MATH1050 Greatest/least element, upper/lower bound

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1 MATH1050 Greatest/ element, upper/lower bound 1 Definition Let S be a subset of R x λ (a) Let λ S λ is said to be a element of S if, for any x S, x λ (b) S is said to have a element if there exists some λ S such that for any x S, x λ x λ Remarks (A) Suppose S has a element Then it is the unique element of S, in the sense below: Suppose λ,λ are elements of S Then λ = λ We may replace the article a in a element by the article the max(s) (B) Notation We denote the element of S by min(s) (C) Terminology We may choose to write S has a element as max(s) exists Et cetera The situation is analogous for element 2 Definition Let S be a subset of R upper x β (a) Let β R β is said to be a(n) bound of S in R if, for any x S, lower x β above (b) S is said to be bounded in R if there exists some β R such that for any x S, below (c) S is said to be bounded in R if S is bounded above in R and bounded below in R Remarks x β x β (A) If β is an upper bound of S, then every number greater than β is also an upper bound of S Therefore S has infinitely many upper bounds It does not make sense to write the upper bound of S is so-and-so The situation is similar for being bounded below (B) Suppose λ is a element of S Then λ is an upper bound of S (How about its converse?) The situation is similar for element and lower bound 3 Example (A) (Well-ordering Principle for Integers) Recall this statement below, known as the Well-ordering Principle for Integers (WOPI): Let S be a non-empty subset of N S has a element There are various re-formulations of the statement (WOPI): (WOPIL) Let T be a non-empty subset of Z Suppose T is bounded below in R by some β Z Then T has a element (WOPIG) Let U be a non-empty subset of Z Suppose U is bounded above in R by some γ Z Then U has a element (The proof for the logical equivalence of (WOPI), (WOPIL), (WOPIG) is left as an exercise) From now on all three of them are referred to as the Well-ordering Principle for Integers 1

2 4 Example (B) S element? element? bounded below in R? bounded above in R? [0,1) 0 Aa nil Ab Yes (by 0) Ac Yes (by 1) Ad [0,+ ) 0 Ba nil Bb Yes (by 0) Bc No Bd (0,+ ) nil Ca nil Cb Yes (by 0) Cc No Cd [0,1) Q 0 Da nil Db Yes (by 0) Dc Yes (by 1) Dd [0,+ ) Q 0 Ea nil Eb Yes (by 0) Ec No Ed [0,1)\Q nil Fa nil Fb Yes (by 0) Fc Yes (by 1) Fd [0,+ )\Q nil Ga nil Gb Yes (by 0) Gc No Gd The detail for the argument here is relatively easy to work out, because each of the sets concerned are constructed using just one interval whose endpoints are rational numbers Where the set concerned has a / element, or is bounded above/below in R, we may just name an appropriate number which serves as a / element of the set or an upper/lower bound of the set in R, and verify that this number satisfies the condition specified in the relevant definition (Aa): [0, 1) has a element, namely, 0 Write S = [0,1) Note that 0 S Pick any x S By definition, 0 x < 1 In particular x 0 It follows that 0 is the element of S Remark The argument for each of (Ba), (Da), (Ea) is similar (Ad): [0,1) is bounded above in R by 1 Write S = [0,1) Pick any x S By definition, 0 x < 1 In particular x < 1 (So x 1 also holds) It follows that S is bounded above by 1 Remark The argument for each of (Ac)-(Gc), (Dd), (F d) is similar Here we focus on arguments which explain why a certain set fails to have a / element or to be bounded above/below in R: (Ab): [0, 1) has no element Write S = [0,1) Suppose S had a element, say, λ (By definition, for any x S, x λ) Then λ S Therefore 0 λ < 1 Define x 0 = λ+1 2 Then 0 λ < x 0 < 1 Note that x 0 S But λ < x 0 So λ would not be a element of S Contradiction arises Hence S has no element in the first place Remark The argument for each of (Ca), (Fa), (Ga), (Db), (Fb) is similar (Bb): [0, + ) has no element Write S = [0,+ ) Suppose S had a element, say, λ (By definition, for any x S, x λ) Then λ S λ 0 Define x 0 = λ+1 Then 0 λ < x 0 2

3 Note that x 0 S But λ < x 0 So λ would not be a element of S Contradiction arises Hence S has no element in the first place Remark The argument for each of (Cb), (Eb), (Gb) is similar (Bd): [0,+ ) is not bounded above in R Write S = [0,+ ) Suppose S were bounded above in R, say, by some β R (By definition, for any x S, x β) Then, since 0 S, we have β 0 Define x 0 = β +1 Then 0 β < x 0 Note that x 0 S But β < x 0 So β would not be an upper bound of S in R Contradiction arises Hence S is not bounded above in R in the first place Remark The argument for each of (Cd), (Ed), (Gd) is similar 5 Definition Let a n n=0 be an infinite sequence in R upper bound (a) Let κ R a n n=0 is said to be of a lower bound n an κ n=0 in R if, for any n N, a n κ bounded above (b) a n n=0 issaidtobe in Rifthereexistssomeκ Rsuchthatforanyn N, bounded below Lemma (1) an κ a n κ Let a n n=0 be an infinite sequence in R Define T(a n n=0) = x R : x = a n for some n N (T(a n n=0) is the set of all terms of a n n=0) The statements below hold: (a) a n n=0 is bounded above in R by β iff T(a n n=0) is bounded above in R by β (b) a n n=0 is bounded below in R by β iff T(a n n=0) is bounded below in R by β Exercise (Word game) 6 Definition Let a n n=0 be an infinite sequence in R an a n+1 a n a n+1 increasing (a) a n n=0 is said to be if, for any n N, decreasing strictly increasing (b) a n n=0 is said to be if, for any n N, strictly decreasing Remarks on terminology (a) a n n=0 is said to be monotonic if a n n=0 is increasing or decreasing an < a n+1 a n > a n+1 (b) a n n=0 is said to be strictly monotonic if a n n=0 is strictly increasing or strictly decreasing Lemma (2) Let a n n=0 be an infinite sequence in R Define T(a n n=0) = x R : x = a n for some n N (T(a n n=0) is the set of all terms of a n n=0) The statements below hold: (a) Suppose a n n=0 is strictly increasing Then T(a n n=0) has no element (b) Suppose a n n=0 is strictly decreasing Then T(a n n=0) has no element Exercise (Word game) 3

4 7 Example (C) For any n N, define a n = (n+1)(n+4) (n+2)(n+3) Define T = x x = a n for some n N (a) a n n=0 is bounded above in R by 1 (Equivalently, T is bounded above in R by 1 ) Let n N We have a n = (n+1)(n+4) (n+2)(n+3) = n2 +5n+4 n 2 +5n+6 = 1 2 n = 1 +5n+6 Hence a n n=0 is bounded above by 1 (b) a n n=0 is strictly increasing Let n N Then a n+1 > a n (c) T has no element a n+1 a n = [(n+1)+1][(n+1)+4] [(n+1)+2][(n+1)+3] (n+1)(n+4) (n+2)(n+3) = (n+2)(n+5) (n+3)(n+4) (n+1)(n+4) (n+2)(n+3) = (n+2)2 (n+5) (n+1)(n+4) 2 (n+2)(n+3)(n+4) = 4 (n+2)(n+3)(n+4) > 0 Suppose T had a element, say, λ Then there would exist some n N such that λ = a n Take x 0 = a n+1 By definition, x 0 S We have x 0 = a n+1 > a n = λ Therefore λ would not be a element of T Contradiction arises It follows that T has no element in the first place (d) T has a element, namely 2 T is bounded below in R (Exercise) 3 4

5 8 Example (D) Let p be a positive prime number Define α = p Let b (α,+ ) Let a n n=0 be the infinite sequence defined recursively by Define T = x R : x = a n for some n N a 0 = b a n+1 = 1 2 (a n + α2 a n ) for any n N The infinite sequence a n n=0 will provide approximations a 0,a 1,a 2,a 3, as close to the irrational number α = p as we like, as the index in the a j s increases It is done as described below algorithmically: Consider the curve C : y = x 2 p on the coordinate plane (C intersects the x-axis at the point ( p,0)) Take M 0 = (a 0,0) = (b,0) For each j N, draw the line l j (whose equation is x = a j ) through M j perpendicular to the x-axis The intersection of l j with C is defined to be K j : its coordinates are given by K j = (a j,a j 2 p) Draw the tangent t j to the curve C at K j (The equation of t j is y = 2a j x (a j 2 +p)) The intersection of t j with the x-axis is defined to be M j+1 = (a j+1,0) y y a j 2 p K j a j 2 p K j 0 α a j x 0 α a j+1 a j x y = x 2 p y = x 2 p y = 2a jx (a j 2 + p) Here are the properties of the infinite sequence a n n=0 and the set T: (a) For any n N, a n > α Remark As a consequence, a n n=0 is bounded below in R by α, and T is bounded below in R by α Note that a 0 = b > α > 0 Suppose it were true that there existed some n N so that a n α Then there would be a smallest N N so that a N > α and a N+1 α (Why? Apply the Well-ordering Principle for Integers) We would have α a N+1 = 1 ) (a N + α2 = a N 2 +α 2 2 2a N a N Since a N > 0, we would have 2αa N a N 2 + α 2 Then (a N α) 2 = a N 2 2αa N + α 2 0 Therefore a N = α But a N > α Contradiction arises Hence, in the first place, we have a n > α for any n N (b) a n n=0 is strictly decreasing Let n N a n > α > 0 Then a n 2 α 2 > 0 also 5

6 Therefore a n+1 a n = 1 2 Hence a n+1 > a n (a n + α2 a n ) a n = 1 2 ( a n + α2 a n ) = a n 2 α 2 2a n < 0 (c) 0 < a n α < b α 2 n for any n N\0 Remark Heuristically speaking, the infinite sequence a n n=0 will descend to as close as α as we like, but it will never reach α Let n N Since a n > α, we have a n α > 0 We also have (d) T has no element a n α = 1 2 Suppose T had a element, say, λ (a n 1 + α2 a n 1 Then there would exist some n 0 N such that λ = a n0 ) α = 1 2 (a n 1 α)+ 1 ( ) α 2 α 2 a n 1 = 1 2 (a n 1 α) α 2 an 1 α a n 1 < 1 2 (a n 1 α) < 1 2 2(a n 2 α) < 1 2 n(a 0 α) = 1 2 n(b α) Now take x 0 = a n0+1 We would have x 0 T and x 0 = a n0+1 < a n0 = λ Contradiction arises Hence T has no element in the first place Further remarks (1) The idea and the calculation will still work even when we do not require p to be a positive prime number; we may allow p to be any positive real number that we like The infinite sequence a n n=0 will provide approximations which descends to α = p as close as we like, but never reaches α (2) How about finding cubic roots of positive real numbers? Suppose p is a positive real number and α = 3 p Suppose b (α,+ ) Define infinite sequence a n n=0 recursively by = b a 0 a n+1 = 1 3 (2a n + α3 a n ) for any n N This infinite sequence a n n=0 will provide approximations which descends to α = 3 p as close as we like, but never reaches α (First draw the picture and formulate the algorithm which are analogous to the ones for the original example on square roots This will give you a feeling on what this infinite sequence is doing Then try to formulate and prove some statements which are analogous to the ones that we have proved for the original example) (3) Can you generalize the idea to finding quartic roots of positive real numbers? Quintic roots? n-th roots? (4) The idea and method described here is a concrete example of the application of Newton s Method (for finding approximate solutions of equations) 6

7 9 Example (E) Let p be a positive prime number Let α = p Let A α = (,α) Q, B α = (α,+ ) Q The properties of A α,b α are stated below: (a) i B α is bounded below by α in R ii A α is bounded above by α in R (b) i A α has no element ii B α has no element The proofs of the statements in (a) are easy and are left as exercises Heuristically the statements in (b) look obviously true, but how to prove them? Below is an argument for the statement in (bii), together with the geometrical idea behind the argument: Suppose it were true that B α had a element, say, λ By definition λ α Since α is irrational, λ > α [We aim to construct a rational number strictly between α and λ, with help from geometry] Consider the curve C : y = x 2 p on the coordinate plane (C intersects the x-axis at the point ( p,0)) Take M = (λ,0) Draw the line l (whose equation is x = λ) through M perpendicular to the x-axis The intersection of l with C is defined to be K: its coordinates are given by K = (λ,λ 2 p) Draw the tangent t to the curve C at K (The equation of t is y = 2λx (λ 2 +p)) The intersection of t with the x-axis is defined to be M = (µ,0) Note that µ = 1 ( λ+ p ) Obtain the desired contradiction by verify that µ is a rational number satisfying 2 λ α < µ < λ y y λ 2 p K λ 2 p K 0 α λ x 0 α µ λ x y = x 2 p y = x 2 p y = 2λx (λ 2 + p) Remarks (1) Can you spot the commonality in the ideas for this example and the one on Newton s Method? (2) Can you generalize the idea and the argument above to establish analogous results in which p is replaced by r, where r is just any positive rational number but r is an irrational number? Or how about n r, where r is a rational number but n r is an irrational number? (3) The pair of sets (A α,b α ) is referred the Dedekind cut for the irrational number α (which is p here) (4) We may define the Dedekind cut for any arbitrary irrational real number in an analogous way: Let τ R Suppose τ is an irrational number Let A τ = (,τ) Q, B τ = (τ,+ ) Q The pair of sets (A τ,b τ ) is called the Dedekind cut for τ It turns out that A τ has no element and B τ has no element However, in such a general situation, the argument will have to rely on the Least-upper-bound Axiom 7

8 10 Appendix: Least-upper-bound Axiom Refer to each of the previous examples in which the (non-empty) subset of R, say, S, under consideration is bounded above in R Inspecting the set S carefully, we will conclude that although S may have no element at all, S seems to be guaranteed an upper bound in R which is of smallest value amongst all upper bounds in R Such a number is called a upper bound of S in R Definition Let S be a subset of R Let σ R upper bound σ is said to be a of S in R if both of the conditions (C1), (C2) below are satisfied: lower bound upper (C1) σ is a(n) bound of S in R lower (C2) σ is the element of the set of all upper lower bounds of S in R The previous examples that we have examples seem to provide evidence in support of the intuitively obvious fact below: Whenever S a non-empty subset of R which is bounded above in R, S will have a upper bound in R You can device as many examples as you like to provide further evidence There is, however, no way to establish this obvious fact without assuming something logically equivalent to it We formulate this intuitively obvious fact in precise mathematical language: Least-upper-bound Axiom for the reals (LUBA) Let A be a non-empty subset of R Suppose A is bounded above in R Then A has a upper bound in R The statement (LUBA) is logically equivalent to the equally obvioius statement: Greatest-lower-bound Axiom for the reals (GLBA) Let A be a non-empty subset of R Suppose A is bounded below in R Then A has a infimum in R (As an exercise in word games, prove that (LUBA), (GLBA) are indeed logically equivalent) In your mathematical analysis course, you will apply the Least-upper-bound Axiom to prove other intuitively obvious results which you have been using without questioning in calculus, such as the Bounded-Monotone Theorem for infinite sequences in R 8