Prove Zorn's Lemma is equivalent to the following statement

[VrhoZna]

Homework Statement

From Introduction to Set Theory Chapter 8.1 exercise 1.4

Prove that Zorn's Lemma is equivalent to the following statement:
For all $(A,\leq)$, the set of all chains of $(A,\leq)$ has an $\subseteq$-maximal element.[/B]

Homework Equations

N/A

The Attempt at a Solution

($\Rightarrow$): Suppose Zorn's Lemma holds and let $(A,\leq)$ be a partially ordered set and let C be its set of chains. It's clear that each element of C is bounded above ($X \subseteq A$ for each $X \in C$) and thus has C has a $\subseteq$-maximal element by Zorn's Lemma.

($\Leftarrow$): Suppose, for all $(A,\leq)$, the set of all chains of $(A,\leq)$ has an $\subseteq$-maximal element. Let ($P, \leq$) be a partially ordered set, C the set of chains of P, X a maximal element of C, and suppose that every chain of P is bounded above; we show that P has a $\leq$-maximal element. Since X is bounded above there exists a $c \in P$ such that $x \leq c$ for all $x \in X$. Now let $y \in \bigcup C = P$ (as the singleton subsets of P are trivially chains of P) such that $y \not\in X$ (if no such y exists then X = P and c is the greatest element of P and hence a $\leq$-maximal element). Then, if $c \leq y$, we have $x \leq y$ for all $x \in X$ and thus $X = X \cup \{y\}$ as $X \cup \{y\}$ is a chain of P and X is a $\subseteq$-maximal element of C, contradicting our choice of y. Hence $y \leq c$ for all $y \in A$ so c is the greatest element of A and thus a $\leq$-maximal element.[/B]

 

[zwierz]

Assume that

For all (A,≤)(A,\leq), the set of all chains of (A,≤)(A,\leq) has an ⊆\subseteq-maximal element.

Then in conditions of the Zorn lemma this maximal chain has an upper bound. This upper bound is a maximal element of $A$.

 

[VrhoZna]

I see now. If $y \in A$ and $c \leq y$, then X is a subset of the chain $X \cup \{y\}$ and so $X = X \cup \{y\}$ and $y \in X$ and we must have y = c.