This is one of the exercises from Terrence Tao's Analysis book:
Let $x\in\mathbb{R}$. Show that there exists an integer $N$ such that $N\leq x<N+1$.
I think that I know how to do this for $N\in\mathbb{N}$. Let's define the set $S=\{N\in\mathbb{N}|N>x\}$. Obviously $S\subseteq\mathbb{N}$ and $S$ is nonempty because the Archimedean property says that there is a natural number $N$ such that $N>x$. Now by well-ordering principle of $\mathbb{N}$, $S$ has a least element, say $N+1$ and hence $N+1>x$. Since $N+1$ is the least element of $S$, then $N\not\in S$ which implies $N\leq x$. All of these together gives $N\leq x<N+1$.
My problem is that I do not know how I can extend my argument so that it works for $\mathbb{Z}$ and not just $\mathbb{N}$.
Best Answer
Your argument is actually false.
The argument fails for $x=-2$, then $S=\mathbb N$, which means that $N$ is not an integer and is also not smaller than or equal to $1$.
However, your argument is very close to what you need it to be. You can set $$S=\{N\in\mathbb Z| N>x\}$$
then proceed with your proof, and use the fact that $S$ is a subset of the set $\{-k, -k+1, -k+2, \dots, 0,1,2,\dots\}$ which is well ordered (and therefore, $S$ has a least element).