Theorem 3

Let E and F be two well-ordered sets. Then at least one of the following two statements is true :

(1) there exists a unique isomorphism of E onto a segment of F;

(2) there exists a unique isomorphism of F onto a segment of E.

Let $\mathfrak{F}$ be the set of mappings of subsets of E into F such that each mapping is defined on a segment of E and is an isomorphism of this segment onto a segment of F. Then the set $\mathfrak{F}$, ordered by the relation “$v$ extends $u$” between $u$ and $v$, is inductive. For if $\mathfrak{G}$ is a totally ordered subset of $\mathfrak{F}$, the union S of the domains of the mappings $u\in\mathfrak{G}$ is a union of segments of E and is therefore itself a segment of E. If $v$ is the least upper bound of $\mathfrak{G}$ in $\Phi(E,F)$ (no. 4, Example 2), then $v(S)$ is the union of the ranges of the mappings $u\in\mathfrak{G}$ and is therefore a segment of F. Finally, for each pair of elements $x$, $y$ of S such that $x<y$ there exists $u\in\mathfrak{G}$ whose domain contains both $x$ and $y$ (because $\mathfrak{G}$ is totally ordered); and since $v(x)=u(x)<u(y)=v(y)$, $v$ is an isomorphism of S onto $v(S)$, and our assertion is proved.

¶ Now let $u_0$ be a maximal element of $\mathfrak{F}$ (no. 4, Theorem 2) and let $S_0$ be the segment of E which is the domain of $u_0$. If we show that either $S_0=E$ or $u_0(S_0)=F$, the theorem will be proved. Let us argue by contradiction and suppose that $S_0\ne E$ and $u_0(S_0)\ne F$. There will then be an element $a\in E$ and an element $b\in F$ such that $S_0=]-\leftarrow,a[$ and $u_0(S_0)]=]-\leftarrow,b[$ (no. 1, Proposition 1). Extend $u_0$ to a mapping $u_1$ of the segment $]-\leftarrow,a]$ into F by putting $u_1(a)=b$; since $u_1$ is an isomorphism of $]-\leftarrow,a]$ onto the segment $]-\leftarrow,b]$, this contradicts the maximality of $u_0$ in $\mathfrak{F}$.

¶ The uniqueness asserted in Theorem 3 is a consequence of the following Lemma :