Munkres

Munkres Introduction to Topology: Section 4 Problem 8

I'm trying out a new format (using LaTeX.css for my solution manuals). Let's see how this goes.

Part A

We first have to demonstrate that $\inf \ \{1/n \ | \ n \ \in \ \mathbb{Z}^{+} \} \ = \ 0$. Let's call this set $A$, and assume that $\inf \ A \ < \ 0$. This implies that there is some $1/n \ \in \ A$ such that $1/n \ \leq \ 0$. Let's assume that this is true for $n \ = \ N$, with $N \ \in \ \mathbb{Z}^{+}$. Now, we are going to prove everything that come next from the axioms of algebraic operations, rather than refering to previous exercies. Why? Because we're hardcore.

If $x \ > \ 0$, then it also follows that $1/x \ > \ 0$.

Assume for $x \ > \ 0$, that $1/x \ < \ 0$. By the order axiom, which states that for $x \ > \ y$ and $z \ > \ 0$, it follows that $x \ \cdot \ z \ > \ y \ \cdot \ z$, we find that $(1/x) \ \cdot \ x \ < \ 0 \ \cdot \ x \ = \ 0$. By definition, $(1/x) \ \cdot \ x \ = \ 1$, so we get $1 \ < \ 0$, which is obviously a contradiction. It follows that $1/x \ > \ 0$.

If $x \ > \ y \ > \ 0$, then it follows that $1/y \ > \ 1/x$.

By the order axiom we used in the previous proof, and knowing that $1/x$ is positive, we get:
$$(1/x) \ \cdot \ x \ > \ (1/x) \ \cdot \ y \ > \ (1/x) \ \cdot \ 0 \ \Rightarrow \ 1 \ > \ y/x \ > \ 0$$ Using the positivity of $1/y$, we get:
$$(1/y) \ \cdot \ 1 \ > \ (1/y) \ \cdot \ (y/x) \ > \ (1/y) \ \cdot \ 0 \ \Rightarrow \ 1/y \ > \ 1/x \ > \ 0$$

Now let's get back to what we were proving. From the first lemma, it can't be true that $1/N \ < \ 0$, so we are left with the case of $1/N \ = \ 0$. By the Archimedean property, $N \ + \ 1 \ \in \ \mathbb{Z}^{+}$. From this, and the previous lemma, it follows that:
$$\frac{1}{N} \ > \ \frac{1}{N \ + \ 1} \ \Rightarrow \ 0 \ > \ \frac{1}{N \ + \ 1}$$ But from the first lemma, $\frac{1}{N \ + \ 1}$ must be positive, so we derive a contradiction. It follows that $\inf \ A \ \geq \ 0$. Assume that $\inf \ A \ = \ s \ > \ 0$. Consider the number $1/s$. There must be some positive integer $m$ greater than $1/s$, as if this weren't the case, then $1/s$ would be an upper bound for $\mathbb{Z}^{+}$, which was proved impossible earlier in the chapter. Thus, we will have:
$$m \ > \ 1/s \ \Rightarrow \ 1/m \ < \ s \ \ \ \ m \ \in \ \mathbb{Z}^{+}$$ This contradicts the fact that $s$ is a least upper bound of $A$, as $1/m \ \in \ A$. it follows that $\inf \ A \ = \ 0$.

Part B

In the second part of this question, we have to prove that for $0 \ < \ a \ < \ 1$, we have $\inf \ \{a^n \ | \ n \ \in \ \mathbb{Z}^{+}\} \ = \ 0$. We call this set $A$. Munkres also gives us a hint. He says we should prove that $(1 \ + \ h)^n \ \geq \ 1 \ + \ nh$, for $h \ = \ (1 \ - \ a)/a$. Firstly, we can observe that on the domain of $0$ to $1$, the range of $h$ spans from $0$ to infinity. We can prove:

The zeroth and first-order terms of $(1 \ + \ h)^n$ are $1 \ + \ nh$ for $n \ \in \ \mathbb{Z}^{+}$.

This is clearly the case for $n \ = \ 1$. We assume it to be true for $n$. We prove it for $n \ + \ 1$. In this case, we have $(1 \ + \ h)^{n \ + \ 1} \ = \ (1 \ + \ h) (1 \ + \ h)^{n}$. We know that $(1 \ + \ h)^{n}$ has zeroth and first order terms of $1 \ + \ nh$. When this is multiplied by $(1 \ + \ h)$, we get $1 \ + \ (n \ + \ 1)h$ as the new zeroth and first-order terms.

The proof of the inequality follows from this. Now, let's expand out the inequality in terms of $a$:
$$(1 \ + \ \frac{1 \ - \ a}{a})^n \ \geq\ 1 \ + \ \frac{n}{a}(1\ - \ a) \ \Rightarrow \ \frac{1}{a^n} \ \geq \ 1 \ + \ \frac{n}{a}(1 \ - \ a) \ \geq \ \frac{n}{a}(1 \ - \ a) \ \Rightarrow \ a^n \ \leq \ \frac{a}{1 \ - \ a} \frac{1}{n}$$ Assume that $\inf A \ \ = \ b \ > \ 0$. It follows that for all $n$:
$$b \ \leq \ a^n \ \leq \ \frac{a}{1 \ - \ a}\frac{1}{n}$$ Which is a contradiction, by Part A. It follows that $\inf A \ \leq \ 0$. Assume $\inf A \ \leq \ 0$ It follows that there must be some $a^N \ \leq \ 0$, for positive $a$ and integer $N$. This is clearly impossible (we can prove that equality never holds in a similar way as we did in exercise one, by showing that $N \ + \ 1$ case yields a contradiction). Thus, we are left with $\inf A \ = \ 0$.