# Classifying the chains of orderable sets' power sets up to isomorphism

Recently, while trying to understand another result, I began to wonder about the following question:

Given some orderable set $$A,$$ what (if anything) can we conclude about the order type or cardinality of a (maximal) $$\subsetneq$$-chain of $$\mathcal{P}(A)?$$

I was initially only entertaining the notion for a countably-infinite set, and my first idea was good (more on that in a moment), but didn't actually turn out to answer the question. For one thing, I realized that I had subconsciously been treating my $$\subsetneq$$-chains as well-ordered, which wasn't intended. For another, it gave me no insight into possible cases in which $$A$$ was orderable, but not well-orderable. (Obviously, I'm talking about $$\mathsf{ZF},$$ here, not $$\mathsf{ZFC}.$$)

The upside is that the initial conclusions I came to seem to hold for all well-ordered chains of $$\mathcal{P}(A),$$ given any well-orderable set $$A,$$ though they didn't provide a full classification of such chains. Edit: Together with the fourth result (which I wasn't able to prove until later), the well-ordered chains of any well-orderable set's power set can be fully classified, with no need for any Choice.

Given a well-orderable set $$A,$$ let $$|A|$$ indicate the cardinal of $$A$$ (necessarily a natural number or an aleph number), and let $$|A|^+$$ indicate the least well-ordered cardinal greater than $$|A|$$ (necessarily a natural/aleph number, respectively, if $$|A|$$ is a natural\aleph number). Then the following hold:

1. For any ordinal $$\alpha<|A|^+,$$ there is a well-ordered chain $$\mathcal{C}$$ of $$\bigl\langle\mathcal{P}(A),\subsetneq\bigr\rangle$$ with order type $$\alpha.$$
2. For any successor ordinal $$\alpha$$ such that $$|A|<\alpha\le|A|^+,$$ there is a well-ordered chain $$\mathcal{B}$$ of $$\bigl\langle\mathcal{P}(A),\subsetneq\bigr\rangle$$ with order type $$\alpha,$$ such that $$\mathcal{B}$$ is maximal among the well-ordered chains of $$\bigl\langle\mathcal{P}(A),\subsetneq\bigr\rangle.$$
3. Given any $$\mathcal{B}\subseteq\mathcal{P}(A)$$ such that $$\mathcal{B}$$ is maximal among the well-ordered chains of $$\bigl\langle\mathcal{P}(A),\subsetneq\bigr\rangle,$$ we have that $$|\mathcal{B}|=|A|+1,$$ and that the order type of $$\mathcal{B}$$ is a successor ordinal.
4. Given any well-ordered chain $$\mathcal{C}$$ of $$\bigl\langle\mathcal{P}(A),\subsetneq\bigr\rangle$$ that is non-maximal among such well-ordered chains, we have that $$|\mathcal{C}|\le|A|,$$ and so the order type of $$\mathcal{C}$$ is some ordinal $$\alpha<|A|^+.$$

My proof outlines follow, then I will come back to my central question.

1. Since $$\alpha<|A|^+,$$ then $$|\alpha|\le|A|$$ by trichotomy of well-ordered sets, so there is an injection $$f:\alpha\to A.$$ Using $$f,$$ we can define $$g:\alpha\to\mathcal{P}(A)$$ by $$g(\beta):=\bigl\{f(\gamma)\mid \gamma\in\beta\bigr\}.$$ Readily, $$g$$ is an injective order-embedding $$\langle\alpha,\in\rangle\to\bigl\langle\mathcal{P}(A),\subsetneq\bigr\rangle;$$ its range is the desired set $$\mathcal{C}.$$

2. Take any successor ordinal $$\alpha$$ such that $$|A|<\alpha\le|A|^+,$$ so that there is some ordinal $$\beta$$ with $$|A|\le\beta<|A|^+$$ such that $$\alpha=\beta\cup\{\beta\}.$$ Since we readily have $$|A|\le|\beta|\le \beta<|A|^+,$$ then $$|\beta|=|A|.$$ Thus, there is a bijection $$f:\beta\to A.$$ Using $$f,$$ we can define $$g:\alpha\to \mathcal{P}(A)$$ by $$g(\gamma):=\bigl\{f(\delta)\mid\delta\in\gamma\bigr\}.$$ Readily, $$g$$ is an injective order-embedding $$\langle\alpha,\in\rangle\to\bigl\langle\mathcal{P}(A),\subsetneq\bigr\rangle;$$ its range is the desired set $$\mathcal{B}.$$

3. Given any such $$\mathcal{B},$$ since $$A$$ is its $$\subsetneq$$-greatest element, then the order type of $$\langle\mathcal{B},\subsetneq\rangle$$ is immediately a successor ordinal. Consider the relation $$\prec$$ on $$A$$ given by $$a\prec b$$ if and only if there is some $$B\in\mathcal{B}$$ such that $$a\in B$$ and $$b\notin B.$$ This is trivially irreflexive, and is readily transitive on $$A$$ by $$\subsetneq$$-orderedness. Given any non-empty $$C\in\mathcal{P}(A),$$ we have $$C\cap A=C\neq\emptyset,$$ so there is a $$\subsetneq$$-least element $$B\in\mathcal{B}$$ such that $$C\cap B\neq\emptyset.$$ By $$\subseteq$$-maximality of $$\mathcal{B},$$ we readily have that $$B\cap C$$ must be a singleton, and its unique element is the $$\prec$$-least element of $$C.$$ Thus, $$\langle A,\prec\rangle$$ is a well-ordering, and moreover, the elements of $$\mathcal{B}$$ comprise the initial segments of said well-ordering. Since $$b\mapsto\{a\in A:a\prec b\}$$ is readily an injection $$A\to\mathcal{B}$$ with range $$\mathcal{B}\setminus\{A\},$$ then $$|\mathcal{B}|=|A|+1,$$ as desired.

4. [Added later] Let $$\prec$$ be any well-ordering of $$A,$$ and let $$\mathcal{C}$$ be any well-ordered chain of $$\bigl\langle\mathcal{P}(A),\subsetneq\bigr\rangle$$ that is non-maximal among such well-ordered chains. If we have $$A\in\mathcal{C},$$ then non-maximality lets us construct a well-ordered chain $$\mathcal{C}'$$ of equal cardinality such that $$A\notin\mathcal{C}';$$ hence, we may assume without loss of generality that $$A\notin\mathcal{C}.$$ Thus, given any $$D\in\mathcal{C},$$ the set $$\bigl\{a\in A:(a\notin D)\wedge\forall E\in\mathcal{C}(D\subsetneq E\implies a\in E)\bigr\}$$ is readily non-empty (dealing separately with the case that $$D$$ is the greatest element of $$\mathcal{C},$$ if any such element exists), and we define $$a_D$$ to be the $$\prec$$-least element of this set. Then the map $$D\mapsto a_D$$ is readily an injection $$\mathcal{C}\to A.$$

Unfortunately, this still doesn't address my original question. If $$A$$ is an orderable (but not well-orderable) infinite set, then $$\mathcal{P}(A)$$ necessarily has chains that aren't well-ordered by $$\subsetneq,$$ since given an order $$\prec$$ on $$A,$$ the map $$a\mapsto\{b\in A\mid b\preceq a\}$$ is an order embedding $$\langle A,\prec\rangle\to\bigl\langle\mathcal{P}(A),\subsetneq\bigr\rangle.$$ Furthermore, if $$A$$ is well-orderable and $$\prec$$ is a well-ordering, then the image of the map $$a\mapsto\{b\in A\mid a\prec b\}$$ is a chain of $$\mathcal{P}(A),$$ but isn't well-ordered.

I suspect that something similar to these results can be concluded in terms of cardinality and/or order type, but I have less experience with order types of orders that aren't well-orders, and have no experience talking about cardinalities that are orderable but need not be well-orderable, so I'm struggling to figure out where to begin. The only thought I've had so far in terms of generalization is to use the Hartogs number instead of the successor cardinal, but I don't yet see how this would help.

If anyone can at least get me started on making such a generalization, point me to some models of $$\mathsf{ZF}$$ to show that it isn't strong enough to let us make such a classification, or even point me to some results along these lines, I'd greatly appreciate it!

Added: I've been able to generalize the first result as follows:

Given any orderable set $$A,$$ any set $$B$$ such that $$|B|<|A|,$$ and any order relation $$\prec$$ on $$A$$ (so on $$B$$), there exists some $$\mathcal{C}\subseteq\mathcal{P}(A)$$ such that $$\langle\mathcal{C},\subsetneq\rangle$$ has the same order type as $$\langle B,\prec\rangle.$$

The proof is straightforward, using the map $$B\to\mathcal{P}(A)$$ given by $$b\mapsto\{a\in A\mid a\preceq b\}.$$

Furthermore, recalling my experience with the Dedekind-cut construction of $$\Bbb R,$$ it's clear that there can be chains of $$\mathcal{P}(A)$$ of cardinality $$\bigl|\mathcal{P}(A)\bigr|,$$ which is a notable departure from the prior results. While the cardinality certainly can't be greater than that, it isn't clear whether that bound is non-strict in all cases. I will keep thinking on it, at any rate.