Munkres Introduction to Topology: Section 11 Problems 5, 6 and 7
I'm trying a new format where I will group relevant problems and their solutions together. In my humble opinion, these three problems are three steps of one problem to show that
a bunch of lemmas (and a principle) are equivalent to each other.
Problem 5
Lemma (Kuratowski). Let $\mathcal{A}$ be a collection of sets. Suppose that for every
subcollection $\mathcal{B}$ of $\mathcal{A}$ that is simply ordered by proper inclusion, the union
of the elements of $\mathcal{B}$ belongs to $\mathcal{A}$. Then $\mathcal{A}$ has an element that
is properly contained in no other element of $\mathcal{A}$.
We are to show that this theorem is implied by Zorn's lemma. Consider some simply ordered collection of sets
$\mathcal{B} \ \subset \ \mathcal{A}$. Now, consider the set:
$$S \ = \ \displaystyle\bigcup_{B \in \mathcal{B}} B$$
By the initial assumption, $S \ \in \ \mathcal{A}$. It is also clear that for any $B \ \in \ \mathcal{B}$,
it will be true that $B \ \subset \ S$, or $B \ = \ S$. Since proper inclusion is the order relation on
$\mathcal{B}$, it is true that $B \ \prec \ S$ or $B \ = \ S$. By definition, $S$ is thus an upper bound on
$\mathcal{B}$. Since any such $\mathcal{B}$ is simply ordered, Zorn's lemma tells us that $\mathcal{A}$ has a
maximal element. By definition, this is an element $A \ \in \ \mathcal{A}$ such that $A \ \prec \ C$ never holds
true for another $C \ \in \ \mathcal{A}$. Since the partial order being considered is proper inclusion, it follows
that there is no $C$ such that $A \ \subset \ C$. Thus, $A$ is an element of $\mathcal{A}$ that is properly contained
in no other element of $\mathcal{A}$ and the proof is complete.
Problem 6
Lemma (Tukey). Let $\mathcal{A}$ be a collection of sets. If $\mathcal{A}$ is of finite type,
then $\mathcal{A}$ has an element has an element that is properly contained in no other element of
$\mathcal{A}$.
Consider some collection $\mathcal{B} \ \in \ \mathcal{A}$ that is simply ordered by proper inclusion.
Let us pick some $B \ \in \ \mathcal{B}$. By the initial assumption, it is true that every finite
subset of $B$ is contained in $\mathcal{A}$. Now, let us consider the set:
$$S \ = \ \displaystyle\bigcup_{B \ \in \ \mathcal{B}} B$$
Let $X$ be a finite subset of $S$. Each $x \ \in \ X$ must, by definition, be contained in some $B \ \in \ \mathcal{B}$. We invoke the axiom of
choice (the existence of a choice function) and assert that we can choose some $B_x \ \in \ \mathcal{B}$ that contains each $x$. Since
there are a finite number of elements $x$, the set of $B_x$ sets is finite, and thus has a largest element in the simple order relation imposed
on $\mathcal{B}$. We call this maximal element $B'$. Since the simple order relation is proper inclusion, every other $B_x$ will be contained
in $B'$. Thus, $X \ \subset \ B'$ (note that this is not necessarily proper inclusion). This means that $X$ is a finite subset of $B'$.
Since each finite subset of $B'$ is
contained in $\mathcal{A}$, it follows that any such $X$ (every finite subset of $S$) is contained in $\mathcal{A}$.
By the initial assumption, it must be true that $S \ \in \ \mathcal{A}$. From this, Kuratowski's lemma implies that there exists some element of $\mathcal{A}$ that
is properly contained in no other element of $\mathcal{A}$. This completes the proof.
Problem 7
Show that the Tukey lemma implies the Hausdorff maximal principle.
Munkres gives us a pretty big hint here. We let $\prec$ be a strict partial order on $A$, and let $\mathcal{A}$ be
the set of subsets of $A$ that are simply ordered by $A$. Munkres tells us to show that $\mathcal{A}$ is of finite type.
First, let us consider some subset $B$ of $A$ such that each finite subset of $B$ belongs to $\mathcal{A}$. This means that
each finite subset of $B$ is simply ordered by the strict partial order. We wish to prove that $B$ is also contained within
$\mathcal{A}$. In order to for this to be true, $B$ must also be ordered by the strict partial order. Let us consider the union
of all order relations defined on each finite subset of $B$. We prove that this is a simple order on $B$.
One can see that either $a \ \prec \ b$ or $b \ \prec \ a$, for any $a$ and $b$ in $B$, as the finite subset $\{a, \ b\}$ of $B$ is contained
within $\mathcal{A}$, and is thus ordered. By definition of the partial order, it must be
true that $b \ \prec \ b$ is not contained in the new order relation, and that for $x \ \prec \ z$ and $z \ \prec \ y$ that $x \ \prec \ y$. Thus,
we have an order relation on $B$, and $B \ \in \ \mathcal{A}$.
We now have to prove that if $B$ is in $\mathcal{A}$, then so will each of its finite subsets. Let $C$ be the order relation on $B$. We define the following
procedure for defining an order relation on any finite subset $S$ of $B$: Take some element of $(x, \ y) \ \in \ C$. If $x$ and $y$ are in $S$, add it to $C_S$,
the order relation on $S$. Otherwise, throw it out. Clearly, all elements of the new order relation will be comparable, and by definition of the partial order, the other
two axioms will hold. Thus, $C_S$ is a simple order relation on $S$.
With the knowledge that $\mathcal{A}$ is of finite type, we can prove the Hausdorff maximum principle. The Hausdorff maximum principle
says that given a set $A$ with a partial order, there exists a maximal simply ordered subset of $A$. Since $\mathcal{A}$ is of finite
type, Tukey's lemma implies that there exists an element $X$ of $\mathcal{A}$ that is properly contained in no other element of $\mathcal{A}$.
Said differently, $X$ is a simply ordered set that is not properly contained in any other simply ordered set. This is the definition of a maximal set, and
thus, Tukey's lemma is equivalent to the maximum principle.
Here is visual representation of what we now know:
$$\text{Hausdorff maximum principle} \ \rightarrow \ \text{Zorn's lemma} \ \rightarrow \ \text{Kuratowski's lemma} \ \rightarrow \ \text{Tukey lemma} \ \rightarrow \ \text{Hausdorff maximum principle}$$
Thus, each of these statements imply one another!