If $X$ is strictly amorphous and $F:X\to [X]^{<\omega}$, is the set $\bigcup\{F(x)\setminus\{x\} : x\in X\}$ finite? Recently I've been thinking about amorphous sets in $\sf{ZF}$ set theory, particularly as related to this currently unanswered question of mine. More specifically though, I've been considering a principle which acts as a sort of strong denial of Choice. Given a set $X$, we let $[X]^{<\omega}$ denote the set of finite subsets of $X$. Given a map $F:X\to [X]^{<\omega}$, we say that $F$ chooses $y$ when there exists $x\in X$ for which $y\in F(x)\setminus\{x\}$. The idea behind this definition is that using only the information about $x$, the function $F$ is able to produce some other objects seemingly arbitrarily. My conjecture is that, whenever $X$ is strictly amorphous, the set of objects chosen by $F$ is finite. Is my conjecture true?
It's worth noting that the converse implication is certainly true. If $X$ is amorphous but not strictly amorphous, we can partition $X$ into finite sets, most of which are nonsingular. The function $F$ which sends $x$ to the corresponding partition will at some point end up choosing almost every element of $X$, which clearly isn't a finite set.
My conjecture can be naturally generalized, in two particular ways. The first is that we can increase the domain from $X$ to $X^n$ for some finite $n$, which is a generalization in the sense that $X=X^1$. The second way is that we could increase the domain from $X$ to being $[X]^{<\omega}$, which is a generalization in the sense that $X$ is isomorphic to the set of its own singleton subsets. If we ask the analogous question for these cases, does the property still hold? Formally, we have the following two generalizations.

Given $n\in\omega$ and $F:X^n\to [X]^{<\omega}$, is the set $\bigcup\{F(p)\setminus p[n] : p\in X^n\}$ finite?


Given $F:[X]^{<\omega}\to [X]^{<\omega}$, is the set $\bigcup\{F(S)\setminus S : S\in[X]^{<\omega}\}$ finite?

My base conjecture would have the interesting consequence of proving that every binary relation on $X$ is definable in the language of equality with only finitely many parameters. This language has quantifier elimination when the domain is infinite (which it is), so the relations on $X$ are extremely simple. I believe my first generalization could be used to prove a similar property about arbitrary $n$-airy relations, albeit proving that seems nontrivial.
 A: The conjecture is true, as well as both of its generalizations. For the base problem, the trick is that for each chosen element, the element is either chosen almost never (for only finitely many $x$), or it's chosen almost always (for all but finitely many $x$). By proving that the cardinality $|F(x)|$ is bounded across all $x$, it follows that only finitely many objects can be chosen almost always, and we can delete these elements to simplify the problem. The remainder of $X$ can be partitioned by letting $F$ recursively make choices and packaging these choices together, and we can prove that this recursive process must be finite so we partition $X$ into finite sets. Finally, we use the strict amorphousness of $X$ to prove that this partition is almost entirely singletons, which can be used to prove that $F$ choses only finitely many objects. The two generalizations can be proven as relatively easy corollaries of the base problem. To begin, we have two lemmas about some fairly well known properties of amorphous sets.
Lemma: Given amorphous $X$, wellorderable $\alpha$, and $f:X \to \alpha$, the image $f[X]$ is finite.
proof: If the image of $f$ were infinite, then a well ordering of the image lets us  partition it into two infinite subsets. The $f$ preimages of these two parts will then partition $X$ into two infinite subsets, a contradiction since $X$ is amorphous.$\Box$
Lemma: If $X$ is amorphous, then $[X]^{<\omega}$ is dedekind finite.
proof: Given a sequence $\{S_n\}_{n\in\omega}$ of sets $S_n\in[X]^{<\omega}$, let $Y=\bigcup_n S_n$ which is either finite or amorphous since $Y\subseteq X$. Define the map $f:Y\to\omega$ via $f(y)=\min\{n : y\in S_n\}$, then $f$ has finite image so we can find $N=\max\{f(y) : y\in Y\}$. It follows that $Y=\bigcup_{n\leq N}S_n$ implying $Y$ is finite, and since each $S_n\subseteq Y$ then $S_n\in \mathcal{P}(Y)$ which is finite, proving that the image of the $S$ sequence is finite. $\Box$

Theorem: Given $X$ strictly amorphous and $F:X\to [X]^{<\omega}$, the set $\bigcup\{F(x)\setminus\{x\} : x\in X\}$ is finite.
proof:
To work out this problem formally, for each $x\in X$ we define the set $B(x):=\{y\in X : x\in F(y)\}$. We're going to take the simplified case where all $x\in X$ have $B(x)$ finite, and handle the general case at the end. Now, define the relation $R=\{(x,y)\in X^2 : y\in F(x)\lor y\in B(x)\}$. Notice that this relation is symmetric, since generally $x\in F(y)\iff y\in B(x)$. We can therefore consider the pair $(X,R)$ as an undirected graph with nodes $X$ and edges encoded by $R$. We moreover notice that each node has only finitely many edges, since the set of edges starting from $x$ is simply $\{(x,y) : xRy\}=F(x)\cup B(x)$, and both both $F(x)$ and $B(x)$ are finite.
Now we're going to partition $X$ into its connected subsets, as defined by the graph $R$. To do this, define $N_0(x)=\{x\}$ and recursively $N_{n+1}(x)=\{y\in X : \exists(z\in N_n(x)), zRy\}$, and finally $N_\omega(x):=\bigcup_{n\in\omega} N_n(x)$. Generally, the set $N_n(x)$ is the set of points reachable from $x$ by traveling across exactly $n$ edges of $R$, so the set $N_\omega(x)$ is the set of all points reachable from $x$. It's evident that the set $P=\{N_\omega(x) : x\in X\}$ forms a partition of $X$. Taking $x$ as constant, we have the map $n\mapsto N_n(x)$ which is a sequence of finite subsets of $X$. Because $[X]^{<\omega}$ is dedekind finite, this sequence has finite image, and thus $N_\omega(x)$ is a finite union of finite sets, which is finite. It follows that $P$ partitions $X$ into finite sets.
Since $X$ is strictly amorphous and $P$ partitions $X$ into finite sets, all but finitely many parts must be singular. Since all $x\in X$ have both $x\in N_\omega(x)$ and also $F(y)\subseteq N_1(x)\subseteq N_\omega(x)$, then $F(x)\subseteq\{x\}$ for all but finitely many $x$. Letting $X_0=\{x\in X : F(x)\setminus\{x\}\neq\emptyset\}$, we have $\bigcup\{F(x)\setminus\{x\} : x\in X\}=\bigcup\{F(x)\setminus\{x\} : x\in X_0\}$. This is a finite union of finite sets, so $F$ can only choose finitely many objects overall.
To handle the general case where $|B(x)|$ is not guaranteed to be finite for all $x\in X$, we first prove that it's finite for all but finitely many. Indeed, the map $x\mapsto |F(x)|$ sends $X\to\omega$ and therefore has finite image, so let $M=\max\{|F(x)| : x\in X\}$. Now let $X_\infty=\{x\in X : |B(x)|\notin\omega\}$, and notice that whenever $B(x)$ is infinite then it's also cofinite, since $X$ is amorphous. Since every $x\in X_\infty$ has $x\in F(y)$ for all but finitely many $y$, and because all $|F(y)|\leq M$, then necessarily $|X_\infty|\leq M$. To convert this into the simplified problem, define $Y=X\setminus X_\infty$ which is strictly amorphous, and define $G(y):=F(y)\setminus X_\infty$ so that $G:Y\to [Y]^{<\omega}$. Notice that all $y\in Y$ have $B(y)$ finite, and also $\{x\in Y : y\in G(x)\}\subseteq B(y)$, so this modification fits the criteria of the simplified problem. It follows that all but finitely many $y\in Y$ have $G(y)\subseteq \{y\}$ and consequently $F(y)\setminus\{y\}\subseteq X_\infty$. Now we just let $X_0=\{x\in X : (F(x)\setminus\{x\})\setminus X_\infty\neq\emptyset\}$, which as established is finite. When we finally union all $F(x)\setminus\{x\}$, there are only finitely many contributions from those $x\in X_0$, and those $x\notin X_0$ can only contribute the finitely many objects in $X_\infty$, thus the union $\bigcup\{F(x)\setminus\{x\} : x\in X\}$ is bounded by a finite union of finite sets, so it's finite.

The above principle can be used to easily imply the two generalizations.
Theorem: Given $X$ strictly amorphous, $n\in\omega$, and $F:X^n\to [X]^{<\omega}$, the set $\bigcup\{F(p)\setminus p[n] : p\in X^n\}$ is finite.
proof: We prove this by induction on $n$. The case $n=0$ is trivial since then $X^0$ is finite, and the case $n=1$ is essentially the previous problem. For $n>1$, decompose $n=m+1$ and define $G(x_0)(x_1,\cdots,x_m)=F(x_0,\cdots,x_m)$ and define $F'(x_0):=\bigcup\{G(x_0)(p)\setminus p[m] : p\in X^m\}$. Because each $x_0\in X$ has $G(x_0):X^m\to [X]^{<\omega}$, inductive premise lets us infer that $F'(x_0)$ is a finite set, so that $F':X\to [X]^{<\omega}$. By applying the previous theorem, we know that $\bigcup\{F'(x_0)\setminus\{x_0\} : x_0\in X\}$ is a finite set. By referring to the definitions of $G$ and $F'$ however, this union is precisely $\bigcup\{F(p)\setminus p[n] : p\in X^n\}$, which proves our claim.
Theorem: Given strictly amorphous $X$ and $F:[X]^{<\omega}\to [X]^{<\omega}$, the set $\bigcup\{F(S)\setminus S : S\in[X]^{<\omega}\}$ is finite.
proof: This follows from the previous theorem by using a clever trick. Let $A=\bigcup\{F(S)\setminus S : S\in [X]^{<\omega}\}$, then $A\subseteq X$ so that $A$ is either finite or amorphous. For each $x\in A$ we let $H(x):=\min\{|S| : x\in F(S)\setminus S\}$, then because $H:A\to \omega$, we must have the image of $H$ bounded, say by $M\in\omega$. Now define $G(p):=F(p[M])$ so that $G:X^M\to [X]^{<\omega}$. By construction of $M$, each $x\in A$ admits $S\subseteq X$ for which $|S|\leq M$ and $x\in F(S)\setminus S$, then we can find $p\in X^M$ having the image $p[M]=S$ and it follows that $G(p)\setminus p[M] = F(S)\setminus S$ and thus $x\in G(p)\setminus p[M]$. This is all just to prove that $A\subseteq \bigcup\{G(p)\setminus p[M] : p\in X^M\}$, and since the latter set is finite (by the previous theorem), then $A$ is also finite.
