I am going through my analysis-notes from when I was a student. I did not solve the following exercise back then and wanted to give it a try:
Determine minimum, maximum, infimum/supremum of the set $C = \{\frac{n-m}{n+m} \vert n,m \in \mathbb{N}_0\}$. Here $\mathbb{N}_0$ denotes the positive naturals
First of all, I rewrote the fractions as $1 – \frac{2m}{n + m}$. Hence, $1$ is an upper bound. Also, $\frac{-2m}{n+m} > -2$, so $-1$ is a lower bound.
By the supremum property of $\mathbb{R}$, I know that both supremum and infimum exist. I think they are 1 and -1. I have trouble prooving this.
Also, these bounds can not be attained (otherwise either $n$ or $m$ has to be zero), so there is no minimum/maximum.
Any hints on how to show that sup(C) = 1 and inf(C) = -1?
Based on Omnomnomnom's hint is my proof that the supremum equals 1:
Suppose that $\alpha < 1$ is an upper bound for $C$, then $\alpha \geq 1 – \frac{2m}{n+m}$ for all $n,m \in \mathbb{N}_0$. However, there exists $n_1 \in \mathbb{N}_0$ such that
$$\frac{1}{n_1 + 1} < \frac{1}{n_1} < \frac{1 – \alpha}{2}.$$
Hence we have that
$$\frac{2}{n_1+1} < 1 – \alpha \Rightarrow 1 – \frac{2}{n_1+1} > \alpha.$$
This is a contradiction ($n = n_1, m = 1$). Therefore, $\operatorname{Sup}(C) = 1$.
Proof that the infimum equals -1:
Suppose $\alpha > -1$ is a lower bound. Then $1 – \alpha < 2$. Therefore $\frac{1 – \alpha}{2} < 1$. There exists $m_1 \in \mathbb{N}_0$ such that $\frac{1}{m_1 +1} + \frac{1 – \alpha}{2} < 1$.
This however implies that
$$\frac{1 – \alpha}{2} < 1 – \frac{1}{m_1 + 1} = \frac{m_1}{m_1 + 1}.$$
Therefore,
$$1 – \alpha < \frac{2m_1}{m_1 + 1}$$
from which it follows that $1 – \frac{2m_1}{m_1 + 1} < \alpha$, which is a contradiction (take $m = m_1, n = 1$).
Best Answer
Hint: Suppose for the purpose of contradiction that $\alpha < 1$ is an upper bound for $C$. Show that there exists an $m,n$ such that $1 - \frac{2m}{n+m} > \alpha$.
Once you have proven this, then you have shown that every upper bound $\alpha$ of $C$ must satisfy $\alpha \geq 1$. In other words, $1$ is the least possible upper bound, which is to say that it is the supremum.
The approach for the infimum is similar.