TITLE: Is it possible that $2^n=3^n$ for some Dedekind-finite cardinal $n\gt0$?
QUESTION [9 upvotes]: 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?

REPLY [5 votes]: 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.