A set being amorphous means every subset of it is either finite or cofinite. AC for finite sets means any family of finite sets has a choice function, i.e. if $f:A\to I$ is surjective and for each $i\in I$, $f^{-1}(i)$ is finite, then there is $s:I\to A$ such that $fs$ is identity on $I$. Over ZF, "every set can be linearly ordered" implies that there is no amorphous set, and it also implies AC for finite sets. But can we have amorphous sets and AC for finite sets at the same time?
Can amorphous set and AC for finite sets coexist
axiom-of-choiceset-theory
Related Solutions
Yes, these sets can exist for all infinite well-ordered cardinals. The construction is essentially the same type. We can even have $\sf DC_{<\operatorname{cf}(\kappa)}$ along with a $\kappa$-amorphous set.
For $\aleph_1$-amorphous sets, those can be linearly ordered. In fact, there can be such a set of reals.
The reason the usual proof doesn't work is that there are different linear orders which are uncountable but every proper initial segment is countable (e.g. $\omega_1$, or $\omega_1\times\Bbb Q$ or $\omega_1\times\Bbb Q\times\Bbb Z$, etc.) which means that the usual proof that relies on "$\Bbb N$ is the unique linear order without a maximal element in which every proper initial segment is finite" does not translate to a general one.
And as for comparability, we can arrange that for every $\kappa$ there is a proper class of pairwise incomparable $\kappa$-amorphous sets. How's that on for size? The idea is relatively simple:
First consider a symmetric extension where we added a $\kappa$-amorphous set for some $\kappa$, let's say using Cohen subsets of some regular cardinal $\lambda$.
Now fix a coding of pairs $(\xi,\zeta)$ of ordinals. In the $\alpha$th stage, if $\aleph_\alpha$ is a regular cardinal, take $\kappa=\aleph_\xi$ and $\lambda=\aleph_\alpha$, and this is our symmetric system. Now take an Easton support product of these forcings, and by genericity argument two amorphous sets coming from different symmetric systems must be incomparable.
(What's clever is that you can use a finite support product and still preserve $\sf ZF$ in the outcome model, at least if you've chosen the obvious symmetric system in the first step above.)
No.
If $A$ is an amorphous set, then its power set is Dedekind-finite. Therefore there is no choice function which chooses from its non-empty subsets. Since if $\mathcal P(X)\setminus\{\varnothing\}$ has a choice function, then $X$ can be well-ordered.
So if $A$ is amorphous, then we have a Dedekind-finite family without a choice function.
This is not an equivalence. For example, in Cohen's first model, the family of co-finite subsets of $A$, the canonical Dedekind-finite set, is Dedekind-finite. And it does not admit a choice function, since $$A\text{ is D-infinite} \iff\{A\setminus X\mid X\text{ is finite}\}\text{ admits a choice function}.$$
So again, in Cohen's model we have a Dedekind-finite family of sets without a choice function.
In general, given any Dedekind-finite set $X$, we can arrange an extension of the universe where $X$ remains Dedekind-finite, and there is a family of sets indexed by $X$ which does not have a choice function.
Best Answer
Here's an elementary proof.
Proposition. If a set is amorphous then there cannot be a choice function on the set of its finite, non-empty subsets.
The proof is by contradiction. Suppose there is an amorphous set $S$ with a choice function $f$ on its finite, non-empty subsets. We define an extension of $f$ covering the infinite subsets.
Let $X$ be an infinite subset of $S$. For each $x \in X$ we define two sets
$A_x = \{y \in X \setminus \{x\} : f(\{x,y\}) = x \}$ and $B_x = \{y \in X \setminus \{x\} : f(\{x,y\}) = y \}$
Exactly one of these will be finite, call it $F_x$.
$A_x$ and $B_x$ can be empty, but for at most one value of $x$ each,say $a$ and $b$ respectively. If $a$ exists define $f(X) = a$. If there is no $a$ but $b$ exists, define $f(X) = b$.
Otherwise, all $F_x \neq \emptyset$ and we can define a function $g:X \rightarrow X$ by $g(x) = f(F_x)$ . Note $g(x) \neq x$ for all $x$.
Consider the sequences of the form $x, g(x), g(g(x)), \dots$ As $X$ has no denumerable subsets these must always eventually start repeating. Call the sets which contain the repeating part of such a sequence loops. Let $\mathscr L$ be the set of loops. The loops are pairwise disjoint and each is finite with at least two elements. So the sets
$\{f(L) : L \in \mathscr L\}$ and $\{g(f(L)) : L \in \mathscr L \}$
are disjoint subsets of $X$ and of the same cardinality as $\mathscr L$. Therefore there are only finitely many loops and we can define $f(X) = f(\bigcup\mathscr L)$ .
In all cases, $f(X) \in X$, so the extended $f$ is a choice function on the set of non-empty subsets of $X$. This is not possible for an amorphous set. So $f$ cannot exist.