Prove by definition (rigorously) that the supremum of $S:=\lbrace{\frac{1}{a} - \frac{3}{b}:a,b \in \mathbb{N}}\rbrace $ is $1$.

Question

I have to prove by using definition that the supremum of the set $S:=\lbrace{\frac{1}{a} - \frac{3}{b}:a,b \in \mathbb{N}}\rbrace $ is $\sup(S)=1$.

I'm stuck in proving that: if $u<\sup(S)=1 $ then $\exists s\in S$ such that $u<s\leq \sup(S)$. I've tried using the archimedean property, using the fact that $u<1$, meaning that, $0<1-u$. Then by Arch. Property (or one of it's corollaries) $\exists a \in \mathbb{N}$ such that $0<\frac{1}{a}<1-u$. Also I can use again the A.P. with $0<1/a$. Then exists $b \in\mathbb{N}$ such that $0<\frac{1}{b}<\frac{1}{a}<1-u$. Now i've got two inequalities. First $0<\frac{1}{a}<1-u$. Second $0<\frac{1}{b}<1-u$. Working with the second, i've got a third inequality: multiplying by -3: $3u-3<-\frac{3}{b}<0$. Then adding first one and third one ive got: $3u-3<\frac{1}{a}-\frac{3}{b}<1-u$. But I'm stuck here. Any hint or help? Thanks in advance.

EDIT: My instructor understands "prove by definition" the following:
i) First prove that $\sup(S)=1$ is an upper bound (i.e. $s\leq \sup(S)=1 \forall s \in S$). (I've already proven that)
ii) Then prove such number (i) IS the least upper bound (i.e. if $u<\sup(S)=1$ then $\exists s \in S$ such that $u<s\leq \sup(S)=1$).

Answer 1

Hint: You can see that the $\sup(S) \leq 1$, because each element is bounded above by 1. Now fix $a=1$ and let $b$ get arbitrarily large.

Answer 2

To make it rigorous, consider the following approach. First, prove as a lemma that $A = \sup S$ iff $A$ is an upper bound for $S$ and for all $\epsilon>0$ there exists some $x \in S$ such that $A - \epsilon < x$.

It suffices to consider only $a=1$, since for any $b \in \mathbb{N}$ and $a >1$, $\frac{1}{a} - \frac 3b < 1 - \frac 3b$. To be completely precise , if $A - \epsilon < x_{a,b}$, then $A - \epsilon < x_{1,b}$.

For fixed $a=1$, the sequence as $b \to \infty$ goes to $1$. We simply wish to show $1 - \epsilon < 1 - \frac 3b$, or $\frac 3\epsilon < b$ which follows from the Archimedian property.