Characterizing infinite Dedekind-finite sets.

Question

Is the following true in $\text{ZF}$?

A set $X$ is an infinite Dedekind-finite set if and only for every bijection $f: X \to X$ there exists a partition $X = C \cup Y$ with $C$ a finite set and $Y$ an infinite Dedekind-finite set such that $f(C) = C$. Moreover, the set $X$ can be completely partitioned into finite sets $C_\iota$ such that $f(C_\iota) = C_\iota$ with $\iota$ ranging over an infinite index set.

My work

This research endeavor came about from my work found here.

I couldn't find this question as a duplicate but before getting bogged down in the details I wanted to just get some quick feedback (please excuse the laziness - I am not an expert in set theory).

Answer 1

This is (mostly) true.

In one direction, given $f:X\rightarrow X$, for each $a\in X$ we can consider the orbit of $a$ under $f$, $$orb_f(a)=\{f^{z}(a): z\in\mathbb{Z}\}.$$ If $X$ is Dedekind-finite, each $orb_f(a)$ must itself be finite, and by definition we have $f(orb_f(a))=orb_f(a)$ and $X=\bigcup_{a\in X}orb_f(a)$.

In the other direction, suppose $X$ is Dedekind-infinite. WLOG let $X=\mathbb{Z}\sqcup Y$. Then consider the self-bijection $f$ which shifts $\mathbb{Z}$ via $z\mapsto 1+z$ and is the identity on $Y$. Clearly we cannot partition $X$ into finite $f$-invariant pieces. However, at the same time we may still be forced to have some finite nonempty $f$-fixed sets: namely, if $Y$ itself is infinite but Dedekind-finite. So there is a slight subtlety there: the strong version of your property fully characterizes Dedekind-finiteness, but the weak version (the guaranteed existence of just one such $C$) doesn't.

Answer 2

Using Noah's answer here is an improved presentation.

---

Title: Decomposing automorphisms of infinite Dedekind-finite sets.

Is the following true in $\text{ZF}$?

Let $X$ be an infinite set.

Here is the decomposition statement,

Proposition 1: If the set $X$ is Dedekind-finite set then for every bijection $f: X \to X$ there exists a partition

$\tag 1 \{ orb_f(a) \mid a\in X\}$

of $X$ into blocks with a finite number of elements such that $f\bigr(orb_f(a)\bigr) = orb_f(a)$ for each block. Moreover, $f$ is a bijective mapping on any invariant (restricted) domain.

To show the 'converse', we state the following,

Proposition 2: If $X$ is Dedekind-infinite set then there exists a bijection $f: X \to X$ and a set $N \subset X$ satisfying $f(N) \subsetneq N$.