(a) Let $A$ be the set of all positive integers $n$ for which this statement is true. It is true for $n=1$ , as $\{1\}$ has only one nonempty subset, namely $\{1\}$ itself with the largest element $1$ , so $1\in A$ . Suppose $n\in A$ . Then every nonempty subset of $\{1,...,n+1\}$ either contains $n+1$ or does not. In the first case, $n+1$ is the largest element of the subset, and in the second case, the subset is also a subset of $\{1,...,n\}$ and, hence, has the largest element. Therefore, $n+1\in A$ , $A$ is an inductive subset of $\mathbb{Z}_{+}$ , and $A=\mathbb{Z}_{+}$ .

(b) Because the statement is proved for all such subsets that are contained within some set $\{1,...,n\}$ , i.e. all bounded from above subsets. But, as it was shown in the text, there are some unbounded subsets as well, for example, the set $\mathbb{Z}_{+}$ itself. To compare to Theorem 4.1, where for every $n\in D$ the smallest element has to be not greater than $n$ , and, hence, be an element of $\{1,\ldots,n\}$ , in our case, the largest element $d$ (if it exists at all) has to be not smaller than any $n\in D$ , which by itself does not imply that there is any set $S_{m}$ such that $d\in S_{m}$ .