I need to prove the above statement for A which is an upper bounded subset of $\mathbb{N}$. Here's what I did:
From the axiom of completness we deduct that the supremum of A exists.
Let $\sup A=a$
Then by definition of supremum: $\exists \varepsilon>0 : a-\varepsilon<x, x\in A $.
Let $ε=1$ then $a-1<x$.
We suppose that $a$ doesn't belong to $A$. Then $x<a$. So in case $\sup A$ doesn't belong to $A$, $a-1<x<a\Rightarrow 0<x<1$ which is a contradiction because $x\in A\leq \mathbb{Z}$
I think this is a good enough proof, however my textbook goes on like this:
$a-1<x<a$. We find $y\in A$ such that: $a-1<x<y<a\Rightarrow 0<y-x<1$ which is a contradiction because both $x,y$ belong to $A$ which is a subset of $\mathbb{N}$
I have two questions. Is my proof not enough? And secondly how does he goes from $a-1<x<y<a$ to $0<y-x<1$