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The University of Sydney School of Mathematics and Statistics

Solutions to Week 2 Tutorial MATH2023: Analysis

Semester 2, 2020

Web Page: http://www.maths.usyd.edu.au/u/UG/IM/MATH2023/ Lecturer: Milena Radnovi´c

1. We can introduce multiplication in the set of natural numbers using the addition and the successor operations. • n · 1 := n;

• n · S(m) := n · m + n. Using Peano axioms, prove that 1 · n = n. Solution: Take A = {n ∈ N | 1 · n = m}. We will apply the induction axiom to prove A = N. First, 1 · 1 = 1, by the definition of multiplication, so 1 ∈ A. Suppose now that n ∈ A, that is 1 · n = n. We have:

1 · S(n) = 1 · n + 1 = n + 1 = S(n), thus S(n) ∈ A. 2. Show that √

1 is an irrational number. 3−5

√ 1 1 is rational, then + 5 = 3 is also a rational number. Solution: If a = √ a 3−5 √ Take 3 = m/n, where m and n are natural numbers, gcd(m, n) = 1. Then m2 = 3n2 , thus m is divisible by 3, i.e. m = 3m1 . That implies 3m12 = n2 , so n is also divisible by 3, which contradicts the condition gcd(m, n) = 1. 3. For each of the following sets, determine the sets of their upper and lower bounds in R. Then find maximum, minimum, supremum and infimum, if they exist in the set of real numbers. (a) (0, 1); (b) Q – the set of rational numbers; (c) ∅ – the empty set;

(d) A = {x ∈ Q | x2 ≤ 3};

(e) B = {x ∈ R | x2 ≤ 3};   1 (f) C = 1 − | n ∈ N ; n Copyright © 2020 The University of Sydney

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(g) D =



 1 |n∈N . n2

Solution: (a) The sets of upper and lower bounds are [1, +∞) and (−∞, 0]. Minimum and maximum of (0, 1) do not exist, sup(0, 1) = 1, inf(0, 1) = 0. (b) The sets of upper and lower bounds are empty. Minimum and maximum of Q do not exist, sup Q = +∞, inf Q = −∞.

(c) The sets of upper and lower bounds of the empty sets are both equal to the whole set of real numbers! Minimum and maximum of the empty set do not exist, sup ∅ = −∞, inf ∅ = +∞. √ (d) The√set of upper bounds of A is [ 3, +∞), and the set of lower bounds √ (−∞, − 3]. Minimum √ and maximum of A do not exist, the supremum is 3, and infimum equals − 3. √ √ (e) The sets are as in (d). B = [− 3, 3], so min B = √ of upper and lower bounds √ inf B = − 3, max B = sup B = 3. (f) The sets of upper and lower bounds are [1, +∞) and (−∞, 0]. The supremum of C is 1, and maximum does not exist since 1 6∈ C. inf C = min C = 0.

(g) The sets of upper and lower bounds are as in (f). sup D = max D = 1. The infimum is 0, and minimum does not exist.

4. (Cantor’s Intersection Theorem) Let In be closed bounded intervals, such that I1 ⊃ I2 ⊃ I3 ⊃ . . . . Prove that the intersection of all the closed intervals I1 , I2 , I3 , . . . is non-empty. You can build your proof along the following steps. Denote In = [an , bn ], A = {a1 , a2 , . . . } and B = {b1 , b2 , . . . }. Show that: (a) an ≤ bm for any m and n.

(b) Any element of the set B is an upper bound of A. (c) A has supremum. (d) sup A ≤ bn for any n.

(e) sup A ∈ In for each n.

(f) The segments I1 , I2 , . . . have non-empty intersection.

In addition, prove the following: (g) B has infimum and inf B ∈ In for each n. (h) sup A ≤ inf B .

(i) The intesection of all segments I1 , I2 , I3 . . . is the segment [sup A, inf B].

Solution: (a) We have an ≤ bn , since they are the left and the right endpoint of segment In . If m < n, from Im ⊃ In , we have am ≤ an and bn ≤ bm , i.e. a m ≤ a n ≤ bn ≤ bm . (b) Follows from (a).

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(c) In (b), we have shown that A bounded from above, thus its supremum exists by the Completeness Axiom. (d) By definition, the supremum is the least upper bound, so the statement follows from (b). (e) From (d), sup A ≤ bn . Since supremum is an upper bound of the set, an ≤ sup A, so sup A ∈ In .

(f) According to (e), sup A ∈ I1 ∩ I2 ∩ I3 ∩ . . . .

(g) Analogously to steps (a)-(e).

(h) From (d), sup A is a lower bound of B, so it is smaller that the infimum of B . (i) Similarly as in (e), we can show that sup B ∈ In for each n, thus [sup A, inf B] ⊂ In . If α < sup A, then there is an element am ∈ A, such that α < am , so α 6∈ Im . It follows that no number smaller than sup A belongs to the intersection of all segments, and the same holds for any number β > inf B. Thus [sup A, inf B] = In . 5. (a) If in the Cantor’s Intersection Theorem (Question 4) we omit the assumption that the intervals are bounded, will their intersection still always be nonempty? (b) What if we omit the assumption that the intervals are closed? Solution: (a) No, for example when  In= [n, +∞). 1 . (b) No, for example when In = 0, n

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