Every well-ordered set has a "Parity" function. The problem is this:
Given a well ordered set $U$, there exists a unique function $$Par: U \rightarrow \mathbb{N}$$ such that $Par(y) = 0$ if $y$ is $0$ or a limit point, and at successor points $x$, $$Par(S(x)) = 1 - Par(x).$$
[Sidenote:] This $Par$ function seems to determine whether the 'distance' of $x$ to a limit point $y$ below $x$ is odd or even. Is this correct? (Not part of the homework, purely a curiosity.)
My idea for the problem is to somehow apply the Transfinite Recursion Theorem. However, to use this, I am required to define an $h: ( U \rightharpoonup \mathbb{N}) \times U \rightarrow \mathbb{N}$ so that $$Par(x) = h(Par|_{seg(x)}, x).$$ My attempts at a construction of such an $h$ seem somewhat "circular", unfortunately. 
Any hints or tips are appreciated. Please no direct answers. :P 
 A: Applications of transfinite recursion often do have a circular feel to them. However, just as with proof by induction, the circularity is only apparent.
Note that the description of $Par(x)$ involves $Par$ of the predecessor of $x$, if there is one. So it's not like, e.g., proving that $\sum_{i=1}^n i = n(n+1)/2$ by induction, where at least the result to be proved isn't "inductive". 
It may help to translate your condition on $h$: $$Par(x) = h(Par|_{seg(x)}, x)$$ into prose:

Find a rule (i.e., $h$) which, when given an element $x\in U$ and a function $f$ defined for all predecessors of $x$, yields a value $h(f,x)$ so that if $f$ is extended to $\hat{f}$ by setting $\hat{f}(x)=h(f,x)$, then if the original $f$ satisfied the conditions given for $Par$, the extended function $\hat{f}$ also satisfies these conditions.

By the "conditions given for $Par$", I mean (1) $f(y)=0$ for $y=0$ or a limit; (2) $f(S(y)) = 1-f(y)$ provided $S(y)$ precedes $x$.
Oh yes, your intuition about the meaning of Par is correct.
