Determine infimum/supremum of the set $\{\frac{n-m}{n+m} \vert n,m \in \mathbb{N}_0\}$

Question

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 $m = m_1, n = 1$).

Answer 1

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.

Answer 2

Your answer is fine.

An option.

1) Supremun.

$1$ is an upper bound .

Consider $a_n:= 1-\frac{2}{n+1}$.

As suggested by Omnomnomnom assume there is a bound $b$ s.t. $b <1$.

$\lim_{n \rightarrow \infty} a_n =1$.

Given $\epsilon$ there is a $n_0$ s.t. for $n\ge n_0$

$|a_n-1| <\epsilon$,

$-\epsilon+1<a_n < \epsilon +1$.

Choose $\epsilon =1-b (>0)$, then

$1-\epsilon =b <a_n$, a contradiction.