How can I find the sup and inf of $m/n$ such that m,n are natural numbers and $m<2n$? In a problem set I found this problem:
Find the $\sup$ and $\inf$ of $\frac mn$ such that $m,n$ are natural numbers and $m<2n$.
All I could do is this:
Since $m$ and $n$ are natural numbers, we have $m/n > 0$.
And since $m<2n$ we have $\frac mn < \frac{2n}n = 2$, thus $\frac mn<2$.
I got stuck on showing that $2$ is the supremum rigorously.
The answer provided in the solutions sheet is not helping at all, it goes like this (in French, translated to English):
Il est clair que $0 < \frac mn < 2$, et on peut facilement en déduire que $\sup = 2$ et que $\inf=0$.
In english:
It is obvious that $0<\frac mn<2$, from this we can easily deduce that  $\sup = 2$ and $\inf=0$
 A: I think what your solution sheet is saying is straight up wrong. We cannot conclude,  just from the inequality $0<\frac mn < 2$, that $\sup =2$. You need some more steps and more assumptions.

To prove that $s$ is the supremum of a set $A$, you need to prove two things.

*

*That $s$ is the upper bound of $A$.

*That any number smaller than $s$ is not the upper bound of $A$.

Let $$A=\left\{\frac mn| m,n\in\mathbb N, m<2n\right\}$$
You already proved that $2$ is the upper bound of $A$. Now, you need to prove that any number smaller than $2$ is not the upper bound of $A$. You can start doing this by saying one of the most common phrases in calculus: **Let $\epsilon > 0$.
Now, you need to prove that $2-\epsilon$ is not an upper bound of $A$. You can do this by proving that there exists some element $a\in A$ such that $a>2-\epsilon$.
To do this, think about what property $a$ needs to fulfill. You know that $a=\frac mn$ for some pair $m,n$. You also know that $a>2-\epsilon$.
You can write this out a bit and get
$$\frac mn > 2-\epsilon\\
m > 2n - \epsilon n$$
and note that the deduction above goes both ways. In other words, if you can find some $m,n\in\mathbb N$ such that $m>2n-\epsilon n$, then you will also have $a=\frac mn > 2-\epsilon$.
However, you have one more limitation, and that is that if you want $\frac mn$ to be an element of $A$, then $m<2n$ must also be true. This means that $m$ must be some value that is greater than $2n-\epsilon n$ and smaller than $2n$.
So, the question remains, can you find such a pair of values?
Hint: Think about what happens to the interval $(2n-\epsilon n, 2n)$ when $n$ is very large.
A: You have already noted that $2$  and $0$ are upper and lower bounds respectively for $S$.
Let's show that $2$ is indeed the supremum of $S$.
. Note that $2n-1<2n$ for any $\mathbb N$ and therefore $\frac{2n-1}{2n}\in S$ for all $n\in \mathbb N$. $\tag 1$
By Archimedean property of the reals, given any $\epsilon\gt 0, $ there exists $N\in \mathbb N$ such that $1/N\lt \epsilon\implies 2-\epsilon< 2-1/N=\color{blue}{\frac{2N-1}{2N}}<2$.  By $(1)$, the quantity in blue is in $S$ and hence by uniqueness of $\sup S$, it follows that $2$ is the supremum of $S$ (For if,suppose on the contrary $x:=\sup S<2$ then by the inequality established earlier, there must be some $m\in \mathbb N$ such that $2-(2-x)=x<\color{red}{\frac{2m-1}{2m}}<2$. But the quantity in red is in $S$ and this means that $S$ has an element larger than $\sup S$, which is a contradiction.).
$\inf S$ is the "greatest" lower bound for $S$. Now, if $\inf S=:y>0$, then by Archimedean property there is an $n\in \mathbb N$ such that $\frac 1n<y$ but $1/n\in S$ and hence $y$ is not a lower bound for $S$. This is a contradiction. Hence $y$ must be $0$.
