Exponentiation of Dedekind cardinals

Question

Question: Let $\mathfrak n$ be an infinite cardinal. In ZF (set theory without the axiom of choice) can either of the implications $$\mathfrak n=\mathfrak n+1\implies2^\mathfrak n=2^{\mathfrak n+1}\implies2^\mathfrak n=2^\mathfrak n+1$$ be reversed? Equivalently, can either of the implications $$\mathfrak n\ge\aleph_0\implies2^\mathfrak n=2^{\mathfrak n+1}\implies2^\mathfrak n\ge\aleph_0$$ be reversed?

Motivation: Idle curiosity.

What I know: The implications can't both be reversed because, if $\mathfrak n=2^\mathfrak m$ is incomparable with $\aleph_0$ (i.e., $2^\mathfrak m$ is a Dedekind-finite infinite cardinal) then $2^\mathfrak m\lt2^\mathfrak m+1$, while $2^{2^\mathfrak m}=2^{2^\mathfrak m}+1$ since $2^{2^\mathfrak m}\ge\aleph_0$ (Russell's theorem).

Question: Is $2^{2^\mathfrak m}=2^{2^\mathfrak m+1}$ if $2^\mathfrak m$ is incomparable with $\aleph_0$?

Answer 1

For the first question, the two implications fail to reverse. For the first implication, just take $\mathfrak{n}$ to be a dually Dedekind infinite, Dedekind finite cardinal. For the second implication, see Theorem 3.5 of the paper Definitions of finite and the power set operation by Howard and Spišiak (which is not published, see this link for a copy).

For the second question, the answer is yes. See the paper Ein Beitrag zur Kardinalzahlarithmetik ohne Auswahlaxiom by Läuchli; for an English version, see Theorem 5.28 of the book Combinatorial Set Theory: With a Gentle Introduction to Forcing (second edition) by Halbeisen.

The key question in this field (proposed by Läuchli) is whether it is provable in $\mathsf{ZF}$ that $2^{2^{\mathfrak{n+1}}}=2^{2^{\mathfrak{n}}}$ (or equivalently, $(2^{2^{\mathfrak{n}}})^2=2^{2^{\mathfrak{n}}}$) for all infinite cardinals $\mathfrak{n}$. I conjecture the answer is no.

Answer 2

ZF doesn't prove that the first implication reverses; i.e. it doesn't prove that $2^{\mathfrak{n}}=2^{\mathfrak{n}+1}\Longrightarrow \mathfrak{n}\geq\aleph_0$. For suppose $X$ is a set which is the disjoint union of an $\omega$-sequence $\left<p_n\right>_{n<\omega}$ of (unordered) pairs $p_n$, and there is no infinite subset of $X$ which meets each pair in at most one element. Let $\mathfrak{n}=|X|$. Then $\aleph_0\not\leq\mathfrak{n}$. But $2^{\mathfrak{n}}=2^{\mathfrak{n}+1}$. For we can define an injective function $\pi:\mathcal{P}(X)\times 2\to\mathcal{P}(X)$ as follows.

Let $A\subseteq X$. We define $\pi(A,0)$ and $\pi(A,1)$.

Suppose first $A$ is infinite. Then there is some $n<\omega$ such that for all $m\in[n,\omega)$, if $A\cap p_m\neq\emptyset$ then $p_m\subseteq A$. Let $n_A$ be the least such $n$. Let $$P_A=\{i\in\omega\bigm|i> n_A\wedge p_i\subseteq A\}$$. Let $$A'=(A\cap \bigcup_{i\leq n_A}p_i)\cup \bigcup_{i\in P_A}p_{i+1}$$. Now define $$\pi(A,0)=A'$$ and $$\pi(A,1)=A'\cup p_{n_A+1}.$$

Now suppose $A$ is finite. Let $m_A$ be the least $m\in\omega$ such that $A\subseteq\bigcup_{i<m}p_i$. (So for example, $A\cap p_{m_A}=\emptyset$.) Now let $$\pi(A,0)=A\cup p_{m_A+1}$$ and $$\pi(A,1)=A\cup p_{m_A+1}\cup p_{m_A+2}.$$

It is straightforward to check that $\pi$ is injective. So we have $|\mathcal{P}(X)\times 2|\leq\mathcal{P}(X)$, so $2^{\mathfrak{n}}=2^{\mathfrak{n}+1}$.

Answer 3

Note that if $\frak m\leq^* n$, then $2^\mathfrak m\leq 2^\mathfrak n$, since given a surjection from $N$ onto $M$, the preimages define an injection between the power sets. So, if $\frak m\leq^* n\leq^* m$, then $2^\mathfrak m=2^\mathfrak n$.

In particular, if there is a Dedekind-finite infinite cardinal, then there is one for which $1+\frak m\leq^* m$. Therefore, if there are any Dedekind-finite cardinals, then the first implication fails to reverse.

Interestingly, $\aleph_0\leq^*\frak m$ is a weaker condition than $1+\frak m\leq^*m$ (so the former does not imply the latter).