Using the Well Ordering principle to show that subsets of integers have a least element Prove: Let $S$ be a nonempty subset of $\Bbb Z$, which is bounded below by the integer $n$ in the sense that $n ≤ x$ for all $x ∈ S$. Use the Well-Ordering Principle for $\Bbb N$ to prove that $S$ contains a smallest member.
I am having a lot of difficulty solving this problem. I know that I need to show that $\exists m \in S$ such that $\forall x \in S$ $n \le m \le x$. So the well ordering principle says that every non-empty subset of $\Bbb N$ has a least element. I thought about using a proof by contradiction. $\forall m \in S$ $\exists x \in S$ such that $m>x$ which implies that $S = \emptyset$. I don't really know where to go from here. Any help would be appreciated.
 A: Define $f\colon S\to\mathbb N$ with $f(x) = x-n$ (or $x-n+1$ to avoid $0$ if necessary). Then, $f$ is order preserving and $f(S)\subseteq \mathbb N$, so there is the least element $y_m\in f(S)$. Let $x_m\in S$ such that $f(x_m) = y_m$. Now, there can't be $x\in S$ such that $x<x_m$, otherwise we would have $f(x)<y_m$. Thus, $x_m$ is minimal in $S$ and therefore the least element.
A: Well I'm not sure what definition of $\mathbb{Z}$ you have at your disposal, but I'm used to the one where $\mathbb{Z}=-\mathbb{N} \cup \{\mathbb{0}\} \cup \mathbb{N}$, where $-\mathbb{N}=\{-x \mid x \in \mathbb{N} \}$. But that was real analysis. Not sure what starting points you've got for abstract. Operating on a definition like the one I gave, it will be useful take advantage of the fact that $-\mathbb{N}$ is indexed by $\mathbb{N}$ therefore you can make inductive arguments on $-\mathbb{N}$.
The general method I have in mind is to show that $Q=\{x \in \mathbb{Z} \mid x \geq n \}$ is "order isomorphic" to $\mathbb{N}$, therefore $Q$ follows the well ordering principle and $S \subseteq Q$, and that's the proof.
Hope that helps!
A: $S \subset \Bbb Z$ and $\forall s \in S, s \ge n $
If $n \le 0$ then in a finite number of iterations you can say:
If $ \exists s = n$ then $s (=n)$ is the minimum element of $S$, else 
If $ \exists s = n + 1$ then $s (=n + 1)$, is the minimum element of $S$, else 
....., else
If $ \exists s = 0$ then $s (=0)$ is the minimum element of $S$, else (you didn't find such an $s$ or in the first place $n > 0$)
$S \subset \Bbb N$ and therefore by the well ordering of $\Bbb N$ it must have a minimum element.
