Let $A = \left\{ \dfrac{m}{m+n} : m,n \in \mathbb{N} \right\} $. I need to calculate $\sup A $ and $\inf A $
Try:
We notice that $\dfrac{m}{m+n} \leq \dfrac{m}{m} = 1 $ so we say that $\sup A = 1 $.
Next, put $A_m = \{ \frac{m}{m+n} : n \in \mathbb{N} \} $ and so $A_1 = \{ \dfrac{1}{1+n} \}$ and since $\dfrac{1}{1+n} > 0 $ for all $n$ then 0 is a lower bound. We prove that it is the greatest lower bound. If not, then we can always find a $n_0$ such that $0 > \dfrac{1}{1+n_0}$ but $n_0 >0$ and so $\frac{1}{1+n_0} > 0$ and thus $0>0$ which is a contradiction, thus $\inf A_1 = 0$. Similarly, we observe that $\inf A_2 = 0$ and so on. Thus
$$ \inf A = \inf \bigcup A_m = \inf ( inf_m A_m ) = 0 $$
Is this correct?
What if replace $\dfrac{m}{m+n}$ with $\dfrac{ m}{|m| + n } $? Do we get the same result if we impose $m \in \mathbb{Z}$. We see that $\sup A$ is still $1$ but now isnt it $\inf A = – \infty$?
Best Answer
Your first line shows only that $\sup A\le 1$; it does not rule out the possibility that $\sup A<1$. However, for any $\epsilon>0$ there is an $m\in\Bbb Z^+$ such that $\frac1m<\epsilon$, and
$$\frac{m}{m+1}=1-\frac1{m+1}>1-\frac1m>1-\epsilon\;,$$
where $\frac{m}{m+1}\in A$, which shows that $\sup A\ge 1$. Combining the inequalities, we have $\sup A=1$.
If instead we take $m=1$, we can use the same argument to show that $\inf A\le 0$: for any $\epsilon>0$ there is an $n\in\Bbb Z^+$ such that $\frac1n<\epsilon$, and
$$\frac1{n+1}<\frac1n<\epsilon\;,$$
where $\frac1{n+1}\in A$. On the other hand, $\frac{m}{m+n}>0$ for all $m,n\in\Bbb Z^+$, so $\inf A\ge 0$, and we conclude that $\inf A=0$. Note that there is no reason to consider cases or to break $A$ into parts.