# Well ordering and maximal chains in power set

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.

• As for your claim at the end, why would $f^{-1}(f(M)\cap K)$ be maximal? You can find counterexamples to that even when $M$ has just two elements (try it!). – Eric Wofsey Dec 22 '18 at 2:26
• Sorry, I wanted to write "$f^{-1}((\{ \emptyset\} \cup f(M) \cup \{M\}) \cap K)$". It will be corrected in the post. – Lucina Dec 22 '18 at 2:33
• If $M$ is a countably infinite set, then the chains of the sort you describe are countable chains, but there are also uncountable chains in $\mathcal P(M)$. For instance, the set of lower Dedekind cuts is an uncountable chain in the power set of the rational numbers. – bof Dec 22 '18 at 2:41
• @Lucina: That doesn't help. The point is that maximality of $K$ in no way guarantees that $f(M)\cap K$ is large, since most elements of $f(M)$ could be incomparable to elements of $K$. If you don't see this, I recommend actually thinking through what could happen if $M$ has two elements. – Eric Wofsey Dec 22 '18 at 2:45
• Are you possibly confusing "well-ordering" with "total ordering?" Because your construction can be done with any total order on $M,$ and any maximal chain on $P(M)$ is a total order on $M.$ – Thomas Andrews Dec 22 '18 at 3:14

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.

Your construction works for any total order on $$M,$$ not just well-orderings.

And given a maximal chain in $$P(M),$$ you get a total order on $$M.$$ These are in 1-1 correspondence.

Start of Proof: Let $$C$$ be a maximal chain in $$P(M).$$ Then $$\emptyset, M\in C,$$ because the chain wouldn't be maximal if they weren't.

If $$x,y\in M$$ we say $$x\leq_C y$$ if and only if $$\forall S\in C$$, if $$y\in S\implies x\in S.$$

To prove:

(1) This is a partial order. Reflexivity and transitivity are easy. The only hard part is proving if $$x\leq y$$ and $$y\leq x,$$ you'd have $$x=y.$$ This is due to the maximality of $$C.$$

(2) This is a total order.

The ultimate reason for this is that any maximal chain in $$P(M)$$ is closed under arbitrary unions and intersections. So the union $$U_x$$ of the elements of $$C$$ which do not contain $$x$$ and the intersection $$V_x$$ of all elements of $$C$$ which do contain $$x$$ are both in $$C$$, and there are no elements of $$C$$ strictly between $$U_x$$ and $$V_x.$$

For (1), this means that if $$x\leq y$$ and $$y\leq x$$ then $$y\in U_x$$ and $$y\not\in V_x.$$ But this means that $$U_x\cup\{y\}$$ is strictly between $$U_x$$ and $$V_x.$$\

For (2), this means that for $$x.y\in M,$$ either $$U_x\subseteq U_y$$ or $$U_y\subseteq U_x$$, since $$U_x,U_y\in C$$ and $$C$$ is a chain. But $$U_x\subseteq U_y$$ means $$x\leq y$$ so either $$x\leq_C y$$ or $$y\leq_C x.$$