A nonempty set of integers bounded above has a greatest element, and a nonempty set of integers bounded below has a least element

Suppose $b$ is an upper bound for $S$. For every $s\in S$ we have $s\le b$, and adding $-s$ gives $0\le b-s$; so $b-s$ is nonnegative and therefore equals $\iota (k)$ for a unique $k\in \mathbb{N}$.

Now suppose instead that $c$ is a lower bound for $S$, and put $S^{\mathrm{\prime }}:=\{-s:s\in S\}$, a nonempty subset of $\mathbb{Z}$. For $s\in S$, adding $-s-c$ to $c\le s$ gives $-s\le -c$, so $-c$ is an upper bound for $S^{\mathrm{\prime }}$.

Put $T:=\{k\in \mathbb{N}:b-\iota (k)\in S\}$. It is nonempty: choosing $s\in S$ and the $k$ of step 1.1 with $\iota (k)=b-s$, we get $b-\iota (k)=b-(b-s)=s\in S$, so $k\in T$.

By well-ordering let $k_0$ be the least element of $T$, and put $m:=b-\iota (k_0)$, so that $m\in S$.

Let $s\in S$. By step 1.1 there is $k\in \mathbb{N}$ with $\iota (k)=b-s$, and then $b-\iota (k)=s\in S$, so $k\in T$ and hence $k_0\le k$.

Applying $\iota$, which preserves the order, gives $\iota (k_0)\le \iota (k)=b-s$; adding $s-\iota (k_0)$ to this inequality gives $s\le b-\iota (k_0)=m$.

So $m\in S$ and $s\le m$ for every $s\in S$: $m$ is a greatest element of $S$. If $m^{\mathrm{\prime }}$ were another one, then $m\le m^{\mathrm{\prime }}$ and $m^{\mathrm{\prime }}\le m$, so $m=m^{\mathrm{\prime }}$ by antisymmetry.

By the first part $S^{\mathrm{\prime }}$ has a greatest element, which has the form $-\mathrm{\ell }$ with $\mathrm{\ell }\in S$; then for every $s\in S$ we have $-s\le -\mathrm{\ell }$, and adding $s+\mathrm{\ell }$ gives $\mathrm{\ell }\le s$. So $\mathrm{\ell }$ is a least element of $S$, and it is unique by antisymmetry as in step 6.1.