# How many Dedekind-finite sets can $\mathbb{R}$ be partitioned into?

Building off Asaf Karagila's answer to my previous question (Can $\mathbb{R}$ be partitioned into dedekind-finite sets?) on partitioning $\mathbb{R}$ into strictly Dedekind-finite sets:

(1) What are the cardinalities (not necessarily well-ordered) $\mathfrak{C}$ such that $\mathbb{R}$ can be partitioned into $\mathfrak{C}$-many strictly Dedekind-finite sets?

(Recall that a strictly Dedekind-finite set is an infinite set $D$ such that every injection $D\rightarrow D$ is a surjection.)

Question (1) is pretty broad, so let me add some interesting specific subquestions:

(2) Can $\mathbb{R}$ be partitioned into $\kappa$-many strictly Dedekind-finite sets for a (well-ordered) cardinal $\kappa$?

My guess is "no," but at present I can't see how to prove it.

(3) Can $\mathbb{R}$ be partitioned into fewer than $2^{\aleph_0}$-many strictly Dedekind-finite sets?

By "fewer" here, I mean "injectible into $2^{\aleph_0}$ (since clearly $2^{\aleph_0}$ surjects onto the size of any partition), but not admitting an injection from $\mathbb{R}$ (demanding no surjection onto $\mathbb{R}$ would be an even stronger demand). Asaf's solution to my previous question was a partition of $\mathbb{R}$ into continuum-many strictly Dedekind finite sets via Cantor-Bernstein; that doesn't seem useful here.

Of course, size behaves weirdly without choice; so in the opposite direction, we can ask:

(4) Can $\mathbb{R}$ be partitioned into $\mathfrak{C}$-many strictly Dedekind-finite sets, where there is no injection $\mathfrak{C}\rightarrow\mathbb{R}$?

There are lots of other questions I could ask about partitioning $\mathbb{R}$ into strictly Dedekind-finite sets, but for now I'll just finish with:

(5) Is there a good source on partitioning $\mathbb{R}$ into "pseudofinite" sets? (Allowing any reasonable notion of "pseudofiniteness".)

• One word on terminology, "strictly" has a sound which is more restrictive than permissive, but [infinite] Dedekind-finite sets are less "strict" than finite sets. As for "pseudofinite", there are a handful of definitions you can find in Truss' paper about finiteness. If I recall correctly that list has been slightly expanded since. However the majority of "pseudofinite" sets cannot be embedded into the real numbers to begin with. Perhaps it's a good idea first get a good grip on the other questions, and before attacking (5), trying to understand what sort of finiteness can happen in $\Bbb R$. – Asaf Karagila Feb 10 '14 at 9:36
• I deleted my answer, I called Eilon and he said that he found a mistake in the proof of the statement I wrote (that it is consistent that there is a Dedekind-finite set $A$ such that $|A\times\omega|=2^{\aleph_0}$). He says it might be true with $\omega_1$ instead, but he doesn't have a proof off hand (and we had to finish the conversation). – Asaf Karagila Feb 10 '14 at 12:29