Hausdorff's maximality principle lemma in Rudin Real and Complex analysis Here's the lemma that is needed before proving the Hausdorff's maximality principle (in Rudin's real and complex analysis appendix):

Lemma Suppose $\mathcal{F}$ is a nonempty collection of subsets of a set $X$ such that the union of every subchain of $\mathcal{F}$ belongs to $\mathcal{F}$. Suppose $g$ is a function which associates to each $A \in \mathcal{F}$ a set $g(A) \in \mathcal{F}$ such that $A \subset g(A)$ and $g(A) - A$ consists of at most one element. Then there exists an $A \in \mathcal{F}$ for which $g(A) = A$.

In the proof, it first defines a tower:

Fix $A_0 \in \mathcal{F}$. Call a subcollection $\mathcal{F}'$ of $\mathcal{F}$ a tower if $\mathcal{F}'$ has the following three properties: 
(a) $A_0 \in \mathcal{F}'$
(b) The union of every subchain of $\mathcal{F}'$ belongs to $\mathcal{F}'$
(c) If $A \in \mathcal{F}'$, then also $g(A) \in \mathcal{F}'$

In the proof, we let $\mathcal{F}_0$ be the intersection of all towers which is also a tower (proof is trivial). Then we go on to define two more collection of sets:

Let $\Gamma$ be the collection of all $C \in \mathcal{F}_0$ such that every $A \in \mathcal{F}_0$ satisfies either $A \subset C$ or $C \subset A$.
For each $C \in \Gamma$, let $\Phi(C)$ be the collection of all $A \in \mathcal{F}_0$ such that either $A \subset C$ or $g(C) \subset A$.

In other words
\begin{align*}
\Gamma &= \{ C \in \mathcal{F}_0 : A \subset C \text{ or } C \subset A \; \text{ for every } A \in \mathcal{F}_0 \}\\
\Phi(C) &= \{ A \in \mathcal{F}_0: A \subset C \text{ or } g(C) \subset A \}
\end{align*}
Now in the proof, it proves that $\Phi(C)$ is a tower by showing that it meets all three properties shown above. The first two properties can be shown easily. For the third property, we do the following:

If $A \in \Phi(C)$ there are three possibilities: Either $A \subset C$ and $A \neq C$, or $A = C$, or $g(C) \subset A$. If $A$ is a proper subset of $C$, then $C$ cannot be a proper subset of $g(A)$, otherwise $g(A) - A$ would contain at least two elements; since $C \in \Gamma$, it follows that $g(A) \subset C$. If $A = C$, then $g(A) = g(C)$. If $g(C) \subset A$, then also $g(C) \subset g(A)$ since $A \subset g(A)$. Thus $g(A) \in \Phi(C)$, and we have proved that $\Phi(C)$ is a tower. The minimality of $\mathcal{F}_0$ implies now that $\Phi(C) = \mathcal{F}_0$, for every $C \in \Gamma$

I am stuck on the last claim in the proof where it says The minimality of $\mathcal{F}_0$ implies now that $\Phi(C) = \mathcal{F}_0$, for every $C \in \Gamma$. What exactly does the minimality mean in this context and how does that lead to the equality? My guess is that $\Phi(C)$ is the smallest tower containing $C$, and since $\mathcal{F}_0$ is also the smallest tower containing $C$, they must be equal?
Also the proof says that

In other words, if $A \in \mathcal{F}_0$ and $C \in \Gamma$, then either $A \subset C$ or $g(C) \subset A$. But this says that $g(C) \in \Gamma$. Hence $\Gamma$ is a tower, and the minimality of $\mathcal{F}_0$ shows that $\Gamma = \mathcal{F}_0$. It follows from the definition of $\Gamma$ that $\mathcal{F}_0$ is totally ordered.

Here, since $\Phi(C) = \mathcal{F}_0$, for any $A \in \mathcal{F}_0$, either $A \subset C$ or $g(C) \subset A$. Since $C \subset g(C)$, we can say that for all $A \in \mathcal{F}_0$, either $A \subset g(C)$ or $g(C) \subset A$, which implies that $g(C) \in \Gamma$. Hence this satisfies property (c) of the definition of tower. Here they again use the minimality condition to show that $\Gamma = \mathcal{F}_0$ which I'm not sure how they did. And how does this result in $\mathcal{F}_0$ being totally ordered?
 A: 
What exactly does the minimality mean in this context and how does that lead to the equality?

Recall that $\mathscr F_0$ is the intersection of all towers. Hence it is a subset of every tower, and is thus the smallest possible tower. But $\Phi(C)$ is also a tower, and a subset of $\mathscr F_0$. Since they are both subsets of the other, $\Phi(C) = \mathscr F_0$.
The same is true for $\Gamma$. By its definition, it is a subset of $\mathscr F_0$, and now you have shown it is a tower, so  $\mathscr F_0$ is a subset of it. So $\Gamma = \mathscr F_0$.
You will encounter this sort of construction a lot. If you have a property on a collection of sets which is preserved under intersection (including infinite intersections), then the intersection of all sets with the property will be the smallest set having that property. Similarly if a property is preserved under unions, then the union of all sets having the property will be the largest set having that property. Two simple examples involving a set $A$ in a topological space:

*

*The union of all open sets within $A$ is the largest open set in $A$. This is the interior of $A$.

*The intersection of all closed sets containing $A$ is the smallest closed set containing $A$. This is the closure of $A$.

As for totally ordered, $\subset$ is already a partial order on $\mathscr F_0$, so to get totally ordered, all that is needed is to show that every two elements of $\mathscr F_0$ are comparable. But $\Gamma$ is by definition every set which is comparable to all sets in $\mathscr F_0$. And since it turns out $\Gamma$ is $\mathscr F_0$, that means every set in $\mathscr F_0$ is comparable to every other.
