Help with the proof of Zorn's Lemma: if $A$ and $B$ are conforming subsets, why must it be that $A\subseteq B$ or $B\subseteq A$?

Question

Here we can find a simple proof of Zorn's Lemma, but I find something that I can't really understand. There is an statement that say: 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.

If I take, by example, the set $X=\{0,1,2,3,4,5,6\}$, I can create two conforming subsets which satisfy the definition but aren't initial segment of each other, like $B=\{0,2,4,5\}$ and $C=\{0,2,3,4\}$.

This proof is similar to another used by Gorodentsev in his book Algebra I, so there is a thing that I am missing here.

---

Definition

We shall say that a subset $A$ of $X$ is conforming if the following two conditions hold:

1. The order $\le$ is a well order of the set A.
2. For every element $x \in A$, we have $x = f(P(A, x))$, where $f$ maps the initial segment to its strict upper bound x.

Definition

If $C$ is a chain in $X$ and $x\in C$, then we define $P(C, x) = \{ y \in C: y < x\}$. A subset of a chain $C$ that has the form $P(C, x)$ is called an initial segment in $C$.

Answer 1

No. You can't.

If $A$ is conforming, then its minimum element is $f(\varnothing)$. So that's unique, say $x_0$; then the next element is the one chosen by $f$ as an upper bound of $\{x_0\}$. So again, that's unique, call that $x_1$, etc. etc.

By transfinite induction we get that if $A$ and $B$ are both conforming, they must have been built by essentially iterating the choice function to a certain length. Since a choice function is a function its output are well-defined, that means that $A\subseteq B$ or $B\subseteq A$.

The reason you think you have two conforming subsets is that you didn't specify the choice function which maps a chain to an upper bound. Once you do, you have to ask yourself, is $f(\{x\mid x>2\})$ going to be equal to $3$ or to $4$?

Answer 2

Probably the error is here:

If I take, by example, the set $X=\{0,1,2,3,4,5,6\}$, I can create two conforming subsets which satisfy the definition but aren't initial segment of each other, like $B=\{0,2,4,5\}$ and $C=\{0,2,3,4\}$.

You need to define the function $f$ first, which assigns a strict upper bound to a set, and then show your set satisfies $x = f(P(B,x))$ for all $x\in B$ to say $B$ is conforming. And similarly for $C$.

That is, $$\begin{align} &\text{for } B & &\qquad \text{for } C \\ & f(\{\}) & = 0 & \qquad \text{same} \\ & f(\{0\}) & = 2 & \qquad \text{same} \\ & \color{red}{f(\{0,2\})} & = \color{red}4 & \qquad \color{red}{f(\{0,2\})} & = \color{red}3 \\ & f(\{0,2,4\}) & = 5 & \qquad f(\{0,2,3\}) & = 4\\ \end{align}$$

The third row shows at least one of $B, C$ can not be 'conforming'.