Assuming $ZF$ itself is consistent, it is consistent that there are sets $D$ which are infinite but cannot be placed in bijection with any of their proper subsets; such sets are called "strictly Dedekind-finite." Consistently, there is even a Dedekind-finite set of reals.

My question is, is it consistent to be able to partition $\mathbb{R}$ into strictly Dedekind-finite sets?

The simplest way to produce strictly Dedekind-finite sets of reals is to use Cohen forcing and take a symmetric submodel of the resulting forcing extension. We can also create $\kappa$-many disjoint strictly Dedekind-finite sets in a similar fashion, for any $\kappa$, without collapsing $\kappa$ (of course, the continuum is bumped above $\kappa$). However, as far as I can see, there is no simple way to adapt this to provide a partition of $\mathbb{R}$ into such sets.

One simple thing I've been able to figure out: suppose $\kappa$ is a (well-ordered) cardinal, $\{D_i: i\in I\}$ is a partition of $\mathbb{R}$ into strictly Dedekind-finite sets, and $\kappa^+$ injects into $\mathbb{R}$. Then there cannot be an injection $I\rightarrow\kappa$. However, letting $\Psi$ be the least ordinal not injectible into $\mathbb{R}$, it is not even clear to me that we cannot partition $\mathbb{R}$ into $\Psi$-many strictly Dedekind-finite pieces; and of course this says nothing about the case when the index set $I$ itself is non-well-orderable.

Hartognumber $\aleph(\mathbb{R})$, see en.wikipedia.org/wiki/Hartogs_number. $\endgroup$ – Joel David Hamkins Feb 9 '14 at 5:01Hartogs, he was French. $\endgroup$ – Asaf Karagila♦ Feb 9 '14 at 5:17Hartogs' number". $\endgroup$ – Asaf Karagila♦ Feb 9 '14 at 5:394more comments