Is it possible that $2^n=3^n$ for some Dedekind-finite cardinal $n\gt0$?

Question

Is it possible that $2^n=3^n$ for some Dedekind-finite cardinal $n\gt0$?

I think the question speaks for itself, but let me try and satisfy the "quality standards" algorithm by padding it.

Yes, I realize that the equality requires a Dedekind-finite cardinal $n$ such that $2^n$ is Dedekind-infinite, and of course such cardinals exist if any infinite Dedekind-finite cardinals exist.

The reason I'm asking is, I want to know what condition I have to put on a (nonzero) cardinal $n$ to ensure that $2^n\lt3^n$, if I don't want to assume the Axiom of Choice. Do I have to say that $n$ is finite, or is it enough to say that $n$ is Dedekind-finite?

Answer 1

Here's a possible consistency result, and a suggestion for a reasonable assumption which work and another one which I conjecture to work as well. (And excuse me in advance for not using $n$ as an infinite cardinal!) $\newcommand{\fp}{\mathfrak p}$

Note that $4^\fp=2^{2\fp}$, so if we have that $2^\fp=2^{2\fp}$ then we have that $2^\fp=3^\fp$ as well. Consider Cohen's first model, with a Dedekind-finite set of reals. There we can find a Dedekind-finite set of reals, with cardinal $\fp$, such that $2^\omega=2^\fp$. Now we have that $$2^\fp=2^\omega=(2^\omega)^2=(2^\fp)^2=2^{2\fp}=4^\fp.$$

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If you want a reasonable condition for $2^n<3^n$, as a start, here's one:

$$\aleph_0\nleq 2^\fp\iff\aleph_0\nleq^*\fp$$ (Namely there is no surjection from $\fp$ onto $\omega$, which is equivalent to saying that $2^\fp$ is Dedekind-finite.)

If $2^\fp$ is Dedekind-finite, then either $3^\fp$ is Dedekind-finite and therefore strictly larger than $2^\fp$; or it is Dedekind-infinite, and therefore definition larger than $2^\fp$ (since we have an inclusion between $2^\fp$ and $3^\fp$, so they must be comparable).

My guess, however, is that the above condition is not optimal, but rather $\fp+1\nleq^*\fp$ is sufficient to prove that $2^\fp<3^\fp$.

Answer 2

Here is another example: Let $\mathrm{fin}(\mathfrak p)$ denote the size of the collection of all finite subsets of $X$, for any $X$ of size $\mathfrak p$. Now, if $\mathfrak p$ is infinite and $\mathrm{fin}(\mathfrak p)$ is Dedekind finite, we are done, because a nice theorem of Läuchli gives us that $(2^{\mathrm{fin}(\mathfrak m)})^{\aleph_0}=2^{\mathrm{fin}(\mathfrak m)}$ for any infinite $\mathfrak m$. (This is shown, for example, in Lemma 4.27 in Halbeisen's book.) But it is consistent that there is an infinite $\mathfrak p$ with $\mathrm{fin}(\mathfrak p)$ Dedekind finite.