Proof of Zorn's lemma from the axiom of choice by good chains

Proof

We argue by contradiction. Let $(P, \le)$ be a nonempty partially ordered set in which each chain has an upper bound. Write $x < y$ for $x \le y$ and $x \neq y$.

1. Suppose that $P$ has no maximal element. For a chain $C$, let $S(C) = \{p \in P : c < p \text{ for all } c \in C\}$, the set of strict upper bounds of $C$.
2. Each $S(C)$ is nonempty. Let $u$ be an upper bound of $C$. Since $u$ is not maximal, some $v \in P$ has $u < v$.
3. For $c \in C$ we have $c \le u < v$, so $c \le v$. If $c = v$, then $v \le u \le v$ and $u = v$, which is false. So $c < v$, and $v \in S(C)$.
4. The sets $S(C)$, for all chains $C$, form a set of nonempty sets. By the axiom of choice, it has a choice function $\sigma$. Put $s(C) = \sigma(S(C))$ and $g(C) = C \cup \{s(C)\}$.
5. Then $s(C)$ is a strict upper bound of $C$. So $s(C) \notin C$, and $g(C)$ is a chain.
6. Let $C$ be a chain. A segment of $C$ is a subset $T \subseteq C$ such that $x \in C$, $y \in T$ and $x \le y$ imply $x \in T$.
7. A segment of a segment of $C$ is a segment of $C$. A union of segments of $C$ is a segment of $C$. Both follow at once from the definition.
8. Call a chain $C$ good when, for each segment $T$ of $C$ with $T \neq C$, the chain $g(T)$ is a segment of $C$.
9. Comparison lemma: if $C_1$ and $C_2$ are good, one of them is a segment of the other. Let $M$ be the union of all sets that are segments of both $C_1$ and $C_2$.
10. By step 7, $M$ is a segment of both $C_1$ and $C_2$. Suppose $M \neq C_1$ and $M \neq C_2$.
11. Since $C_1$ and $C_2$ are good, $g(M)$ is a segment of both. So $g(M) \subseteq M$ by the definition of $M$. This contradicts $s(M) \notin M$ from step 5.
12. Hence $M = C_1$ or $M = C_2$, and $C_1$ is a segment of $C_2$ or $C_2$ is a segment of $C_1$. This proves the lemma.
13. Let $U$ be the union of all good chains. Let $x, y \in U$, with $x \in C$ and $y \in D$ for good chains $C, D$. By step 12, one of $C, D$ contains the other. So $x$ and $y$ lie in one chain and are comparable. Thus $U$ is a chain.
14. Each good chain $D$ is a segment of $U$. Let $x \in U$, $y \in D$ and $x \le y$. Pick a good chain $C$ with $x \in C$.
15. By step 12, $C \subseteq D$, so $x \in D$; or $D$ is a segment of $C$, and then $x \in D$ because $x \in C$, $y \in D$ and $x \le y$.
16. The chain $U$ is good. Let $T$ be a segment of $U$ with $T \neq U$. Pick $u \in U \setminus T$ and a good chain $D$ with $u \in D$.
17. Let $t \in T$. If $u \le t$, then $u \in T$, since $T$ is a segment of $U$; this is false. As $U$ is a chain, $t < u$. Since $D$ is a segment of $U$ (step 14), $t \in D$.
18. So $T \subseteq D$. Then $T$ is a segment of $D$, because it is a segment of the larger set $U$. Also $T \neq D$, because $u \in D \setminus T$.
19. As $D$ is good, $g(T)$ is a segment of $D$. As $D$ is a segment of $U$, step 7 shows that $g(T)$ is a segment of $U$. So $U$ is good.
20. Let $w = s(U)$ and $V = g(U) = U \cup \{w\}$. By step 5, $V$ is a chain and $w > x$ for all $x \in U$.
21. The set $U$ is a segment of $V$. Indeed, if $x \in V$, $y \in U$ and $x \le y$, then $x \neq w$, since $w \le y$ would give $w \le y < w$. So $x \in U$.
22. The chain $V$ is good. Let $T$ be a segment of $V$ with $T \neq V$. If $w \in T$, then each $x \in U$ has $x \le w$, so $x \in T$, and $T = V$, which is false. So $T \subseteq U$, and $T$ is a segment of $U$.
23. If $T = U$, then $g(T) = V$ is a segment of $V$. If $T \neq U$, then $g(T)$ is a segment of $U$ by step 19, hence of $V$ by steps 21 and 7.
24. So $V$ is a good chain, and $V \subseteq U$ by the definition of $U$. Then $w \in U$, which contradicts $w \notin U$ from step 5.
25. Hence the assumption in step 1 is false, and $P$ has a maximal element.