Problem Solving Answers to Problem Set 8

11 July 2012

1. Let C be a nonempty closed bounded subset of the real line and f : C → C be a non-decreasing continuous function. Show that there exists a point p ∈ C such that f(p) = p. (A set is closed if its complement is a union of open intervals. A function g is non-decreasing if g(x) ≤ g(y) for all x ≤ y.) Answer: Let

a = inf C, b = sup C,

and let

[a, b] \ C = ⋃(c_{i}, c_{i+1})

where the disjoint open intervals are arranged in order of decreasing size. Then we can extend the function f : C → C to a continuous non-decreasing function f : [a, b] → [a, b] by defining f(x) linearly on (a_{i}, b_{i}) from f(a_{i}) to f(b_{i}) (the end-values being already defined since a_{i}, b_{i} ∈ C). We know that the extended function has a fixed point. This is trivial unless f(a) > a and f(b) < b. Otherwise let

c = inf{x ∈ [a, b] : f(x) ≤ x}.

It is clear by continuity that f(c) ≤ c. Also if f(c) < c then again by continuity we can find d < c such that f(d) < d. It follows that

f(c) = c.

If c ∈ C then all is done. Suppose

c ∈ (a_{i}, b_{i}).

Since a_{i} < c it follows from the definition of c that

f(a_{i}) > a_{i}.

(For c was the smallest point such that f(c) ≤ c.) But

f(a_{i}) ≤ c,

since f is non-decreasing. Since f(a_{i}) ∈ C, it follows that

f(a_{i}) ≤ a_{i}.

This is a contradiction. So c must be in C, and the result is proved.

2. Modified Let P (x) be a polynomial with integer coefficients

of the form

P (x) = x^{n} + a_{n−1}x^{n−1} + ⋯ + a_{2}x^{2} + a_{1}x + a_{0}. Show that if the degree n is odd, the constant term a_{0} is odd, and P (x) has an odd number of odd coefficients, then P (x) has at least one irrational real root. Answer: Apologies for the absurdly incorrect problem originally set here. Since n is odd, there are an odd number of real roots (including any multiplicities). By Gauss’ Lemma, any rational roots are in fact integers. Since a_{0} is odd, no even integer can be a root. Since the number of odd coefficients is odd, no odd integer can be a root. Hence there must be an irrational real root.