(a) See Exercise 13 of §3.

(b) $0$ is a lower bound, as $n>0$ implies $1/n>0$ (Exercise 2(i)), but for every $x>0$ there is $n>1/x>0$ ($\mathbb{Z}_{+}$ is unbounded), therefore, $1/n<x$ (Exercise 2(j)). That is, every positive number is not a lower bound for the set, and $0$ is the greatest lower bound.

(c) Using hint, if $0<a<1$ , then $1/a>1$ (Exercise 2(j)), and $h=(1-a)/a$ $=1/a-1>0$ , hence, if we show that for a positive $h$ , $(1+h)^{n}\ge1+nh$ , we can use this fact to argue that for every $x>0$ there exists $n$ such that $(1+h)^{n}=(1/a)^{n}=1/a^{n}>1/x$ (all we need is to find $n$ such that $1+nh>1/x$ , or $n>(1/x-1)/h$ , which always exists as $\mathbb{Z}_{+}$ is unbounded), and, therefore, $a^{n}<x$ , implying there is no positive lower bound. To show that $(1+h)^{n}\ge1+nh$ for all positive integer $n$ we can use the induction principle again: $(1+h)^{1}=1+1\cdot h$ , and $(1+h)^{n}\ge1+nh$ implies $(1+h)^{n+1}\ge(1+nh)(1+h)$ $=1+(n+1)h+nh^{2}$ $>1+(n+1)h$ . Now, since $a>0$ , by induction, $a^{n}>0$ for all $n\in\mathbb{Z}_{+}$ , and $0$ is the greatest lower bound for the set.