How to prove well-foundedness of Cantorian Ordinals I am probably misrepresenting history but the classical naive definition
of ordinals was as equivalence classes of well-ordered sets under
order-preserving bijections. The collection of such ordinals is well-ordered
under the order defined by order-preserving injections, being well-ordered
is the whole point of ordinals.
This approach was superseded by the familiar vonNeumann approach which
provides a canonical representation of the equivalence-class based
ordinals by using ordinals defined by insert-our-favourite-definition-of
an-ordinal as witnesses. 
In the vonNeumann approach to ordinals there is not much to prove
because well-foundedness is essentially built into the definition
or inherited from regularity.
My question relates to how Cantor and his cohort proved the ordinals
were well-founded before the vonNeumann approach took over. How do
you show every nonempty set of ordinals has a minimum when dealing
with equivalence classes.
Translating I think the theorem becomes - Let $\mathcal{B}$
be a nonempty set of well-ordered sets then there exists a well-ordered
set $M$ and an $A\in\mathcal{B}$ such $M\simeq A$ and for every
$B\in\mathcal{B}$ there is an order-preserving injection from $A$
to $B$.
Can someone give me a clue for a proof, or tell me if the question
even makes sense? I believe Cantor would have proved this without
the axiom of choice?
 A: I don't know how Cantor would have done it, but we can "locally" construct canonical representatives by embedding everything from $\mathcal{B}$ into an upper bound, such as the supremum. So, there exists


*

*A well-ordered set $\sup(\mathcal{B})$

*For each $A \in \mathcal{B}$, an order preserving map $I_A : A \to \sup(\mathcal{B})$ identifying $A$ with an initial segment of $\sup(\mathcal{B})$.

*$\sup(\mathcal{B}) = \cup_A \{ \mathrm{image}(I_A) \mid A \in \mathcal{B} \}$


By this device, we can construct a new set
$$\mathcal{B}' = \{ \mathrm{image}(I_A) \mid A \in \mathcal{B} \} $$
If we are asking questions depending only on order types, then we can answer questions about $\mathcal{B}$ by translating them into questions about $\mathcal{B}'$.
(I've omitted dealing with technicalities regarding the case where $\mathcal{B}$ has equivalent objects, since they are irrelevant to the given question, and IMO they are relatively straightforward but tedious)

The supremum can be constructed by the following device.
Recall that any two well-ordered sets are comparable. Furthermore, if $A \leq B$, then there is a (unique!) order preserving map $I_{AB}: A \to B$ identifying $A$ with an initial segment of $B$. 
First, let $S$ be the disjoint union; that is,
$$ S = \{ (A,a) \mid a \in A \wedge A \in \mathcal{B} \} $$
and define a preordering
$$ (A,a) \leq (B,b) \Longleftrightarrow
\begin{cases} I_{AB}(a) \leq b & A \leq B
\\ a \leq I_{BA}(b) & B \leq A \end{cases} $$
Then we can take $\sup(\mathcal{B})$ to be the equivalence classes of $S$, where  $x \equiv y$ iff $x \leq y$ and $y \leq x$.
$\leq$ will be a well-order on $\sup(\mathcal{B})$.
A: For any well-orders $C,D$ let $C\sim D$ mean that $C,D$ are order-isomorphic.
Let $<_C$ be a well-order on $C.$ An initial segment of $C$ is $\{p\in C:p<_C  q\}$ for some $q\in C,$ well-ordered by $<_C.$
Let $\mathbb B$ be a non-empty family of well-ordered sets. For each $C\in \mathbb B,$ if no initial segment of $C$ is order-isomorphic  to a member of $\mathbb B$ then   let $f(C)=C.$ Otherwise let $f(C)=\{p\in C:p<_Cq\}$ where $q$ is the $<_C$-least $y\in C$ such that some $D\in \mathbb B$ satisfies $D\sim \{p\in C:p<_Cy\}.$
$(\bullet).$  If $X $ is an initial segment of $f(C)$ then $X$ is also an initial segment of $C,$ and no $E\in  \mathbb B$ satisfies $E\sim X.$
Claim. $f(C) \sim f(D)$ for all $C,D\in \mathbb B.$
Proof. Suppose not. WLOG there is an  order-isomorphism $g:f(D)\to X$ where $X$ is an initial segment of $f(C). $ But there exists $E\in \mathbb B$ and an order-isomorphism $\psi:E\to f(D).$ Then the composite function $(g\psi):E\to X$ is an order-isomorphism, contrary to $(\bullet).$
So take some (any) $C\in \mathbb B$ and some (any) $A\in \mathbb B$ such that $A\sim f(C).$ Then for every $D\in \mathbb B,$ either $A \sim D=f(D)$ or  $A$ is order-isomorphic to  the initial segment $f(D)$ of $D.$  
