This question is continued from: Explanation of Proof of Zorn's lemma in Halmos's Book

From the Book:

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.

Let $f$ be a choice function for $X$, that is, $f$ is a function from the collection of all non-empty subsets of $X$ to $X$ such that $f(A) \in A$ for all $A$ in the domain of $f$. For each set $A$ in $\mathscr{X}$, let $\hat{A}$ be the set of all those elements $x$ of $X$ whose adjunction to $A$ produces a set in $\mathscr{X}$; in other words, $\hat{A} = \{ x \in X: A \cup \{ x \} \in \mathscr{X} \}$. Define a function $g$ from $\mathscr{X}$ to $\mathscr{X}$ as follows: if $\hat{A} — A \neq \varnothing$, then $g(A) = A \cup \{ f(A' - A) \}$; if $\hat{A} - A = \varnothing$, then $g(A) = A$. It follows from the definition of $\hat{A}$ that $\hat{A} — A = \varnothing$ if and only if $A$ is maximal. In these terms, therefore, what we must prove is that there exists in $\mathscr{X}$ a set $A$ such that $g(A) = A$. It turns out that the crucial property of $g$ is the fact that $g(A)$ (which always includes $A$) contains at most one more element than $A$.

Now, to facilitate the exposition, we introduce a temporary definition. We shall say that a subcollection $\mathscr{J}$ of $\mathscr{X}$ is a tower if

  1. $\varnothing \in \mathscr{J}$,
  2. if $A \in \mathscr{J}$, then $g(A)\in \mathscr{J}$,
  3. if $\mathscr{C}$ is a chain in $\mathscr{J}$, then $\bigcup_{A \in \mathscr{C}} A \in \mathscr{J}$.

Towers surely exist; the whole collection $\mathscr{X}$ is one. Since the intersection of a collection of towers is again a tower, it follows, in particular, that if $\mathscr{J}_0$ is the intersection of all towers, then $\mathscr{J}_0$ is the smallest ower. Our immediate purpose is to prove that the tower $\mathscr{J}_0$ is a chain.

Question: Aren't the definitions of a tower and the set $\mathscr{X}$ the same? $\mathscr{X}$ is surely a tower but was it necessary to introduce this definition of towers, or could we just consider the set of sets that had the properties of $\mathscr{X}$ and then take their intersection to get Jo?

  • $\begingroup$ It is all here. This author from India has filled in all the missing gaps. arxiv.org/pdf/1207.6698.pdf $\endgroup$ – user67321 Mar 18 '13 at 16:46

The definitions of a tower and $\mathscr{X}$ are not quite the same.

Recall that $\mathscr{X}$ was posited to be some collection of subsets of a set $X$ which satisfy three properties:

  1. $\mathscr{X}$ is nonempty;
  2. every subset of an element of $\mathscr{X}$ is in $\mathscr{X}$; and
  3. the union of every chain of sets in $\mathscr{X}$ is in $\mathscr{X}$.

The main differences between this and and the definition of a tower are two-fold:

  • A tower is some subcollection of the collection $\mathscr{X}$ we began with; being a tower is somewhat dependent on the collection $\mathscr{X}$ we began with, and different "choices" of $\mathscr{X}$ may lead to different towers.
  • Condition (2) is quite different. While for $\mathscr{X}$ we demand that all subsets of a set in $\mathscr{X}$ are also in $\mathscr{X}$, to be a tower a collection $\mathscr{J}$ must satisfy that for each $A \in \mathscr{J}$ some other particular set related to $A$, which he denotes $g(A)$, must also belong to $\mathscr{J}$; furthermore, $g (A)$ is actually a superset of $A$, and not necessarily a subset.

Whether or not it was necessary to introduce towers is a more philosophical question, but their introduction does facilitate the remainder of the proof because we get a smallest tower, which he denotes $\mathscr{J}_0$. It turns out that $\mathscr{J}_0$ is very closely related to the object of our search -- a maximal set in $\mathscr{X}$.


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