Math 104, Summer 2019 PSET #2 (Due Tuesday 7/2/2019)

Math 104, PSET #2 Daniel Suryakusuma, 24756460

Math 104, Summer 2019 PSET #2 (due Tuesday 7/2/2019)

Problem 6.2. Show that if α and β are Dedekind cuts, then so is α + β = {r1 + r2 : r1 ∈ α and r2 ∈ β}.

Solution. This result is explicitly given in the main text of Ross section 6 as the deﬁnition of addition in R. However, we’ll back-track and verify this result. Suppose α, β are Dedekind cuts. Then we can explicitly write α = {r1 ∈ Q | r1 < s1} and β = {r2 ∈ Q | r2 < s2}. To show (α + β) :={r1 + r2 | r1 ∈ α, r2 ∈ β} is a Dedekind cut, we simply refer to the deﬁnition of (requirements for) a Dedekind cut.

(i) To see (α + β) 6= Q, simply consider elements ξ1, ξ2 ∈ Q but with ξ1 ∈/ α and ξ2 ∈/ β . By our deﬁnitions, (ξ1 + ξ2) ∈/ (α + β), but (ξ1 + ξ2) ∈ Q, hence (α + β) 6= Q. To see (α + β) 6= {}, consider that α, β are nonempty hence r1, r2 exist, and thus r1 + r2 ∈ (α + β) exists. To see this inclusion explicitly, see (ii). (ii) Let r1 ∈ α, r2 ∈ β. Then we have for all s1, s2 ∈ Q, that (s1 < r1) =⇒ s1 ∈ α and s2 < r2 =⇒ s2 ∈ β. Adding these inequalities, we have [(s1 + s2) < (r1 + r2)] =⇒ (s1 + s2) ∈ (α + β) per our deﬁnition of (α + β). Because s1 ∈ Q, s2 ∈ Q =⇒ (s1 + s2) ∈ Q, we satisfy requirement (ii). (iii) We already have that α, β each contain no largest rational. In other words, ∀r1 ∈ α and ∀r2 ∈ β, we have the existence of some c1, c2 with r1 < c1 and r2 < c2. Because these all exist, c1 + c2 ∈ Q. Adding the inequalities, for all (r1 + r2) ∈ (α + β), we must have (r1 + r2) < c1 + c2, and hence there exists some (c1 + c2) ∈ Q larger than (r1 + r2) ∈ (α + β) for all choices r1 ∈ α, r2 ∈ β. We conclude (α + β) has no greatest rational.

As we have satisﬁed all requirements from the deﬁnition, we conclude (α + β) is a Dedekind cut.

Problem 6.3. 1. Show α + 0∗ = α for all Dedekind cuts α. 2. We claimed, without proof, that addition of Dedekind cuts satisﬁes property A4. Thus if α is a Dedekind cut, there is a Dedekind cut −α such that α + (−α) = 0∗. How would you deﬁne −α?

Solution. (1) Recall that we defined α ≤ β ⇐⇒ α ⊆ β. Hence to show α = β, we choose to show α ⊆ β and α ⊆ β (as opposed to the equivalent inequality argument). ∗ ∗ We define 0 in the natural way: 0 :={r0 ∈ Q | r0 < 0}, or in other words the set of all negative rationals, where 0∗ corresponds to the real number 0. ∗ Let r1 ∈ α and r0 ∈ 0 . From above, we have r2 < 0. Hence r1 + r0 < r1, and by our definition of α,(r1 + r0) ∈ α. Thus from above, we have α + 0∗ ⊆ α. ∗ Let r1, r2 ∈ α with r1 < r2. Then r1 − r2 < 0 =⇒ (r1 − r2) ∈ 0 . Then add (using the result of Problem 6.2 ∗ ∗ above) r2 ∈ α and (r1 − r2) ∈ 0 to get: r2 + (r1 − r2) = r1 ∈ α, so α + 0 ⊆ α. We have shown α ⊆ α + 0∗ ⊆ α, so we necessarily have α + 0∗ = α as required.

(2) Now we need to deﬁne some Dedekind cut −α so that α + (−α) = 0∗. As in Rudin, we propose

(−α) := { p | ∃r>0 [−p − r∈ / α] } . That is, some rational number less than −p is not in α. Although not prompted, we prove this claim To show this, we ﬁrst show (−α) ∈ R as a Dedekind cut and then α + (−α) = 0∗. (i) If s∈ / α and p = −s−1, then −p−1 ∈/ α, and hence p ∈ (−α). So (−α) is non-empty. If q ∈ α, then −q∈ / (−α), so (−α) 6= Q. (ii) Fix p ∈ (−α) and r > 0, so −p − r∈ / α. Such an example exists by (i) above as we showed (−α) nonempty. Consider (q < p) =⇒ (−q−r > −p−r) =⇒ −q−r∈ / α =⇒ q ∈ (−α). We have shown that q < p =⇒ q ∈ (−α), as required. r r (iii) Take any p ∈ (−α), r > 0 as above. Fix t := p + 2 . Then t > p, and −t − 2 = −p − r∈ / α, so t ∈ (−α). But r t = p + 2 > p, and this holds true for all p ∈ (−α), so (−α) has no maximal element.

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Thus we conclude (−α) ∈ R; that is, it is a Dedekind cut. To see α + (−α) ⊂ 0∗, consider that (r ∈ α, s ∈ (−α) =⇒ (−s∈ / α) =⇒ (r < −s) =⇒ (r + s < 0).

∗ ∗ −v To see inclusion in the other direction (α+β ⊃ 0 ), let v ∈ 0 and set w := 2 . Then w > 0, and by the Archimedean property of Q, there is an integer n with nw ∈ α but with (n + 1)w∈ / α. Deﬁne p := −(n + 2)w. Then surely, (−p − w∈ / α) =⇒ p ∈ (−α). Moreover, v = nw + p ∈ α + (−α). Hence α + (−α) ⊃ 0∗. Because 0∗ ⊂ α + (−α) ⊂ 0∗, we conclude that α + (−α) = 0∗, which validates our proposition.

Problem 7.3. For each sequence below, determine whether it converges and, if it converges, give its limit. No proofs are required. 1/n (f) sn = 2 (−1)n (i) n 1 2 (r) 1 + n 3n (q) n! Solution. If we provide the limit, it follows by deﬁnition that the sequence converges. (f) lim 21/n = 1 (−1)n (i) lim n = 1 1 2 2 1 (r) Does not converge as 1/n is known not to converge: 1 + n = 1 + n + n2 n n n 3 √ 3 3 (q) n! ≈ n n , so we deduce lim n! = 0. 2πn( e )

Problem 7.4. Give examples of (a) A sequence (xn) of irrational numbers having a limit lim xn that is a rational number. (b) A sequence (rn) of rational numbers having a limit lim rn that is an irrational number.

Solution. The problem does not ask for proof√ or veriﬁcation, so we simply state our examples and their limits. n (a) Very simply, we see that the sequence xn := n2 , surely a sequence of rational numbers, has a limit of (converges to) 0.

(b) A classical example of this is the ratio between ﬁbonacci numbers, where F := 0,F := 1,F := F + F . √ 0 1 n n−1 n−2 Fn+1 1+ 5 Let us deﬁne rn := for n > 0. It is a known fact that lim rn = , the golden ratio. Surely, it is a sequence Fn 2 of rational numbers, as it is the ratio between natural numbers; and we know rn ∈/ Q.

1 Problem 8.2. (e) Determine the limit of the sequence sn = n sin n and prove your claim.

Solution. Suppose we do not know that lim anbn = lim an lim bn and that we are allergic to this fact. 1 2 Deﬁne sn := n sin n. Let > 0 and N := . We will show s := lim sn = 0. Consider that (n > N) by deﬁnition of N 2 gives n > which results in: 2 1 1 n > = · 2 > · | sin n| (2 > 1 ≥ | sin n|, ∀n) 1 =⇒ > · | sin n| ( > 0 =⇒ n > N > 0) n

| sin n| 1 =⇒ > = sin n − 0 . |n| n

2 1 1 As we have shown (n > ) =⇒ n2 sin n − 0 < , we thus have lim n sin n = 0, as desired.

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Problem 8.5. 1. Consider three sequences (an), (bn) and (sn) such that an ≤ sn ≤ bn for all n and lim an = lim bn = s. Prove lim sn = s. This is called the “squeeze lemma.” 2. Suppose (sn) and (tn) are sequences such that |sn| ≤ tn for all n and lim tn = 0. Prove lim sn = 0.

Solution. (1.) Fix > 0. To show lim sn = s, it suﬃces to show for our given that for some n,

s − < sn < s + =⇒ |sn − s| < at which point we have that sn converges to s as desired. Because we are given lim an = lim bn = s, by deﬁnition of limit and convergence, we have some Na with (n > Na) =⇒ (s − < an) and likewise, we have some Nb with (n > Nb) =⇒ (bn < s + ), where we drop the absolute values because our hypothesis ensures ∀n that an ≤ sn ≤ bn. Because we have a ﬁnite amount (2) of conditions, we can simply assert

(n > max{Na,Nb}) =⇒ (s − < an ≤ sn ≤ bn < s + ) , where the middle inequality is given in the hypothesis, and the strict inequalities at the sides are from the fact sa, sb converge to s. Subtracting sn across the inequality, we get − < sn − s < + which is equivalent to 0 ≥ |sn − s| < , precisely the deﬁnition that gives limn→∞ sn = lim sn = s, and we are done.

Solution. (2.) Given (sn), (tn) sequences with |sn| ≤ tn is equivalent to −tn ≤ sn ≤ tn for all n. Given lim tn = 0, it easily follows that lim −tn = 0, and the squeeze lemma (above) asserts that lim sn = 0 (for free). √ 2 1 Problem 8.8. (b) Prove that lim[ n + n − n] = 2 . Solution. First we notice: √ √ p ( n2 + n − n)( n2 + n + n) (n2 + n) − n2 n n2 + n − n = √ = √ = √ . n2 + n + n n2 + n + n n2 + n + n 1 Fix > 0 and let N := 1+2 2 . Then n > N implies: ( 1−2 ) −1   1 1 + 22 1 1 + 22 1   n > 2 =⇒ − 1 > =⇒ > 1 +  1+2  1 − 2 n 1 − 2 n 1−2 − 1 2 − (1 − 2) r 1 =⇒ > 1 + 1 − 2 n r 2 r 1 n2 n n =⇒ > 1 + + 1 = + + 1 − 2 n n2 n2 n √ 2 n2 + n + n =⇒ > 1 − 2 n 1 − 2 1 n =⇒ = − < √ 2 2 n2 + n + n p 1 =⇒ − < n2 + n − 2

p 1 =⇒ ( n2 + n − n) − < , 2 √ 2 1 as required to show lim n + n − n = 2 .

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Problem 8.9. Let (sn) be a sequence that converges. 1. Show that if a ≤ sn for all but finitely many n, then a ≤ lim sn. 2. Show that if sn ≤ b for all but finitely many n, then lim sn ≤ b. 3. Conclude that if all but finitely many sn belong to [a, b], then lim sn belongs to [a, b].

Solution. (1.) We want to show that (a ≤ sn) for all but finitely many n implies a ≤ lim sn. Because there is only a finite number of n where a > sn, it is easy to see there exists N1 ∈ N with (n > N1) =⇒ (sn ≥ a). Because sn converges, let s := lim sn. Suppose for contradiction that s < a. Then we can fix := a − s > 0 and choose some N ≥ N1. Then because lim sn = s, we have:

(n > N) =⇒ |sn − s| < = a − s

⇐⇒ |sn − s| < (a − s)

=⇒ −(a − s) < (sn − s) < (a − s)

Where the right inequality sn − s < a − s implies sn < a for all n > N, which contradicts our hypothesis that a ≤ sn for all but ﬁnitely many n. Then it must be so that our supposition was wrong that s < a, and instead, we have s ≥ a as required.

(2.) Now we want to show that if sn ≤ b for all but finitely many n, then lim sn ≤ b. We proceed analogously as in part (1) above. Because sn converges, let s := lim sn. Because there is only a finite number of n where sn > b, it is easy to see there exists N2 ∈ N with (n > N2) =⇒ (sn ≤ b). Suppose for contradiction that s > b. Then we can fix := s − b > 0 and choose some N ≥ N2. Then because lim sn = s, we have:

(n > N) =⇒ |sn − s| < = s − b

⇐⇒ |sn − s| < s − b

⇐⇒ −(s − b) = b − s < (sn − s) < s − b

=⇒ b < sn, hence for all n > N, we have b < sn, but this is a contradiction to the given statement that the set of such n is finite. Hence our supposition s < b is false, and we must have s ≥ b, as desired. (3.) Suppose that all but a finite number of sn belong to [a, b]. Let U ⊂ {n} be this finite subset of n for which sn ∈/ [a, b]. Surely, U satisfies the hypotheses of (1) and (2); that is, because only a finite many n have the property: (a > sn) ∪ (sn < b), parts (1) and (2) above assert that a ≤ lim sn ≤ b. Hence

lim sn ∈ [a, b], which was to be shown.

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3 3 an+4an a +4a Problem 9.3. Suppose lim an = a and lim bn = b, and sn = 2 . Prove lim sn = 2 carefully, using bn+1 b +1 the limit theorems.

Solution. We have a := lim an, b := lim bn. Consider:

3 an + 4an lim sn = lim 2 bn + 1 3 lim an + 4an = 2 (Ross Thm 9.6) lim [bn + 1] 3 lim an + lim [4an] = 2 (Ross Thm 9.3) lim (bn) + lim (1) 3 (lim an) + lim (4) · lim (an) = 2 (Ross Thm 9.4, exponentiation as iterated mult.) (lim bn) + lim (1) a3 + (lim 4)a = (substituting known quantities a, b.) b3 + (lim 1) a3 + 4a = , (lim 4 = 4, lim 1 = 1) b2 + 1 precisely as required.

Problem 9.9. Suppose there exists N0 such that sn ≤ tn for all n > N0.

1. Prove that if lim sn = ∞ then lim tn = ∞.

Solution. Fix some M > 0. Because lim sn = +∞, we have the existence of some N ≥ N0 with sn > M, ∀n > N. Then for all n > N ≥ N0, we have M < sn ≤ tn, which shows lim tn = +∞.

2. Prove that if lim tn = −∞ then lim sn = −∞.

Solution. Like in (a), ﬁx some M < 0. Because lim tn = −∞, we have some N ≥ N0 with tn < M, ∀n > N. Then for all n > N ≥ N0, we have sn ≤ tn < M, which shows lim sn = −∞.

3. Prove that if lim sn and lim tn exist, then lim sn ≤ lim tn.

Solution. We are given the existence of some N0 past which, for all n > N0, we have sn ≤ tn. If lim tn = −∞, we have proven the desired result in (2). If lim sn = ∞, we have proven this in (1). Surely, we cannot have tn converging but sn → +∞, as this would contradict sn ≤ tn in the hypothesis. Correspondingly, sn cannot converge if tn → −∞, as seen by the symmetric argument. Because we are given s, t exist as deﬁned above, it only remains to prove lim sn ≤ lim tn where lim s, lim t 6= ±∞.

We have shown in 8.9 part (1) (letting a := 0) that tn ≥ sn =⇒ (tn − sn) ≥ 0 for all n > N0 (with only ﬁnite points n ≤ N0 otherwise) implies lim(tn − sn) ≥ 0. And by the limit theorems, this implies lim sn ≤ lim tn ≥, as desired.

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sn+1 Problem 9.12. Assume all sn 6= 0 and that the limit L = lim exists. sn 1. Show that if L < 1, then lim sn = 0. Hint: Select a so that L < a < 1 and obtain N so that |sn+1| < a|sn| n−N for n ≥ N. Then show |sn| < a |sN | for n > N. 1 2. Show that if L > 1, then lim |sn| = ∞. Hint: Apply (a) to the sequence tn = ; see Theorem 9.10. |sn|

Solution. (1) Notice that because we do not know if |sn|, |sn+1| themselves converge, we cannot apply the usual limit theorem. We proceed as suggested by the hint. Suppose L := lim sn+1 < 1. Because L is a limit of absolute sn values and all sn 6= 0, we have 0 < L.

Fix a with 0 < L < a < 1, and let :=(a − L) > 0. Then because L exists, there exists some N for which

|sn+1| (n ≥ N) =⇒ − L < = (a − L) |sn| |s |s | ⇐⇒ −(a − L) < n+1 − L < a − L ⇐⇒ 2L − a < n+1 < a sn |sn|

⇐⇒ (2L − a)|sn| < |sn+1| < a|sn|, where the right inequality holds for all n ≥ N.

We claim n−N |sn| < a |sN | for all n > N and show this formally by induction (although redundant because convergence guarantees all n > N). i Consider the subset U ⊂ N of indicies i for which we have |sN+i| < a |sN |. We have shown |sN+1| < a|sN | above, k so 1 ∈ U. Assume k ∈ U so that |sN+k| < a |sN |. Recall that 0 < L < a < 1 and that convergence to L guarantees that

(n > N) =⇒ |sn+1| < a|sN |.

So letting n := N + k + 1 > N, we immediately have |sN+(k+1)| < a|sN+k|. Using consequences of ordered ﬁeld axioms and k ∈ U, we have k (|sN+k+1|)(|sN+k|) < (a|sN+k|) a |sN | .

Because sn > 0 for all n, we can divide across the inequality to get

k |sN+k+1| < a |sN |, and we have k + 1 ∈ U. Because we showed k ∈ U =⇒ k + 1 ∈ U, we have U = N by the induction axiom. Hence n−N we have |sn| < a |sN | as desired, for all n > N. Then from the above, we have 0 < |sn| < an−N , for all n > N. Consider that |sN | lim an lim an−N = , lim aN as the denominator is constant and the numerator is known to converge to 0 for 0 < a < 1. Then by the limit theorems, we have |s | lim 0 = 0 ≤ lim n ≤ lim an−N = 0. |sN | By the squeeze lemma we have proven, this asserts that lim |sn| = 0. Because the denominator is constant as N is |sN | ﬁxed, we conclude that lim |sn| = 0 =⇒ lim sn = 0, which was to be shown.

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Solution. (2) Now we show that if L > 1, then lim |sn| = ∞. Per custom, we follow the provided hint and let 1 tn := . Let R := 1/L. Because L > 1, then 0 < R < 1. That is, |sn|

1 |sn| |tn+1| 0 < = lim = lim < 1. lim sn+1 |sn+1| |tn| sn

|tn+1| By our deﬁnitions, lim |t | = R. Fix some 0 < a < R < 1. Using R instead of L and tn instead of sn, part n (i) above gives us that there exists some N with (n > N) =⇒ 0 < |sn| = |tN | < an−N . Because these are all |sN | |tn| positive, by consequences of axioms of an ordered ﬁeld, we have the following for the reciprocals:

|t | ∀n > N, n > aN−n |tN | N |tn| a > n |tN | a

N |tn| a 1 Because for n > N, is greater than n , and it is known that for all n > N, 0 < a < 1 =⇒ n > 1, then we |tN | a a conclude aN 1 lim = aN lim = aN · (+∞) = +∞, an an which was to be shown.

an Problem 9.15. Show limn→∞ n! = 0 for all a ∈ R.

an Solution. Using our result in 9.12 above, we can see that sn := n! 6= 0 for all n. Moreover, we have that

n+1 sn+1 a n! a L := lim = lim · n = lim = 0 < 1 =⇒ lim sn = 0, sn (n + 1)! a n + 1 by the limit theorems, precisely as required.