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Math 7350 HW 3 Solutions: Ordered Fields, Cluster Points, and Piecewise Linear Functions, Assignments of Mathematics

University of Memphis (U of M)Mathematics

Solutions to homework assignment 3 for math 7350, a college-level mathematics course. The solutions cover topics such as ordered fields, cluster points, and piecewise linear functions. Proofs and examples to help students understand these concepts.

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Math 7350 Homework Assignment 3 Fall 2004

Due October 14.

1. Let R(x) be the set of rational functions f(x) = P(x)

Q(x)where Pand Qare polynomials

with real coefficients and Qis not identically zero. Define an order ≤by f≤gif

∃K > 0 : ∀x≥K:f(x)≤g(x).

(a) Show that with this order, R(x) is an ordered field. [You may assume it is a

field, so you just need to check that ≤is a total order which respects + and ×.]

(b) Show that R(the set of constant functions) has an upper bound, but no least

upper bound in R(x).

2. Show that the set of (real) cluster points of a sequence (xn) is a closed set.

3. (a) Show that there are at most 7 sets that can be obtained by applying the oper-

ations −and ◦iteratively to a set S(including Sitself). [Hint: Show that if F

is closed, F◦− ⊆F, and if Uis open, U−◦ ⊇U.]

(b) Give an example where all 7 sets in (a) are distinct.

4. A function f:R→Ris upper semi-continuous at a∈Rif ∀ε > 0 : ∃δ > 0 : ∀x:|x−

a|< δ ⇒f(x)≥f(a)−ε. Show that fis upper semi-continuous for all x∈Riff

f−1[(a, ∞)] is open for all a∈R.

5. A continuous function g: [a, b]→Ris called piecewise linear iff there exists a subdi-

vision a=x0< x1<· · · < xn=bsuch that gis linear on [xi, xi+1]. If f: [a, b]→R

is a continuous function and ε > 0, show that there is a piecewise linear function

g: [a, b]→Rsuch that |f(x)−g(x)|< ε for all x∈[a, b].

Discover Assignments of Mathematics University of Memphis (U of M)

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Math 7350 Homework Assignment 3 Fall 2004

Due October 14.

Let R(x) be the set of rational functions f (x) = P Q^ ((xx)) where P and Q are polynomials with real coefficients and Q is not identically zero. Define an order ≤ by f ≤ g if ∃K > 0 : ∀x ≥ K : f (x) ≤ g(x). (a) Show that with this order, R(x) is an ordered field. [You may assume it is a field, so you just need to check that ≤ is a total order which respects + and ×.] (b) Show that R (the set of constant functions) has an upper bound, but no least upper bound in R(x).
Show that the set of (real) cluster points of a sequence (xn) is a closed set.
(a) Show that there are at most 7 sets that can be obtained by applying the oper- ations −^ and ◦^ iteratively to a set S (including S itself). [Hint: Show that if F is closed, F ◦−^ ⊆ F , and if U is open, U −◦^ ⊇ U .] (b) Give an example where all 7 sets in (a) are distinct.
A function f : R → R is upper semi-continuous at a ∈ R if ∀ε > 0 : ∃δ > 0 : ∀x : |x − a| < δ ⇒ f (x) ≥ f (a) − ε. Show that f is upper semi-continuous for all x ∈ R iff f −^1 [(a, ∞)] is open for all a ∈ R.
A continuous function g : [a, b] → R is called piecewise linear iff there exists a subdi- vision a = x 0 < x 1 < · · · < xn = b such that g is linear on [xi, xi+1]. If f : [a, b] → R is a continuous function and ε > 0, show that there is a piecewise linear function g : [a, b] → R such that |f (x) − g(x)| < ε for all x ∈ [a, b].

Math 7350 Homework 3 Solutions Fall 2004

Let R(x) be the set of rational functions f (x) = P Q^ ((xx)) where P and Q are polynomials with real coefficients and Q is not identically zero. Define an order ≤ by f ≤ g if ∃K > 0 : ∀x ≥ K : f (x) ≤ g(x). (a) Show that with this order, R(x) is an ordered field. [You may assume it is a field, so you just need to check that ≤ is a total order which respects + and ×.] If f ≤ g and h ≤ k, then ∃K 1 : ∀x ≥ K 1 : f (x) ≤ g(x) and ∃K 2 : ∀x ≥ K 2 : h(x) ≤ k(x). Let K = max{K 1 , K 2 }, then if x ≥ K, f (x) + h(x) ≤ g(x) + k(x), so f + h ≤ g + k. Thus ≤ respects +. Similarly, if f ≤ g and 0 ≤ h, then ∃K 1 : ∀x ≥ K 1 : f (x) ≤ g(x) and ∃K 2 : ∀x ≥ K 2 : 0) ≤ h(x). Let K = max{K 1 , K 2 }, then if x ≥ K, f (x)h(x) ≤ g(x)h(x), so f h ≤ gh. Thus ≤ respects ×. Similarly, if f ≤ g ≤ h then ∃K 1 : ∀x ≥ K 1 : f (x) ≤ g(x) and ∃K 2 : ∀x ≥ K 2 : g(x) ≤ h(x). Let K = max{K 1 , K 2 }, then ∀x ≥ K : f (x) ≤ h(x) and so f ≤ h. Thus ≤ is transitive. If f 6 = g, then h = f − g 6 = 0. Write h = P Q^ ((xx)). Then P 6 = 0. Pick K larger than all the roots of P (x) and Q(x). Then the sign of P (x) and Q(x) does not change for all x > K. Hence either f − g ≤ 0 or 0 ≤ f − g, but not both. Hence either f ≤ g or g ≤ f , but not both. Thus f ≤ g and g ≤ f imply f = g, and for any f and g, either f ≤ g or g ≤ f. Thus ≤ is a total order. (b) Show that R (the set of constant functions) has an upper bound, but no least upper bound in R(x). f (x) = x is an upper bound, since for any constant function g(x) = c, we can pick K = c and then ∀x ≥ K : g(x) = c ≤ x = f (x). There is no least upper bound, since if f is an upper bound, then for all c, f ≥ c + 1, so f − 1 ≥ c. But then f − 1 is an upper bound and f − 1 < f.
Show that the set of (real) cluster points of a sequence (xn) is a closed set. Let c ∈ R be a point of closure of the set of cluster points of (xn). Then for all m > 0, there is a cluster point y of (xn) with y ∈ (c − (^) m^1 , c + (^) m^1 ). Then there are infinitely many n such that |xn − y| < (^) m^1. Inductively define nm to be the smallest n with n > nm− 1 and |xn − y| < (^) m^1. Then |xnm − c| < (^) m^2. But then c is a cluster point (limit point) of (xnm ), and so is a cluster point of (xn).
(a) Show that there are at most 7 sets that can be obtained by applying the oper- ations −^ and ◦^ iteratively to a set S (including S itself). [Hint: Show that if F is closed, F ◦−^ ⊆ F , and if U is open, U −◦^ ⊇ U .] If F is closed, then F is a closed set containing F ◦. But then F must contain the smallest such set, F ◦−. Similarly, if U is open, then U is an open subset of U −, so must be a subset of the largest such set U −◦.

Math 7350 HW 3 Solutions: Ordered Fields, Cluster Points, and Piecewise Linear Functions, Assignments of Mathematics

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Math 7350 Homework Assignment 3 Fall 2004

Math 7350 Homework 3 Solutions Fall 2004