Let $A = \{\frac{m}{n}; m,n \in \mathbb{N} \ \text{and} \ m <2n\}$. Show that $\sup A = 2$.

Question

Let $A = \{\frac{m}{n}; m,n \in \mathbb{N} \ \text{and} \ m <2n\}$. Show that $\sup A = 2$.

I showed that

a) Given $\frac{m}{n} \in A$ then $m < 2n \Rightarrow \frac{m}{n} < 2.$

I am not able to show that

b) For all $\epsilon > 0,$ exists $\frac{m}{n} \in A$ such that $2 - \epsilon < \frac{m}{n}$.

I'm trying to use the property: givem $a,b \in \mathbb{R}$, with $a>0$,$\exists n \in \mathbb{N}$ such that $na>b.$

Answer 1

Hint: Consider the numbers of the form $2-\dfrac1n$.

Answer 2

If $\epsilon > 0$ then there exist $n \in N$ s.t $\frac 1 n < \epsilon$. We can let $m = 2n -1 < 2n$. So $2 - \frac m n = \frac 1 n < \epsilon.$

.... Or we can start from $2 - \epsilon < \frac m n$ and get $\frac {2n-m} n < \epsilon$ and figure if $m = 2n - 1$ that will give us $\frac 1 n < \epsilon $ and therefore ...