# Long chains of amorphous cardinalities

An amorphous set is an infinite set that cannot be partitioned into 2 infinite subsets. An amorphous cardinality is the cardinality of an amorphous set. Working in $$\sf ZF$$, it is consistent that amorphous sets exist.

Amorphous sets necessarily don't have a lot of structure. For example, for an amorphous set $$A$$, there can't be an infinite well-ordered chain of subsets of $$A$$, since the differences of consecutive elements in such a chain can be used to partition the set. However, if we are instead given a chain of cardinalities below $$A$$, $$(\mathfrak c_i)_{i \in \mathbb N}$$ such that $$\mathfrak c_i < \mathfrak c_j < |A|$$ for $$i < j$$, this argument does not work, because we can't uniformly choose representatives (and injections) for each cardinality.

To me, the existence of "long" chains of cardinalities below a set implies that it is "big". If there is such a sequence of length $$\kappa$$, then even if there isn't a literal subset of $$A$$ with cardinality $$\kappa$$, $$A$$ does in a sense contain at least $$\kappa$$ elements. I am interested in knowing whether amorphous sets can be arbitrarily "big" according to this intuition.

I have 3 questions of increasing strength:

A monotonic sequence of cardinalities of length $$\alpha$$ is a function $$s : \alpha \to \sf Card$$ such that $$s(\alpha) < s(\beta)$$ for $$\alpha < \beta$$.

1. Given an ordinal $$\alpha$$, is it consistent that there exists a monotonic sequence of amorphous cardinalities of length $$\alpha$$?

2. Is it consistent that for every ordinal $$\alpha$$ there exists a monotonic sequence of amorphous cardinalities of length $$\alpha$$?

3. Is it consistent that there is a monotonic class sequence $$F : \sf Ord \to \sf Card$$ of amorphous cardinalities?

• You’re more likely to get a positive answer here if you use surjective comparison $<^*$ instead. Mar 28 at 12:04
• If you replace amorphous with merely infinite Dedekind finite, then you can hope for a positive answer. The argument of my answer doesn't work at all in this case, and really used the amorphous attribute. Mar 28 at 15:50
• @JoelDavidHamkins Is there an obvious reason to keep asking about cardinalities instead of sets themselves if we are talking about dedekind finite sets? I don't see a reason a dedekind finite set can't have a long chain of subsets. Mar 28 at 16:28
• Yes, it definitely can. For example, there can be infinite Dedekind finite sets of reals, which have a linear order, and consequently many cuts into proper subsets. These are all different cardinalities, and so also a chain of cardinalities. But also subsets can have incomparable cardinalities. Mar 28 at 18:02
• Whereas below an amorphous cardinal, the cardinalities are linearly ordered, of type $\omega+\omega^*$ as I explain in my answer. Mar 28 at 21:01

If $$A$$ is amorphous, then every subset of $$A$$ is either finite or cofinite in $$A$$. Since every cardinal below $$A$$ is determined by a subset of $$A$$, it follows from this that the cardinals below $$A$$ are exactly the finite cardinals and the cardinalities of the cofinite-in-$$A$$ sets. These cardinalities are exhaustive and distinct (since otherwise we would form a countably infinite subset of $$A$$), and thus form a chain of length $$\omega+\omega^*$$. This order type has $$\alpha$$-sequences for exactly those ordinals $$\alpha$$ at most $$\omega+n$$ for some finite $$n$$; that is, $$\alpha$$ below $$\omega\cdot 2$$. So the answer to question (1) is no.
The answer to questions (2) and (3) are similarly no, since there can be no $$\omega\cdot2$$th amorphous set in any chain of cardinalities.
The longest well-ordered chain of amorphous cardinalities has order type $$\omega+\omega$$, since if $$A$$ is amorphous, then so is $$A+n$$ for any finite $$n$$. So we can start with an amorphous set and systematically add increasing finite sets to it to make the second copy of $$\omega$$ — but the limit will not be amorphous. Every amorphous cardinality is thus part of a maximal chain of amorphous cardinalities of order type $$\omega+\omega^*+\omega$$, which is to say, $$\omega+\mathbb{Z}$$. And furthermore, every chain of amorphous cardinalities is a subchain of such a chain.