Question on a step of "A Simple Proof of Zorn's Lemma" by Lewin

Solution 1:

We first show that $U$ is a chain. It’s not hard to see that $U$ is a chain. Suppose that $x,y\in U$; there are conforming sets $A_x$ and $A_y$ such that $x\in A_x$ and $y\in A_y$. Lemma $1$ implies that one of these sets is a subset of the other, so without loss of generality suppose that $A_x\subseteq A_y$. Then $x,y\in A_y$, and $A_y$ is a chain, so either $x\le y$, or $y\le x$.

Now we show that $U$ is well-ordered by $\le$. Suppose that $\varnothing\ne S\subseteq U$; we want to show that $S$ has a $\le$-least element. Fix $s\in S$; there is some conforming set $A$ such that $s\in A$. $A$ is well-ordered by $\le$, so let $a=\min\{x\in A\cap S:x\le s\}$; I claim that $a=\min S$. Let $x\in S$; if $x\in A$, then certainly $a\le x$, so suppose that $x\in S\setminus A$. There is some conforming $A_x$ such that $x\in A_x$, and clearly $A_x\nsubseteq A$, so $A$ is an initial segment of $A_x$. Since $x\in A_x\setminus A$, it’s clear that $a<x$ in this case as well and hence that $a=\min S$.

It remains to be shown that if $x\in U$, then $x=f(P(U,x))$. Let $x\in U$; there is a conforming set $A_x$ such that $x\in A_x$. By definition $x=f(P(A_x,x))$, so we’re done if we can show that $f(P(U,x))=f(P(A_x,x))$. It suffices to show that $P(U,x)=P(A_x,x)$. Clearly $P(A_x,x)\subseteq P(U,x)$, so we need only show that $P(U,x)\subseteq P(A_x,x)$, i.e., that if $y\in U$ and $y<x$, then $y\in A_x$. Suppose, then, that $y\in U$ and $y<x$; there is a conforming set $A_y$ such that $y\in A_y$. If $A_y\subseteq A_x$, then certainly $y\in A_x$, and we’re done. Otherwise, $A_x$ is an initial segment of $A_y$, and again $y\in A_x$, since $y<x$.