Is it consistent for a Dedekind-infinite set $A$ to satisfy the strict inequality $2^{|A|}<2^{\aleph_0}$?

Question

Is there a model $M$ of ZF in which there is a Dedekind-infinite set $A$ and an injective function $f:\mathcal{P}(A)\rightarrow \mathbb{R}$ such that there is no injective function in the other direction?

Answer 1

If $A$ is any infinite set such that $A$ maps onto $\omega$, then $2^{\aleph_0}\leq 2^{|A|}$.

To see this, let $f\colon A\to\omega$ be a surjection, then for any $X\subseteq\omega$, $Y=\{a\in A\mid f(a)\in X\}$ is determined uniquely by $X$, and so defines the injection witnessing $2^{\aleph_0}\leq 2^{|A|}$.

Now, if $2^{|A|}\leq 2^{\aleph_0}$, then $A$ can be understood as a set of real numbers. Not trivially, but every infinite set of reals maps onto $\omega$, so much holds without any choice at all. In particular, if $A$ is any infinite set of reals, then $2^{|A|}\geq 2^{\aleph_0}$.

Of course, in the case of a Dedekind-infinite set, we literally have an injection from $\omega\to A$ which in turn can be mapped into a surjection from $A$ onto $\omega$ (map the range to its preimages, everything else goes to $0$). So certainly this is not possible.