"A Simple Proof of Zorn's Lemma" article. (explanation of a step)

Question

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:

1. The order $<$ is a well order of the set A.
2. 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 $P(A, z) = P(B, y)$. (???)

Answer 1

$P(A,z)\subseteq P(B,y):$

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, $\alpha\in P(A,z).$

Answer 2

Here I want to clarify part of the @Matematleta's proof which took some of my time.

On the other hand, if $<$, then since $\in (,)$, we have $\in (,)$, which is a contradiction.

We assume that $<$.

1) From $\in A\setminus (,)$, we obtain $z\in A \land [z\notin B\lor z\geqslant y]$

2) The second OR-case $z\geqslant y $ implies contradiction: $z\geqslant y >\alpha>z$. That means that $z\notin B$.

3) $z \in A \land z\notin B$ implies $z\in A\setminus B$. y is the least element of $A\setminus B$. So $y\leqslant z$.

4)Then $z<\alpha<x\leqslant z$ - contradiction.

That means that the assumption $<$ is false.

p.s. Proof of the 4th proposition: $\alpha < x$ here is from $\alpha\in P(A,x)$. In it's turn, this is from $\alpha\in B$ and $\alpha\notin B\setminus P(A,x)$. Proposition $\alpha\notin B\setminus P(A,x)$ is true because $\alpha\in B\setminus P(A,x)$ leads to a contradiction: $y\leqslant \alpha< y$.

Answer 3

Actually this proof is not correct!

Suppose that X is a partial order under ≤. Let Low(A) and Up(A) be respectively the sets of all strict lower bounds and all strict upper bounds of a subset A of X; and let P(A, x) be all strict predecessors of x in A, i. e. A ∩ Low({x}).

Let g a choice function that assigns to every non-empty set one of its elements; and let f be a function that assigns to every chain C in X the strict upper bound f(C) = g(Up(C)).

Now suppose that x and y ∈ X are incomparable (not x ≤ y and not y ≤ x) and that A = {x} and B = {y} and that g(the union of A and Up(A)) = x and g(the union of B and Up(B)) = y. So x = f(P(A,x)) and y = f(P(B,y)).

Now the sets A = {x} and B = {y} are conforming as defined in Lewin’s paper, because both are well-ordering and the second condition holds for both. But we supposed that x and y are incomparable; so are the sets A = {x} and B = {y}. So the whole lemma in the bottom of the first page of the paper is incorrect.