Let $M$ be a set and "$\le$" a well-ordering of $M$. For $x \in M$ define:
$$ M_{\le x} := \{ y \in M \ \vert\ y \le x \} $$
The map
$$ f : M \to \mathcal{P}(M) \ ,\ x \mapsto M_{\le x}$$
is injective and has the following property:
$$ x \le y \Leftrightarrow f(x) \subseteq f(y)$$
Furthermore the set $\{ \emptyset \} \cup f(M) \cup \{M\}$ is a maximal chain in $\mathcal{P}(M)$ (ordered by "$\subseteq$").
My question is: Does every maximal chain in $\mathcal{P}(M)$ derive this way? Is there a bijection between the well-orderings of $M$ and the maximal chains in its power set? I was not yet able to proof that.
I would be super glad if this was even provable without the axiom of choice.
This would give a short proof of the well-ordering theorem by Hausdorffs maximal principle. Also, if $(M,\le)$ is a partial ordered set and $K$ is a maximal chain in $\mathcal{P}(M)$, I think that $f^{-1}((\{\emptyset\} \cup f(M) \cup \{M\} ) \cap K)$ (where $f$ is now defined via the partial ordering "$\le$") is a maximal chain in $M$. So this would give a short proof of the maximal principle via well-ordering theorem. What do you think? I hope there is no obvious mistake…
Edit: Well, there was an obvious mistake and as you pointed out, maximal chains in $\mathcal{P}(M)$ really dont need to relate to well-orderings of $M$. My original goal was to characterize the statement "$M$ is well-ordable" to a statement like "the maximal principle applies to $M$", whatever this should mean. I know the proof about maximal elements/chains in the set of partial well-orderings (well-orderings on subsets of $M$) being equivalent to well-orderings of $M$ but this doesnt seem useful to construct a maximal chain in $M$ if $M$ is partial ordered.
Best Answer
No. Indeed, let $A\subset\mathcal{P}(M)$ be any chain that is not well-ordered (such a chain exists as long as $M$ is infinite; for instance you can just take an infinite descending sequence where you remove one element at a time). By Zorn's lemma, $A$ is contained in some maximal chain $B$ of $\mathcal{P}(M)$, which is not well-ordered since it contains $A$. Since all the chains you describe are well-ordered, this $B$ cannot be of that form.