Section 8*: Problem 6 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

(a) Show that there is no function $h:\mathbb{Z}_{+}\rightarrow\mathbb{R}_{+}$ satisfying the formula $\begin{aligned} & h(1)=3,\\ & h(i)=[h(i-1)-1]^{1/2} & & \mbox{for }i>1. \end{aligned}$ Explain why this example does not violate the principle of recursive definition.

(b) Consider the recursion formula $\begin{aligned} & h(1)=3,\\ & h(i)=\left\{ \begin{aligned} & [h(i-1)-1]^{1/2} & & \mbox{if }h(i-1)>1\\ & 5 & & \mbox{if }h(i-1)\le1 \end{aligned} \right\} & & \mbox{for }i>1. \end{aligned}$ Show that there exists a unique function $h:\mathbb{Z}_{+}\rightarrow\mathbb{R}_{+}$ satisfying this formula.

(a) $h(2)=\sqrt{2}<2$ , $h(3)<\sqrt{2-1}=1$ , so that $h(3)-1<0$ , and $h(4)$ is not well defined. It does not violate the principle, because there is no well defined $\rho$ function. If we try to define $\rho$ similar to Exercise 5, i.e. $\rho(f:\{1,...,n\}\rightarrow\mathbb{R}_{+})=\sqrt{f(n)-1}$ , then the definition does not work for $f$ such that $f(n)<1$ .

(b) Here there is a well-defined $\rho$ , namely, $\rho(f:\{1,...,n\}\rightarrow\mathbb{R}_{+})=\begin{cases} \sqrt{f(n)-1} & \mbox{if }f(n)>1\\ 5 & \mbox{if }f(n)\le1 \end{cases}$ , which together with $a_{0}=3$ allows us to apply Theorem 8.4.

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Section 8*: Problem 6 Solution

Section 8*: Problem 6 Solution