A possible ${\sf (ZF)}$-theorem in the spirit of the $3$-set-lemma

Question

The number $3$ plays an interesting role in the following statement:

$\newcommand{\S}{\sf(S_3)}\S$ Let $X$ be a non-empty set and let $f:X \to X$ be fixpoint-free (that is $f(x) \neq x$ for all $x\in X$). Then there are subsets $X_1, X_2, X_3 \subseteq X$ with $X_1\cup X_2\cup X_3 = X$ and $$X_i \cap f(X_i) = \emptyset$$ for $i \in \{1,2,3\}$.

There are easy examples showing that one cannot get by using $2$ subsets only. Statement $\S$ can be proved using the axiom of choice. (Although the full force of (AC) is not needed as the strictly weaker Boolean Prime Ideal Theorem (BPI) is sufficient, but it appears that the $\S$ is not a theorem of $\newcommand{\ZF}{{\sf (ZF)}}\ZF$.)

The motivation of this question is to create a theorem of $\ZF$ in the spirit of $\S$.

Question. Is the following statement a theorem of $\ZF$?

There is a cardinal $\kappa_0 \geq 3$ with the following property: If $X$ is a set with more than $1$ element and $f:X\to X$ is fixpoint-free, then there is a ${\frak S}\subseteq {\cal P}(X)$ with $\bigcup{\frak S} = X$, an injection $\iota:{\frak S}\to \kappa_0$, and the property that $S\cap f(S) = \emptyset$ for all $S\in {\frak S}$.

If the above statement is not a theorem of $\ZF$, then I am wondering about the following special case in $\newcommand{\N}{\mathbb{N}}\N$:

If $f:\N\to\N$ is fixpoint-free, then there is a finite set ${\frak S}\subseteq {\cal P}(\N)$ with $\bigcup{\frak S} = \N$ and $S\cap f(S) = \emptyset$ for all $S\in {\frak S}$.

(Note that if we drop the finiteness requirement on ${\frak S}$, then ${\frak S} = \big\{\{n\}:n\in\mathbb{N}\big\}$ is a trivial solution.)

Answer 1

The "special case in $\mathbb N$" is a theorem of ZF:

Theorem (ZF). For any well-orderable set $X$ and any fixpoint-free function $f:X\to X$ there is a family $\mathfrak S\subseteq\mathcal P(X)$ such that $|\mathfrak S|\le3$ and $\bigcup\mathfrak S=X$ and $S\cap f(S)=\varnothing$ for all $S\in\mathfrak S$.

Proof. I will show that the graph $G=(X,E)$ with vertex set $X$ and edge set $E=\{\{x,f(x)\}:x\in X\}$ is $3$-colorable. Let $\{C_i:i\in I\}$ be the set of all connected components of $G$. (Of course each $C_i$ is $3$-colorable, being a unicyclic connected graph.) Fix a well-ordering of $X$.

If $C_i$ contains no odd cycle, let $x_i$ be the least vertex in $C_i$. There is a unique proper red-blue vertex coloring of $C_i$ such that $x_i$ is red.

If $C_i$ contains a (necessarily unique) odd cycle, Let $x_i$ be the least vertex in $C_i$ which is not a cut vertex. Then there is a unique proper red-blue-green coloring of $C_i$ such that $x_i$ is the only green vertex and $f(x_i)$ is red.

I don't know about your general statement, but it may be worth mentioning that it's not provable in ZF if we strengthen it by requiring the cardinal $\kappa_0$ to be an aleph:

Theorem. (ZF) Suppose that for any set $X$ and any fixpoint-free involution $f:X\to X$ there is a well-orderable family $\mathfrak S\subseteq\mathcal P(X)$ such that $\bigcup\mathfrak S=X$ and $S\cap f(S)=\varnothing$ for all $S\in\mathfrak S$. Then every collection of $2$-element sets has a choice function.

Proof. We may assume that the $2$-element sets are disjoint, and then they are the orbits of an involution.