Prove that Hausdorff's Maximal Principle is equivalent to the following: If $(A,\preceq)$ is a partially ordered set and $B$ is a chain of $A$, then $A$ has a maximal chain $\mathfrak{C}$ such that $B\subseteq\mathfrak{C}$.
Proof: Let $(A,\preceq)$ and $B$ is a chain of $A$. let's consider $\mathfrak{F}$ as the collecction of all the chains of $A$ that continent to $B$ ordered by inclusion, by Hausdorff's Maximal principle, $\mathfrak{F}$, has a maximal chain $\mathfrak{C}$; let $K=\bigcup \mathfrak{C}$, and it will be shown that $K$ is a maximal chain of $A$ that continent to $B$:
Let $x,y \in K$, then $x\in D$ and $y \in E$, for some $D,E\in \mathfrak{C} $, since $\mathfrak{C}$ is a chain, $D\subseteq E$, or $E\subseteq D$, then $x,y\in D$ or $x,y\in E$, in both cases $x,y$ are comparable.
from the above it is concluded that $K$ is upper bound of $A$, and by Zorn's Lemma, $\mathfrak{F}$ has a maximal element. which is precisely $K$.
Hence $B\subseteq \mathfrak{C}$.
I would like to know if the proof is completely correct, I have doubts about the other equivalence, I would like to know if it is possible to receive a suggestion.
Best Answer
The proof uses Zorn's lemma incorrectly. In fact, you shouldn't be using Zorn's lemma at all. But in any case, to apply Zorn's lemma you need to show that every chain has an upper bound, and you've only showed that one chain has an upper bound.
Instead, you need to use the following fact:
Then you move directly from verifying that $K$ is an upper bound of $\frak C$ to $K$ being a maximal element of $\frak F$.
Alternatively, use Zorn's lemma directly by showing that the union of a chain of chains in $A$ is itself a chain in $A$, and therefore by Zorn's lemma there exists a maximal element.