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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 = fr1 + r2 : r1 2 α and r2 2 βg. 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 = fr1 2 Q j r1 < s1g and β = fr2 2 Q j r2 < s2g. To show (α + β) :=fr1 + r2 j r1 2 α; r2 2 βg 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 2 Q but with ξ1 2= α and ξ2 2= β . By our definitions, (ξ1 + ξ2) 2= (α + β), but (ξ1 + ξ2) 2 Q, hence (α + β) 6= Q. To see (α + β) 6= fg, consider that α; β are nonempty hence r1; r2 exist, and thus r1 + r2 2 (α + β) exists. To see this inclusion explicitly, see (ii). (ii) Let r1 2 α; r2 2 β. Then we have for all s1; s2 2 Q, that (s1 < r1) =) s1 2 α and s2 < r2 =) s2 2 β. Adding these inequalities, we have [(s1 + s2) < (r1 + r2)] =) (s1 + s2) 2 (α + β) per our definition of (α + β). Because s1 2 Q; s2 2 Q =) (s1 + s2) 2 Q, we satisfy requirement (ii). (iii) We already have that α; β each contain no largest rational. In other words, 8r1 2 α and 8r2 2 β, we have the existence of some c1; c2 with r1 < c1 and r2 < c2. Because these all exist, c1 + c2 2 Q. Adding the inequalities, for all (r1 + r2) 2 (α + β), we must have (r1 + r2) < c1 + c2, and hence there exists some (c1 + c2) 2 Q larger than (r1 + r2) 2 (α + β) for all choices r1 2 α; r2 2 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 :=fr0 2 Q j r0 < 0g, or in other words the set of all negative rationals, where 0∗ corresponds to the real number 0. ∗ Let r1 2 α and r0 2 0 . From above, we have r2 < 0. Hence r1 + r0 < r1, and by our definition of α,(r1 + r0) 2 α. Thus from above, we have α + 0∗ ⊆ α. ∗ Let r1; r2 2 α with r1 < r2. Then r1 − r2 < 0 =) (r1 − r2) 2 0 . Then add (using the result of Problem 6.2 ∗ ∗ above) r2 2 α and (r1 − r2) 2 0 to get: r2 + (r1 − r2) = r1 2 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) := f p j 9r>0 [−p − r2 = α] g : 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) 2 R as a Dedekind cut and then α + (−α) = 0∗. (i) If s2 = α and p = −s−1, then −p−1 2= α, and hence p 2 (−α). So (−α) is non-empty. If q 2 α; then −q2 = (−α), so (−α) 6= Q. (ii) Fix p 2 (−α) and r > 0, so −p − r2 = α. Such an example exists by (i) above as we showed (−α) nonempty. Consider (q < p) =) (−q−r > −p−r) =) −q−r2 = α =) q 2 (−α). We have shown that q < p =) q 2 (−α), as required. r r (iii) Take any p 2 (−α); r > 0 as above. Fix t := p + 2 . Then t > p, and −t − 2 = −p − r2 = α, so t 2 (−α). But r t = p + 2 > p, and this holds true for all p 2 (−α), so (−α) has no maximal element. 1 Math 104, PSET #2 Daniel Suryakusuma, 24756460 Thus we conclude (−α) 2 R; that is, it is a Dedekind cut. To see α + (−α) ⊂ 0∗, consider that (r 2 α; s 2 (−α) =) (−s2 = α) =) (r < −s) =) (r + s < 0): ∗ ∗ −v To see inclusion in the other direction (α+β ⊃ 0 ), let v 2 0 and set w := 2 . Then w > 0, and by the Archimedean property of Q, there is an integer n with nw 2 α but with (n + 1)w2 = α. Define p := −(n + 2)w. Then surely, (−p − w2 = α) =) p 2 (−α). Moreover, v = nw + p 2 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 p 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 proofp 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 : p 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 2= 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 > · j sin nj (2 > 1 ≥ j sin nj; 8n) 1 =) > · j sin nj ( > 0 =) n > N > 0) n j sin nj 1 =) > = sin n − 0 : jnj n 2 1 1 As we have shown (n > ) =) n2 sin n − 0 < , we thus have lim n sin n = 0, as desired. 2 Math 104, PSET #2 Daniel Suryakusuma, 24756460 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 jsnj ≤ 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 + =) jsn − sj < 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 8n that an ≤ sn ≤ bn. Because we have a finite amount (2) of conditions, we can simply assert (n > maxfNa;Nbg) =) (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 ≥ jsn − sj < , precisely the definition that gives limn!1 sn = lim sn = s, and we are done. Solution. (2.) Given (sn); (tn) sequences with jsnj ≤ 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). p 2 1 Problem 8.8. (b) Prove that lim[ n + n − n] = 2 . Solution. First we notice: p p p ( n2 + n − n)( n2 + n + n) (n2 + n) − n2 n n2 + n − n = p = p = p : n2 + n + n n2 + n + n n2 + n + n 1 Fix > 0 and let N := 1+2 2 .
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