Need clarification of "conforming subsets" in “A Simple Proof of Zorn's Lemma” by J. Lewin I am having difficulties with the definition of a "conforming subset" in J. Lewin's A Simple Proof of Zorn's Lemma:
The definition includes the statement: "For every element of $x\in A$, we have $x=f(P(A,x))$" where $f$ is a choice function that "assigns to every chain in $X$, a strict upper bound" and $P(A,x) = \{y\in A ~|~ y \lt x\}$.
My Question: Why is the choice function necessary here? Isn't $x$, by definition, a strict upper bound of the initial segment (and chain) $P(A,x)$?   
 A: The conforming subsets are unique for each order type, and their exact elements depends on the choice of the choice function.
The point of this definition is to hide the transfinite recursion, but it is exactly that. A transfinite recursion. It means that we start with $\varnothing$, which is a chain, and use the choice function to select an upper bound of this chain. Then we have some $\{x\}$, where $x$ was the upper bound chosen for $\varnothing$; so it is either that $x$ is maximal, or the chain has a strict upper bound, say $y$, so we have the chain $\{x,y\}$, etc.
Now you can still ask, why do we need the choice function? In general there is no unique choice of upper bounds. And we will generally need to make infinitely many of these choices. 
When looking at this from the recursion point of view, the choice function is needed for the recursion process to be well-defined to begin with. Which is necessary for the conforming subset definition to be useful (i.e., to produce a chain, rather than just a subset).
