Long chains of amorphous cardinalities

Question

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?

Answer 1

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.