I'm reading the proof of Zorn's lemma from Halmos' Naive Set Theory. There is a small detail (page 63) that I can't understand. What Halmos essentially writes is the following.
Let $X$ be a set partially ordered by $\preccurlyeq$ such that every chain in $X$ has an upper bound. Let $^1$ $\mathcal S:=\{\bar s(x):x\in X\}$. Let $\mathcal X$ be the set of all (and only) the chains in $X$. It is clearly seen that for any $\mathcal C\in \mathcal X$ (i.e., $\mathcal C$ is a chain in $X$) there is an $a\in X$ such that $\mathcal C\subseteq\bar s(a)\in\mathcal S$.
Then he makes makes the following comment.
Since each set in $\mathcal X$ is dominated by some set in $\mathcal S$, the passage from $\mathcal S$ to $\mathcal X$ cannot introduce any new maximal elements.
Question: What is meant by this comment? Does it mean that any maximal element of $\mathcal X$ belongs to $\mathcal S$? But this is not necessarily true! So I must me misinterpreting his comment. Can you please elaborate what Halmos means?
$^1$ For any $x\in X$, the initial segment of $x$ is defined as $\bar s(x):=\{a\in X:a\preccurlyeq x\}$.
On further thought, that is a bit sloppy. If we consider the four element Boolean algebra, it has two maximal chains but only one maximal initial segment (namely the whole thing).
What is true is that maximal elements in $\mathcal{X}$ yield maximal elements in $\mathcal{S}$, since every chain has an upper bound: if $A$ is a maximal chain and $a$ is an upper bound of $A$ (which exists by assumption), then $\overline{s}(a)$ is an initial segment containing $A$ and hence is maximal in $\mathcal{S}$.
(Why is $\overline{s}(a)$ maximal in $\mathcal{S}$ if $A$ is maximal in $\mathcal{X}$? Suppose $\overline{s}(b)\supsetneq\overline{s}(a)$. Then $b>a$. But this means $A\cup\{b\}$ is a chain properly containing $A$, which can't happen.)
Rephrasing the above, and this is really the point, we have:
This is what we really want for this proof.