I would like to prove the following claim:
If $A \approx \alpha$, where $\alpha$ is some ordinal, then $A$ is well-orderable
I have no experience with a proof like this, but I imagine it amounts to constructing a set of ordered pairs $R'$ equipped with the following properties:
- $R' \subseteq A \times A$
- $R'$ is irreflexive $\iff$ $\forall x \in A [x\not R' x$]
- $R'$ is transitive $\iff$ $\forall x,y,z \in A [xR'y \land yR'z \rightarrow xR'z]$
- $R'$ satisfies trichotomy $\iff$ $\forall x,y \in A [xR'y \lor yR'x \lor x=y]$
- $R'$ is well-founded $\iff$ $\forall X \big[X \neq \emptyset \land X \subseteq A \rightarrow \exists y \big(y \in X \land \neg \exists z (z \in X \land z R' y)\big)\big]$
The above definitions are all taken from my book: The Foundations of Mathematics: By Kenneth Kunen
To start, assume that $A \approx \alpha$ $\iff$ there exists a bijection $f$ such that $f: \alpha \xrightarrow[\text{onto}]{1-1}A$
Additionally, because $\alpha$ is an ordinal, it is well ordered by $\in$.
(Not sure I can do this) In particular then, imagine an $\in_\alpha \subseteq \alpha \times \alpha \quad$ i.e. $\in_\alpha$ is the set of ordered pairs that well-orders $\alpha$.
Consider the following function with its accompanying mapping rule:
$g: \in_\alpha \to A \times A$
where $\forall \langle \alpha_0, \alpha_1 \rangle \in \ \in_\alpha: \quad$ $g\big(\langle\alpha_0,\alpha_1 \rangle \big)= \langle f(\alpha_0),f(\alpha_1) \rangle$
My belief is that $g[\in_\alpha]$ is effectively the $R'$ that I initially aimed to construct...i.e. $\quad R' = \text{ran}(g)$
It seems like properties 2-5 can all be proven by contradiction:
E.G.
Assume $R'$ is not irreflexive. Therefore, $\exists x \in A [xR'x]$. Let $\langle x,x \rangle \in R' \iff \langle x,x \rangle \in \text{ran}(g)$
By definition of $g$, this means that $\exists \delta \big( \delta \in \alpha \land f(\delta)=x \land g( \langle \delta, \delta \rangle) =\langle x,x \rangle\big)$.
But this implies that $\langle \delta, \delta \rangle \in \ \in_\alpha$, which is a contradiction because $\in_\alpha$ is irreflexive.
Does this general strategy work? Cheers~