MthT 430 Projects Chap 8b - Intermediate Value Property Remarks

1.

Let f be a continuous function on [0,1] such that

· f(0) > 0, · f(1) < 1.

Draw a graph of several (not too complicated) continuous functions, f, satisfying the two properties to decide whether there is always an x, 0 < x < 1, such that f(x) = x.

Now prove that there is an x, 0 < x < 1, such that f(x) = x.

See also Spivak, Chapter 7, Problem 11.

Remark: Apply the Intermediate Value Theorem to the function F(x) = f(x) - x.

2.

Suppose that

· f and g are continuous functions on [0,1] · f(0) > g(0), · f(1) < g(1).

Draw graphs of several pairs of continuous functions, f, g, satisfying the three properties to decide whether there is always an x, 0 < x < 1, such that f(x) = g(x).

Now prove that if f, g, are continuous there is an x, 0 < x < 1, such that f(x) = g(x).

Remark: Apply the Intermediate Value Theorem to the function F(x) = f(x) - g(x).

3.

Suppose we are working with a number system (such as the rational numbers Q) which satisfies (P1 - P12), but does not satisfy (P13-LUB); id est , there is a non empty set A of numbers, A is bounded above, but A does not have a least upper bound. For this A, let

BA º {b | b     is an upper bound for A.}

Define

f(x) = ì ï í ï î 1, x Î BA, -1, x Ï BA.

Show that f is continuous at all x, but does not satisfy the Intermediate Value Property (IVP).

Remark: To show that f is continuous everywhere, consider two cases:

·

b Î B_A: f(b) = 1. Since b is not the least upper bound, there is a d > 0 such that b - d Î B_A. Then f º 1 on (b -d, ¥).

·

b Ï B_A: Since b is not an upper bound for A, There is a d > 0, such that b + d Î A. Then f º -1 on (-¥, b + d).

Thus NOT (P13-LUB) implies NOT (P13-CFIVP). This shows that (P13-CFIVP) implies (P13-LUB) as stated in chap8b.tex.

See http://www.math.uic.edu/ lewis/mtht430/chap8b.htm#CFIVP

Remark: If A, B_A, are the sets described in the previous problem, assume that A is a subset of [0, 1/2). We can use a folding string argument to construct a nondecreasing sequence of finite binary expansions s_n = ._binc₁¼c_n such that, for every n, s_n is not an upper bound for A, but s_n + (1/2ⁿ) is an upper bound for A. To do this, at each step select the left interval [a_n,m_n) iff the midpoint m_n is an upper bound for A.

It follows that B_A is exactly the set of upper bounds for the sequence {s_n}. Now conclude that the lim_{n ®¥}s_n does not exist in the number system where B lives !

Thus NOT (P13-LUB) implies NOT (P13-BIN) and NOT (P13-BISHL). Either (P13-BIN) or (P13-BISHL) implies (P13-LUB).

There are other arguments to show the equivalence of (P13-LUB), (P13-CFIVP), (P13-BIN), and (P13-BISHL).

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On 06 Nov 2007, 21:47.