I have been reading Halmos's book on naive set theory on my own and have got stuck in the Chapter on Zorn's lemma. The 2nd and 3rd paragraphs are not very clear to me. Here is the text:
Zorn's lemma. If $X$ is a partially ordered set such that every chain in $X$ has an upper bound, then $X$ contains a maximal element.
Proof. The first step is to replace the abstract partial ordering in $X$ by the inclusion order in a suitable collection of sets. More precisely, we consider, for each element $x \in X$, the weak initial segment $\bar{s}(x)$ consisting of $x$ and all its predecessors. The range $\mathscr{S}$ of the function $\bar{s}$ (from $X$ to$\wp(X)$) is a certain collection of subsets of $X$, which we may, of course, regard as (partially) ordered by inclusion. The function $\bar{s}$ is one-to-one, and a necessary and sufficient condition that $\bar{s}(x)\subseteq \bar{s}(y)$ is that $x\leq y$. In view of this, the task of finding a maximal element in $X$ is the same as the task of finding a maximal set in $\mathscr{S}$. The hypothesis about chains in $X$ implies (and is, in fact, equivalent to) the corresponding statement about chains in $\mathscr{S}$.
Let $\mathscr{X}$ be the set of all chains in $X$; every member of $\mathscr{X}$ is included in $\bar{s}(x)$ for some $x \in X$. The collection $\mathscr{X}$ is a non-empty collection of sets, partially ordered by inclusion, and such that if $\mathscr{C}$ is a chain in $\mathscr{X}$, then the union of the sets in $\mathscr{C}$ (i.e., $\bigcup_{A \in \mathscr{C}}A$) belongs to $\mathscr{X}$. Since each set in $\mathscr{X}$ is dominated by some set in $\mathscr{S}$, the passage from $\mathscr{S}$ to $\mathscr{X}$ cannot introduce any new maximal elements. One advantage of the collection $\mathscr{X}$ is the slightly more specific form that the chain hypothesis assumes; instead of saying that each chain $\mathscr{C}$ has some upper bound in $\mathscr{S}$, we can say explicitly that the union of the sets of $\mathscr{C}$, which is clearly an upper bound of $\mathscr{C}$, is an element of the collection $\mathscr{X}$. Another technical advantage of $\mathscr{X}$ is that it contains all the subsets of each of its sets; this makes it possible to enlarge non-maximal sets in $\mathscr{X}$ slowly, one element at a time.
Now we can forget about the given partial order in $X$. In what follows we consider a non-empty collection $\mathscr{X}$ of subsets of a non-empty set $X$, subject to two conditions: every subset of each set in $\mathscr{X}$ is in $\mathscr{X}$, and the union of each chain of sets in $\mathscr{X}$ is in $\mathscr{X}$. Note that the first condition implies that $\varnothing\in\mathscr{X}$. Our task is to prove that there exists in $\mathscr{X}$ a maximal set.
and the proof continues...
In the 2nd paragraph:
'Since each set in $\mathscr{X}$ is dominated by some set in $\mathscr{S}$, the passage from $\mathscr{S}$ to $\mathscr{X}$ cannot introduce any new maximal elements.'
Here I am able to prove that every maximal element of $\mathscr{S}$ has to be a maximal element of $\mathscr{X}$, but surely there can be maximal elements of $\mathscr{X}$ which are not maximal elements of $\mathscr{S}$ and hence 'extra'- please explain..
In the 3rd para- The author considers a set $\mathscr{X}$ with the given properties, and states that the problem of finding a maximal element in $X$ is equivalent to finding a maximal set in $\mathscr{X}$. How come?
Detailed but simple answers would be much appreciated.