Equal number of finite and infinite subsets implies amorphous

Question

We work in $\sf ZF$.

An amorphous set is a set that cannot be partitioned into $2$ disjoint infinite sets.

If $A$ is an amorphous set then it has an equal number of finite subsets and infinite subsets, with the bijection $x \mapsto x^c$. Is the opposite also true? That is, given a nonempty set $x$ with an equal number of infinite and finite subsets, is it amorphous?

I showed that it is enough to show that $2^A$ is Dedekind finite: define $B$ as the set of finite subsets of $A$ and $C$ as the set of cofinite subsets of $A$. If there exists a bijection $f:B \to [A]^{\not<\omega}$, then define $g : B \cup C \to \mathcal P (A)$ as $x \mapsto f(x)$ if $x \in B$ and $x^c$ otherwise. $g$ is a bijection, so if $\mathcal P (A)$ is Dedekind finite, $B \cup C = \mathcal P (A)$.

Answer 1

Referring to Theorem to 4.21, Proposition 4.22 In L. Halbeisen's book as explained in the comment:

Let $A$ be an infinite set.

Let $\text{fin}(A)$ be the finite subsets of $A$, $\text{cof}(A)$ the cofinite subsets of $A$. Write $\mathcal P(A)=\text{fin}(A)\sqcup\text{cof}(A)\sqcup \text{binf}(A)$, where $\text{binf}(A)$ is the rest.

If $|\mathcal P(A)\setminus \text{fin}(A)|=|\text{fin}(A)|$ then $|\mathcal P(A)|=2|\text{fin}(A)|$.

Now assume $A$ is not amorphous (so $\text{binf}(A)\ne\emptyset$) and $|\mathcal P(A)\setminus \text{fin}(A)|=|\text{fin}(A)|$. Let $\sigma\colon\mathcal P(A)\setminus \text{fin}(A)\to\text{fin}(A)$ be a bijection. Let $\tau\colon \mathcal P(A)\setminus\text{cof}(A)\to\text{fin}(A)$ be defined by $$ \tau(x)=\begin{cases} \sigma(A\setminus x)&x\in\text{fin}(A)\\ \sigma(x)&x\in\text{binf}(A) \end{cases} $$ Then $\tau$ is a bijection. Let $u\in\text{binf}(A)$. Then $\tau(u)\in\text{fin}(A)$, so $\tau(u)\not\in\tau(\text{fin}(A))$. Therefore $\tau\rvert_{\text{fin}(A)}$ is an injective map from $\text{fin}(A)$ onto a proper subset of itself. Therefore $\text{fin}(A)$ is Dedekind infinite (i.e. $\aleph_0\le|\text{fin}(A)|$.)

Now consider the proof of Prop 4.22. The restriction that $n$ not be a power of two is only used in the previous-to-last paragraph, and it's used to show that under the assumptions of the proposition (which permit amorphous!) $\aleph_0\le|\text{fin}(A)|$. Once that is shown, what's described in the last paragraph is applicable to any $n$, including powers of $2$. In the present case we have $|\mathcal P(A)|=n|\text{fin}(A)|$ where $n=2$. The last paragraph describes how to modify the proof of Theorem 4.21 (which is fairly complicated) to get a contradiction by injecting $\text{Ord}$ into $\text{fin}(A)$. So for any infinite $A$ for which $\aleph_0\le|\text{fin}(A)|$, it follows that $|\mathcal P(A)|\ne n|\text{fin}(A)|$ for any $n\ge 1$.