(a) By induction. If there is only one element in $A$ then it is the largest. Suppose there is a largest element for any finite set of cardinality $n-1$ . If $A$ has $n$ elements, and $x\in A$ , then $A-\{x\}$ has the largest element $y$ (by induction), and $y$ is the largest element of $A$ if $x<y$ , otherwise $x$ is the largest element of $A$ .

(b) The bijection is constructed by induction: if $A$ has $n$ elements and for all ordered finite sets of cardinality $n-1$ the fact is proved, we simply take $b$ , the largest element of $A$ , and extend the order preserving bijection $f$ of $A-\{b\}$ with $\{1,\ldots,n-1\}$ (the existence of which is assumed by the induction hypothesis) by letting $f(b)=n$ .

A more interesting approach is to define $f$ without induction: $f(a)=$ the cardinality of $S_{a}=\{x\in A|x\le a\}$ . $S_{a}$ is a non-empty subset of $A$ ($a\le a$ ), so, it has cardinality form $1$ to $n$ (Corollary 6.6). Further, if $a<b$ , then $x\le a$ implies $x\le b$ , and $b\not\le a$ , i.e. $S_{a}\subsetneq S_{b}$ , and $f(a)<f(b)$ (Corollary 6.6). Overall, we have an injective function $f$ from $A$ to $S_{n+1}$ , but we also have a bijective function $g$ from $A$ to $S_{n+1}$ ($A$ has cardinality $n$ ), so that $f$ must be bijective ($f\circ g^{-1}$ is a bijective correspondence of $S_{n+1}$ with the image set of $f$ ).