so α + β 6= ?. Similarily, since α (Q, there exists z 2 Q with z2 = α and w 2 Q with w2 = β. We want to show that z + w2 = α + β. That is, we want to show that z + w 6= x + y for all x 2 α and y 2 β. By our lemma (problem 2), for all x 2 α, since x 2 α and z2 = α, x < z and similarily, for all y 2 β, y < w. By a problem in homework 2, this implies that x + y < z + w for all x 2 α and all y 2 β and therefore x + y 6= z + w and z + w2 = α + β. (c2) Let z 2 α + β and w < z. Then z = x + y for some x 2 α and some y 2 β. Write w = x + (w − x). Since w = x + (w − x) < z = x + y we get w − x < y by the rules of an ordered field. Therefore, since β is a cut, w − x 2 β and we have written w as a sum of an element of α and an element of β, so w 2 α + β by definition. (c3) Let z 2 α + β, show there exists u 2 α + β with w < u. Let z = x + y where x 2 α and y 2 β. Since β is a cut, y isn't its largest element, so there exists v 2 β with y < v. Then z = x + y < u = x + v but u is also the sum of an element of α and an element of β so is in α + β by definition. ∗ c) Check that addition of cuts is commutative and associative in R . Show that α + β = β + α. Notice that β + α = fy + x j y 2 β and x 2 αg by definition and this is equal to fx + yj x 2 α and y 2 βg = α + β because addition is commutative in Q. We also need to show that (α + β) + γ = α + (β + γ) which again follows directly from the fact that (x + y) + z = x + (y + z) in Q. d) Prove that if α < β then for any cut γ, α + γ < β + γ. ∗ If α < β in R , then α ( β as subsets of Q and therefore, for all x 2 α; x 2 β as well. Let z 2 α + γ. Therefore z = a + g for some a 2 α and some g 2 γ.