There is an old result due to Mycielski and Sierpiński, and popularized in a Monthly article by Taylor and Wagon ([A Paradox Arising from the Elimination of a Paradox](https://doi.org/10.1080/00029890.2019.1559416); see also [this MO answer](https://mathoverflow.net/a/22935)), that can be stated as follows: in [Solovay's model](https://en.wikipedia.org/wiki/Solovay_model), one can partition $2^\omega$ into more than $2^\omega$ nonempty disjoint sets. Or if we let LM denote the axiom that “all subsets of $\mathbb{R}$ are Lebesgue-measurable,” then we can state this: ZF + LM proves $|\mathbb{R}| < |\mathbb{R}/\mathbb{Q}|$. Taylor and Wagon refer to this phenomenon as the **division paradox**.


This result is sometimes used as an argument in favor of the axiom of choice and against LM, but something bothers me about drawing this conclusion. If $A$ and $B$ are sets, then standardly we have the following definition.



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> We say that $|A|<|B|$ if there is an injection from $A$ to $B$ but no bijection from $A$ to $B$.
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But suppose we make the following definition.



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> W say that $B$ **outnumbers** $A$ if there is an injection from $A$ to $B$ but no surjection from $A$ to $B$.
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In the presence of the axiom of choice, $B$ outnumbers $A$ if and only if $|A|<|B|$, but without the axiom of choice they are not equivalent. As a sanity check on whether outnumbering is a reasonable concept, note that [it can be shown](https://cs.nyu.edu/pipermail/fom/2019-June/021567.html) that if $C$ outnumbers $B$ and $B$ outnumbers $A$, then $C$ outnumbers $A$. The question I have is this:



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> In Solovay's model, let $B$ be a partition of $A$ into nonempty disjoint sets. Can $B$ outnumber $A$?
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If the answer is no, then I would be inclined to interpret the division paradox as telling us that in the absence of the axiom of choice, our intuitions about cardinalities are unreliable, and we should be using the concept of outnumbering to compare sizes of sets. In particular, the division paradox gives us no compelling reason to reject Solovay's model.



Partitioning a set of cardinality $\kappa$ into more than $\kappa$ disjoint subsets
Well, by definition (more or less), if $B$ is a partition of $A$, then there is a surjection from $A$ onto $B$. So it is impossible, in general, for a partition of a set to outnumber that set.


The Division Paradox tells us that our intuition, which usually tells us that injections and surjections tell "the same story" with regards to cardinality, is really reliant on the axiom of choice, in a very significant way.


Note that you don't need to go as far as Mycielski and Sierpinski and the Solovay model, by the way, to get "big partitions" of the real numbers. Given any model of $\sf ZFC$ there is a symmetric extension given by adding Cohen real in which there is a "paradoxical partition" of the reals. (See <http://karagila.org/2020/countable-sets-of-reals/> for details.)


