I am reading the second part of the prolouge of Spivak's Calculus. In the text, he proves the Well-Ordering Principle. Here is a sentence from the book:
Suppose that the set A has no least member. Let B be the set of natural numbers $n$ such that $1,...,n$ are all not in A. Clearly 1 is in B (because if 1 were in A, then A would have a smallest member).
My first question is, if 1 was in the set A, how could it have a smallest member/element; isn't smallest a relative term, requiring at least two elements by which we compare the one to the other? With this line of reasoning, wouldn't a set have to have at least two elements in order to have a least element?
MY second question is, does the set B contain all the natural numbers that are not in A?
First question: A smallest element of $A$ is an element $x\in A$ such that $x\le y$ holds for all $y\in A$ (or equivalently: $x<y$ for all $y\in A\setminus\{x\}$). Some care must be taken and we distinguish between smallest and minimal if the set is not totally orderd (i.e. there may exist elements $a,b$ such that neither $a=b$ nor $a<b$ nor $a>b$), but for the natural numbers this is not necessary. Anyway, the empty set has no minimal element because it does not have any element at all. But a one element set $A$ does have a minimal element because this one element is $\le$ all elements of $A$ (including itself, or alternatively: it - vcuously - is $<$ than all other elements.
Second question: No. Suppose $A=\{4,8,12,16,\ldots\}$, then $B=\{1,2,3\}$ whereas $\mathbb N\setminus A=\{1,2,3,5,6,7,9,\ldots\}$. To put what the text says into different words that do not require "$\ldots$": $B$ contains exactly those natural numbers $n$ for which $x\in A$ implies $x>n$. $B=\{\,x\in\mathbb N\mid \forall y\in A\colon x<y\,\}$