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.

  • $\begingroup$ 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!). $\endgroup$ Commented Dec 22, 2018 at 2:26
  • $\begingroup$ Sorry, I wanted to write "$f^{-1}((\{ \emptyset\} \cup f(M) \cup \{M\}) \cap K)$". It will be corrected in the post. $\endgroup$
    – Lucina
    Commented Dec 22, 2018 at 2:33
  • $\begingroup$ 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. $\endgroup$
    – bof
    Commented Dec 22, 2018 at 2:41
  • $\begingroup$ @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. $\endgroup$ Commented Dec 22, 2018 at 2:45
  • 1
    $\begingroup$ 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 $M.$ $\endgroup$ Commented Dec 22, 2018 at 3:14

2 Answers 2


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.

  • $\begingroup$ +1: For an alternative that requires slightly less Choice (only the Ordering Principle), suppose that $M$ is infinite, take any ordering of $M,$ as well as the inverse order. Both will correspond to maximal chains (as pointed out in Thomas Andrews's answer), but at most one is a well-order. $\endgroup$ Commented Apr 9, 2022 at 17:55

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 $y\leq_C x.$


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