Every element in well-ordered set can uniquely be expressed as $y = S^n(x)$

Question

Let $U$ be a well-ordered set. If $y \in U$ then $y$ can be expressed uniquely on the form,

$y = S^n(x)$

Where $x$ is either the least element of $U$ or a limit point, $n \in \mathbb{N}$ and $S$ is a recursive function defined by

$S^0(x) = x, \ S^{i+1}(x) =S(S^i(x))$

I wish I could show some effort I'm not even sure where to start. The result feels intuitively very good since well-ordered sets behaves in a good manner and feels in my mind somewhat similar to $\mathbb{N}$ so it's maybe not a big surprise that one could express every element via a recursive successor function, but how do I go about proving this?

Should I go for a contraction by assuming there is some $y \in U : y \neq S^{n}(x)\ \forall \ n \in \mathbb{N} $? Can I invoke the Transfinite Recursion Theorem?

For uniqueness, I guess I should to something like this: Let $y\in U:y = S^{n}(x) \land y = S^{n'}(x) $ and then somehow derive that we must have that $n=n'$.

I would really appreciate any help on this.

Answer 1

Prove this by transfinite induction. (I suppose that $S(x)$ returns the successor of $x$, otherwise you might want to actually define what $S$ is in the question.)

If $y$ is the least element or a limit ordinal, then certainly this is true since $y=S^0(y)$.

Now suppose that $y$ is not the least element nor it is an ordinal. What is it?

To show uniqueness, show with a similar argument, that $S$ is injective (and therefore $S^n$ is injective too).

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On a side remark, I gave my students a similar exercise last semester. This is truly something that should be done after talking about von Neumann ordinals and ordinal arithmetic. Doing that before these two topics is difficult, filled with bad and ad hoc notations and terminology, and unclear.

After knowing what is the von Neumann ordinal assignment, what is a successor ordinal, and the basics of ordinal addition, this becomes an incredibly straightforward exercise.

Answer 2

We have the successor function

$\quad S: x \mapsto \text{LeastElement(}\{y \gt x\}\text{)}$

defined on all elements of $U \setminus \{\text{max(}U\text{)}\}$.

Let $L$ denote the elements of $U$ that do not have an immediate predecessor, i.e. elements of $U$ not in the range of $S$.

Let $y$ be any element in $U$ and define

$\tag 1 C_y = \{ x \in U \, | \, \text{There exists an integer } n \ge 0 \text{ such that } y = S^n(x) \}$

Since $y \in C_y$, the set $C_y$ is non-empty and has a least element that must also belong to $L$.

So $y$ has the form $S^n(\alpha)$ with $\alpha \in L$.

Exercise: For each $\alpha \in L$ define

$\tag 2 L_\alpha = \{ S^n(\alpha) \, | \, \text{integer } n \ge 0\}$

Show that the family $(L_\alpha)_{\, \alpha \in L}$ of sets is a partition of $U$.

Hint: Show that if $\alpha, \beta \in L$ then $S^n(\alpha) = S^m(\beta)$ implies that $\alpha = \beta$.

We conclude that for any element $y \in U$ there corresponds a unique $\alpha \in L$ and $n \ge 0$ such that

$\tag 3 y = S^n(\alpha)$