There is a short proof of Zorn's lemma by J.W.Lewin:
A Simple Proof of Zorn's Lemma.pdf
Summary:
I don't understand this step:
"Therefore if z is the least member of
$A\setminus P(B, y)$, we have $P(A, z) = P(B, y)$."
I'm interested in both inclusions from left to right and from right to left.
A subset A of X is conforming if the following two conditions hold:
- The order $<$ is a well order of the set A.
- For every element $x\in A$, we have $x = f(P(A, x))$.
Theorem
If A and B are conforming subsets of X and $A\neq B$, then one of these two sets is an initial segment of the other.
Proof We may assume that $A\setminus B\neq\emptyset$. Define x to be the least member of
$A\setminus B$. Then $P(A, x) \subseteq B$. We claim that $P(A, x) = B$. To obtain a contradiction,
assume that $B\setminus P(A, x)\neq\emptyset$, and define y to be the least member of
$B\setminus P(A, x)$. Given any element $u\in P(B, y)$ and any element $v\in A$ such that
$v < u$, it is clear that $v \in P(B, y)$. Therefore if z is the least member of
$A\setminus P(B, y)$, we have $P(A, z) = P(B, y)$. (???)
Best Answer
$P(A,z)\subseteq P(B,y):$
if $\alpha\in P(A,z),$ then $\alpha\in A$ and $\alpha<z$ so $\alpha$ is in $A$ but $\textit{not}$ in $A\setminus P(B,y)$ (by definition of $z$), so in fact $\alpha\in P(B,y).$
$P(B,y)\subseteq P(A,z):$
if $\alpha\in P(B,y),$ then $\alpha\in B$ and $\alpha<y$ so $\alpha\in P(A,x)$ (by definition of $y$). Now, $\alpha \neq z$ (because $z\notin P(B,y)$). On the other hand, if $z<\alpha$, then since $\alpha\in P(B,y),$ we have $z\in P(B,y)$, which is a contradiction. Therefore, we have that $\alpha<z$ and $\alpha\in A.$ That is, $\alpha\in P(A,z).$