I'm right now covering Section 4 of Topology by James R. Munkres, 2nd edition, and am stuck with the following problems in the exercise set after Section 4:
Problem 8(c):
Show that given $a$ with $0 < a < 1$, we have $\inf \{ a^n | n \in Z^+ \} = 0$. Hint: Let $h = (1-a)/a$, and show that $(1+h)^n \geq 1+nh$.
Now I can show that for any $h > 0$, we have $(1+h)^n \geq 1+nh$, but how do I use this hint to show what we've been asked in the problem?
Problems 9(a), (b), and (c):
How to prove these facts using only the axioms of the section and using only the preceding problems in the exercise set?
As David Holden said in the comments, if $h=\frac{1-a}a=\frac1a-1$, then $\frac1a=1+h$, and for $n\in\Bbb Z^+$ you know that $\frac1{a^n}=\left(\frac1a\right)^n=(1+h)^n\ge 1+nh$ and therefore $a^n\le\frac1{1+nh}$. What happens to $\frac1{1+nh}$ as $n\to\infty$?
$9$(b) asks you to show that if $x\notin\Bbb Z$, then there is exactly one $n\in\Bbb Z$ such that $n<x<n+1$; in other words, every non-integer lies between exactly one pair of successive integers. Suppose first that $x>1$. Let $A=\{n\in\Bbb Z_+:x<n\}$
By the Archimedean property $A\ne\varnothing$, and then by Theorem $4.1$ $A$ has a least element $m$; let $n=m-1$. By hypothesis $x\ne n$, so $x\ne n$. If $x<n$, then $n\in A$ with $n<m=\min A$, which is absurd, so we must have $n<x<n+1$. If $0<x<1$, there’s nothing to prove: just take $n=0$. That leaves only the case $x<0$. In that case use Exercise $2$(c) to conclude that $0<-x$; then use what we’ve just proved to conclude that there is an $n\in\Bbb Z$ such that $n<-x<n+1$. Finally, use Exercise $2$(d) a couple of times to complete the proof.
$9$(c) asks you to prove that if $x-y>1$, then there is at least one $n\in\Bbb Z$ such that $y<n<x$. If $x\in\Bbb Z$, what can you say about $x-1$? If $x\notin\Bbb Z$, apply $9$(b).
$9$(d) asks you to prove that if $y<x$, then there is a rational number $z$ such that $y<z<x$. HINT: If $y<x$, then $x-y>0$. Show that there is an $n\in\Bbb Z_+$ such that $nx-ny>1$ and apply $9$(c).