Prove that every well-ordered proper class has an initial segment order isomorphic to the ordinal numbers, ON.
I have a plan to prove this but it uses a recursive definition and induction which I do not have enough experience with to construct this proof. So any help or advice would be greatly appreciated!
1) Let W be a well-ordered proper class
2) Recursively define a function $f:ON \to W$ so that $\beta$ maps to the least element of W not in $f[\beta]$
3) Use induction to show that this is an order embedding onto an initial segment
Yes the idea is correct.
Recall that an ordered class $W$ is well-ordered if every non-empty set has a least element. But equivalently this is the same as saying that every non-empty subclass has a least element. We define $f$ by recursion.
Suppose that $f$ was defined for every $\alpha<\beta$. Then $f[\beta]$ is a set, therefore $W\setminus f[\beta]$ is a proper class, and in particular non-empty. Let $f(\beta)$ be the least element of $W\setminus f[\beta]$.
The proof that this is an embedding from $\sf Ord$ into $W$ whose range is an initial segment is similar, we use transfinite induction and our induction hypothesis is this:
For every $\alpha<\beta$, $f[\alpha]$ is an initial segment of $W$, and if $\gamma<\delta<\alpha$ then $f(\gamma)<f(\delta)$.
Then the above definition of $f$ makes it almost trivial that this is true for $\beta+1$, and for limit stages it's as simple.