(a) Consider $\mathbf{x}^{(n)}=(\underbrace{1,\ldots,1}_{n},2,1,1,\ldots)$ , $n\ge0$ . Then, $\mathbf{x}<\mathbf{x}^{(n)}$ iff $1=x_{n+1}=x_{n+2}=\ldots$ . Therefore, the section $A_{n}=S_{\mathbf{x}^{(n)}}$ contains exactly those sequences that have arbitrary values of the first $n$ coordinates, and all $1$ ’s after that. The order of the elements of the section $A_{n}$ is the "reverse dictionary". Indeed, the case $n=0$ is trivial, and for $n\ge1$ , let $f:A_{n}\rightarrow(\mathbb{Z}_{+})^{n}$ be defined by $f(x_{1},\ldots,x_{n})=(x_{n},\ldots,x_{1})$ . Then, as it is easy to see, $f$ is a bijection preserving the order (in $A_{n}$ we compare all coordinates one-by-one from right to left, and in the dictionary order we compare from left to right).

(b) For every sequence $\mathbf{x}\neq(1,1,\ldots)$ define $f(\mathbf{x})=j\in\mathbb{Z}_{+}$ such that $1=x_{j+1}=x_{j+2}=...$ and $x_{j}\neq1$ . Additionally, define $f((1,1,\ldots))=0$ . Note, that $f(\mathbf{x})<f(\mathbf{y})$ implies $\mathbf{x}<\mathbf{y}$ (but not vice versa). For every nonempty subset $B$ of sequences, consider the set of all $J=\{f(\mathbf{x})|\mathbf{x}\in B\}$ . $J\subset\mathbb{Z}_{+}$ is nonempty, and, hence, has the smallest element $n$ . Consider the set $M$ of all sequences $\mathbf{x}\in B$ such that $f(\mathbf{x})=n$ . $M$ is a subset of the section $A_{n}$ defined in (a), which has the same type as the well-ordered set $\mathbb{Z}_{+}^{n}$ . Hence, there is the smallest element $\mathbf{m}\in M\subset B$ . For $\mathbf{x}\in B$ , either $f(\mathbf{x})>n=f(\mathbf{m})$ , and, hence, $\mathbf{m}<\mathbf{x}$ , or $f(\mathbf{x})=f(\mathbf{m})$ and $\mathbf{m}\le\mathbf{x}$ .