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 ∈ α and y < x, the y ∈ α and (c3) α doesn’t have a largest element (more formally, for every p ∈ α, there exists q ∈ α with p < q). Which of the following are cuts? Show your work (as in give why they pass/fail specific conditions).

∗ a) 2 = {x ∈ Q | x < 2} This is a cut. ∗ ∗ ∗ ∗ (c1) 1 ∈ 2 so 2 6= ∅ and 3 ∈/ 2 so 2 6= Q. (c2) If x ∈ 2∗, then x < 2 by definition. Therefore, if y < x, then y < x < 2 so y ∈ 2∗ ∗ p+p p+2 2+2 p+2 ∗ (c3) If p ∈ 2 , then p < 2, so p = 2 < 2 < 2 = 2, so there exists q = 2 ∈ 2 larger than p.

b) α = {x ∈ Q | x ≤ 2} This is not a cut since it has a largest element, namely 2.

2 c) β = {x ∈ Q | x < 2 or x < 0} This is a cut. (c1) −1 ∈ β so β 6= ∅ and 4 ∈/ β so β 6= Q. (c2) Let x ∈ β and y < x. If y < 0, then y ∈ β. If 0 ≤ y < x, then since both sides of the inequality are positive, y < x is equivalent to y2 < x2 < 2, so y ∈ 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 ∈ β.

2 d) γ = {x ∈ Q | x ≤ 2 or x < 0}

This is a cut. In fact√ it is the same exact√ cut√ as in√ part c) since the only difference between the set (−∞, 2) and (−∞, 2] is 2 and 2 ∈/ Q.

2. Prove that if α is a cut and p ∈ α and q∈ / α, then p < q. This Lemma will be useful for subsequent problems.

Proof by contradiction: Assume that α is a cut, p ∈ α, q∈ / α but q ≤ p. If q = p, then q ∈ α and we have a contradiction, so assume q < p. However, since α is a cut, it satisfies condition (c2), and since p ∈ α and q < p, q ∈ 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 ∈ Q let ∗ ∗ q = {x ∈ Q | x < q} ∈ R .

∗ ∗ ∗ a) Show that 2 < 3 as elements in R . 2∗ = (−∞, 2) and 3∗ = (−∞, 3). Let x ∈ 2∗. Therefore x < 2 < 3 so x ∈ 3∗ so ∗ ∗ ∗ ∗ ∗ ∗ 2 ⊂ 2 as subsets of Q. In addition, 2.5 ∈ 3 and 2.5 ∈/ 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 ∈ R , one of α < β, β < α, α = β must hold. Start with the assumption that there exists x ∈ α with x∈ / β. What does this imply about all y ∈ β? An ordered set is S such that 1. for all s, t ∈ S, one of s < t, t < s, or s = t holds and 2. for all s, t, u ∈ S, if s < t and t < u, then s < u. (These two properties are called triality and transitivity respectively.)

Let α, β ∈ R and assume that there exists x ∈ α with x∈ / β. Then for all y ∈ β, y < x by the lemma in problem 2. By property (c2) for α, since y < x and x ∈ α, y ∈ 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 ∈ α and x∈ / β, α 6= β and therefore β ( α as subsets of Q ∗ and therefore β < α in R . Similarily, if there exists z ∈ β with z∈ / 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 ∈ α, x∈ / β so by the lemma for problem 2, for all y ∈ β, y < x for all x ∈ α. However, by condition (c2), this im- plies y ∈ α for all y ∈ β which is a contradiction. Therefore α = β which proves triality.

∗ Assume that α, β, γ ∈ R with α < β and β < γ. Therefore, as subsets of Q, α ( β and β ( γ. Let x ∈ α. Then x ∈ β by the first inclusion and x ∈ γ by the second inclusion and therefore α ⊆ γ. Since α ( β there exists y ∈ β with y∈ / α. By the second inclusion, y ∈ γ, so there exists y ∈ γ with y∈ / 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 = {x + y | x ∈ α and y ∈ β}.

a) Show how this definition works for 2∗ + 3∗ and demonstrate that 2∗ + 3∗ = 5∗. By definition, 2∗ + 3∗ = {x + y | x < 2 and y < 3} Since x < 2 and y < 3 implies ∗ ∗ ∗ ∗ x + y < 5 (previous homework), 2 + 3 ⊆ 5 as subsets in Q. Additionally, if z ∈ 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∗ and y ∈ 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 ∈ α and since ∅ ( β, there exists y ∈ β. Therefore, x + y ∈ α + β, so α + β 6= ∅. Similarily, since α (Q, there exists z ∈ Q with z∈ / α and w ∈ Q with w∈ / β. We want to show that z + w∈ / α + β. That is, we want to show that z + w 6= x + y for all x ∈ α and y ∈ β. By our lemma (problem 2), for all x ∈ α, since x ∈ α and z∈ / α, x < z and similarily, for all y ∈ β, y < w. By a problem in homework 2, this implies that x + y < z + w for all x ∈ α and all y ∈ β and therefore x + y 6= z + w and z + w∈ / α + β. (c2) Let z ∈ α + β and w < z. Then z = x + y for some x ∈ α and some y ∈ 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 ∈ β and we have written w as a sum of an element of α and an element of β, so w ∈ α + β by definition. (c3) Let z ∈ α + β, show there exists u ∈ α + β with w < u. Let z = x + y where x ∈ α and y ∈ β. Since β is a cut, y isn’t its largest element, so there exists v ∈ 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 = {y + x | y ∈ β and x ∈ α} by definition and this is equal to {x + y| x ∈ α and y ∈ β} = 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 ∈ α, x ∈ β as well. Let z ∈ α + γ. Therefore z = a + g for some a ∈ α and some g ∈ γ. Since α < β, a ∈ β and therefore, z is written as a sum of an element of β and an element of γ and therefore z ∈ β + γ and α + γ ⊆ β + γ as subsets in Q. Since α ( β, there exists b ∈ β with b∈ / α. In fact there does exist g ∈ γ such that b + g∈ / α + γ, but g can not be chosen arbitrarily. I’m sorry for getting so sloppy here, but if we choose g such that lub(γ) − g + lub(β) − b < lub(β) − lub(α) we will be safe. Notice that lub of a cut only exists for rational cuts (we don’t have R yet), so I have really only shown this for rational cuts.

∗ ∗ e) EXTRA CREDIT Prove that the cut 0 is the additive identity in R by show- ing containment in both directions.

∗ ∗ Let α ∈ R any cut. We will first show that α + 0 ⊆ α. Let z ∈ α + 0∗ = {x + y | x ∈ α, y ∈ 0∗}. Since y ∈ 0∗, this means that y < 0 by the definition of a rational cut. Therefore, x + y < x as rational numbers, and since x ∈ α and α satisfies condition (c2), z = x + y ∈ α and therefore α + 0∗ ⊆ α. Now show that α ⊆ α + 0∗. Let z ∈ α, we want to show there exists x ∈ α and y ∈ 0∗ such that z = x + y. Since z ∈ α with α a cut, by condition (c3) we know there exists some w ∈ α such that z < w. Therefore, z − w < 0, so w − z ∈ 0∗. Since z = w + (z − w) with w ∈ α and (z − w) ∈ 0∗, z ∈ α + 0∗ and α ⊆ α + 0∗. ∗ ∗ ∗ Therefore, α+0 = α for all α ∈ R and R satisfies condition A4 to be a field (existance of an additive identity).

5. Let α be a cut. Define −α = {x ∈ Q | there exists t∈ / α with t < −x}.

a) Illustrate how this definition would work for −(2∗) on the number line and demon- strate that −(2∗) = (−2)∗. ∗ ∗ By definition −(2 ) = {x ∈ Q | ∃ t∈ / 2 with t < −x} = {x ∈ Q | ∃ t ≥ 2 with t < −x} = {x ∈ Q | ∃ t st 2 ≤ t < −x} = {x ∈ Q | ∃ t st x < −t ≤ −2} but since the name t was arbitrary, this is the same as {x ∈ Q | ∃ s st x < s ≤ −2}. By letting ∗ s = −2, we see that this is just {x ∈ Q | x < −2} = (−2) . All that probably makes more sense when drawn out on an number line, but that’s kind of hard to do in TeX.

b) Prove that −α is a cut.

(c1) Since α 6= Q, there does exist t∈ / α and since Q has no upper bound, there exists y with t < y. Therefore −y ∈ −α and −α 6= ∅. Since α 6= ∅, there exists y ∈ α and by the lemma from problem 2, y < t for all t∈ / α, so −y∈ / −α and −α 6= Q. (c2) Let x ∈ −α and y < x (so −x < −y), then there exist t such that t∈ / α and t < −x < −y, so y ∈ −α. t+−p (c3) Let p ∈ −α, show there exists q ∈ −α with p < q. Let −q = 2 Since t < −p, t+t t+−p −p−p t = 2 < 2 = −q < 2 = −p and therefore q ∈ −α. Since −q < −p, we have q > p and p is not the largest element of −α. Since p was arbitrary, there doesn’t exists a largest element of −α.

c) Why did we not define the additive inverse of α as {x ∈ Q | − x∈ / α}? (Hint: try the wrong definition for 2∗.) This is the wrong definition because if we did it for 2∗, we would get the set (−∞, −2] which isn’t a cut. The complicatedness of the definition was all to exclude the endpoint.

d) EXTRA CREDIT We will prove that −α is the additive inverse of α for rational cuts. Namely, prove that if x ∈ q∗ and y ∈ −(q∗), then x + y ∈ 0∗, and therefore q∗ + (−(q∗)) ⊆ 0∗. Next show that for any z ∈ 0∗, there exist x ∈ q∗ and y ∈ −(q∗) such that x + y = z, and therefore 0∗ ⊆ q∗ + (−(q∗)). Conclude that q∗ + (−(q∗)) = 0∗.

first for q∗ and (−q)∗ (so x < q and y < −q implies x + y < q + y < q + −q < 0) now for −(q∗)x < q and y st ∃t∈ / q∗ with t < −y so y < −t and x + y < q + y < q + −t but t∈ / q∗ so q < t so −t < −q and x + y < q + −q = 0. other direction, let z ∈ 0∗, so z < 0 and pick x ∈ q∗ such that q + z < x < q, then show −x + z < −q, so −x + z ∈ (−q)∗ 6. Multiplication of cuts is significantly more annoying because the product of two negative numbers is positive. If 0∗ < α and 0∗ < β, define multiplication of cuts by α · β = {xy| x ∈ α, y ∈ β, x, y > 0} ∪ {z ∈ Q| z ≤ 0}.

a) Illustrate how this definition works for 2∗ ·3∗ on the number line and demonstrate that 2∗ · 3∗ = 6∗. ∗ ∗ 2 · 3 = {xy | 0 < x < 2 and 0 < y < 3} ∪ {z ∈ Q| z ≤ 0} by the definition above. Since y > 0, we can multiply 0 < x < 2 by y to get 0 < xy < 2y and y < 3 implies 2y < 6 and we see that if w ∈ 2∗ · 3∗, then either 0 < w < 6 or w ≤ 0 which means ∗ ∗ ∗ that 2 · 3 ⊆ 6 as subsets in Q. In addition, if w ∈ 6∗, then w < 6. If w ≤ 0, then w ∈ 2∗ · 3∗ by definition. If 0 < w, w w find  > 0 sufficiently small that 0 < 2− < 3. Then w = (2 − ) 2− which is a product of a positive element of 2∗ and a positive element of 3∗ so is in 2∗ · 3∗ and therefore 5∗ ⊆ 2∗ · 3∗ and the two are therefore equal.

b) EXTRA CREDIT Prove that α · β is a cut.

(c1) Let α and β as above. Then for all z < 0, z ∈ α · β so ∅ 6= α · β. To show that α · β 6= Q, use the fact that α 6= Q to find w ∈ Q with w∈ / α and w > 0 and t∈ / β with the same condition. Then, by our lemma, for all x ∈ α, x < w and for all y ∈ β, y < t. Multiplying both sides of the first inequality by y > 0 we get xy < wy and multiplying both sides of the second by w > 0 gives wy < wt, so have that xy < wt for all x ∈ α and y ∈ β, so wt 6= xy for any x ∈ α, y ∈ β and wt > 0, so wt∈ / α · β. (c2) Let w ∈ α · β and t < w. Show t ∈ α · β. If t ≤ 0, then t ∈ α · β by definition, so assume that 0 < t < w. Since w ∈ α · β, w > 0, there must exist x > 0, x ∈ α and y > 0, y ∈ β with w = xy. Show there exists x0 ∈ α and y0 ∈ β with t = x0y0. We have 0 < t < xy and for any x0 < x, x0 ∈ α and same for y0, so we will need to pull a trick like we did with part a). We have t = xy −  for some  > 0 (they’re both rational t numbers, so this is perfectly ok). We want to write t = (x − δ) x−δ for some 0 < δ < x t such that x−δ < y. Therefore, we are looking for t < y(x − δ) = xy − yδ. That is  0 0 xy −  < xy − yδ so choose δ < min{ y , x} will work. Therefore, we have t = x y 0 0 t where 0 < x = x − δ < x is in α and 0 < y = x0 < y so is in β so we are done. (c3) Let w ∈ α · β, show there exists u ∈ α · β with w < u. Again, if w < 0, there exists w < w/2 < 0 with w/2 ∈ α · β by definition, so assume that 0 ≤ w. If w = 0, ∗ ∗ then since 0 ( α and 0 ( β, there exists x ∈ α and y ∈ β with x ≥ 0 and y ≥ 0. If either x or y equal zero, use condition (c3) to find an x > 0 or y > 0, so we find that 0 = w < xy ∈ α · β, so 0 ∈ α · β by (c2). Therefore, assume that w ∈ α · β with w > 0 and therefore, there exists x ∈ α and y ∈ β with x, y > 0 and w = xy. Since α is a cut, there exists x00 ∈ α with x < x00 and y00 ∈ β with y < y00. By the same arguement as with (c1), this implies that w = xy < x00y00 with x00y00 ∈ α·β by definition.

c) Define multiplication of negative cuts using the fact that if α < 0∗, the −α > 0∗. Use this to define (−2)∗ · 3∗ and show that (−2)∗ · 3∗ = (−6)∗ = −6∗. Let α < 0∗ and β > 0∗. Then 0∗ < −α so (−α) · β is a cut defined above. Then we define α · β to be the cut −((−α) · β). Similarily, if both α and β are negative cuts, we can define α · β as (−α) · (−β). Therefore, the definition of (−2)∗ · 3∗ is as −(−(−2)∗ · 3∗). However, similar to 5a), ∗ ∗ ∗ ∗ −(−2) = {x ∈ Q | ∃ s st x < s ≤ 2} = {x ∈ Q | x < 2} = 2 , so −(−(−2) · 3 ) = −(2∗ · 3∗). Therefore, by part a) of this problem, −(−(−2)∗ · 3∗) = −(6∗) which, in an arguement identical to that in 5a) is equal to (−6)∗. The fact that (−6)∗ = −6∗ is 5d), the extra credit problem which has at least the outline of a proof. ∗ 7. Least Upper Bound Property Let E ⊂ R any non-empty bounded subset, and [ let σ = α. α∈E a) Prove that σ is a cut. (c1) Since E is non-empty, there exists at least one α ∈ E a cut which is, itself, non- S empty. Therefore, there exists some x ∈ α and x ∈ α∈E α = σ, so σ is non-empty. ∗ Since E 6= R and is bounded there exists some cut β such that α ≤ β for all α ∈ E and some z∈ / β. Since α ⊆ β for all α ∈ E and z∈ / β, z∈ / α for all α ∈ E. Therefore, z∈ / σ and σ 6= Q. (c2) Let x ∈ σ and y < x. Since x ∈ σ, there exists some α ∈ E with x ∈ α (by definition of union). Since α is a cut, it satisfies condition (c2) and therefore y ∈ α and hense y ∈ σ (again by definition of union). (c3) This is nearly identical to (c2) we just need to use the property (c3) on the cut α rather than (c2)

b) Prove that α ≤ σ for all α ∈ E. Let α ∈ E. Then by definition of union, for all x ∈ α, x ∈ σ and therefore α ⊆ σ as ∗ subsets of Q and hense α ≤ σ as elements in R .

c) Prove that for any cut γ with γ < σ, there exists a cut β ∈ E with γ < β. ∗ Let γ ∈ R with γ < σ. Therefore, γ ( σ as subsets of Q and there exists x ∈ σ with x∈ / γ. Since x ∈ σ, there exists some β ∈ E such that x ∈ β. Since x ∈ β and x∈ / γ, γ < β by our proof of triality in 3c) which completes this problem.

∗ d) Conclude that σ is the least upper bound of E and therefore R has the least upper bound property. We have shown that α ≤ σ for all α ∈ E (part b)) and therefore, σ is an upper bound on the set E. In addition, we have shown that any element less than σ is not an upper bound on E (part c)). Therefore, by definition, σ is the least upper bound of the set ∗ E. Since our set E was an arbitrary bounded subset of R , we now can conclude that ∗ R has the Least Upper Bound Property!!!!

∗ ∗ ∗ Hopefully this homework helps make sense of why Q = {q | q ∈ Q} ⊆ R as a sub-ordered ∗ field and why R has the least upper bound property. The final step in the construction of R is ∗ constructing an isomorphism (one to one onto map that perserves +, · and <) from R → R that ∗ 2 √involves mapping q to q and identifies a cut like α = {x ∈ Q| x < 2 or x < 0} with the√ number √2. Note that this necessarily means that 1 < 1.4 < 1.41 < 1.414 < 1.4142 < ··· < 2 and 2√< ··· < 1.4143 < 1.415 < 1.42 < 1.5 < 2 which is what we are used to for our usual definition of 2 in R. A number like π which is difficult to describe in terms of algebra can be defined to be the cut that is in between all of 3.14 < π < 3.15, 3.141 < π < 3.142, 3.1415 < π < 3.1416, 3.14159 < π < 3.14160, 3.141592 < π < 3.141593,....