# Is denying “Paradoxical Partitioning” equivalent to accepting the Axiom of Choice?

In this accepted answer (to this question here, from a couple of years ago) it has been noted:

One more point to make about the paradoxical decomposition to more parts than elements [...] currently we do not know of any model of ZF+$\lnot$AC where such decomposition does not exist. Namely, as far as we know, in all models where choice fails there is some set which can be partitioned into more parts than elements.

(Just for clarification: From the context of the above references the comparison "more parts than" is obviously meant in terms of the cardinality of the suitable partition being strictly larger than the cardinality of the suitable initial set itself.)

Now, I find the negation or denial of such "paradoxial partitioning" interesting; i.e. the statement (proposition, "$\lnot$PP"):

"Each partition of each given set has cardinality less than, or at most equal to, the cardinality of the given set."

And I like to further explore how this suggested "proposition $\lnot$PP" (being considered along with ZF, of course) relates to "the standard set theory including Axiom of Choice", ZFC. Therefore

My questions:

Are there any models of ZFC known (or could there be any such models, in principle) in which there is some set which can be partitioned into more parts than elements ?

And:

Are there any models of ZF known (or could there be any such models, in principle) in which there is no set which can be partitioned into more parts than elements, and which is not also a model of ZFC ?

(And just for reference:
Is there a conventional or concise way of expressing the suggested "proposition $\lnot$PP" in terms of standard notation, such as used in the sources linked above? Has it perhaps been discussed already, by some other name? ...)

• I would note that the statement you called $\neg PP$ is stronger than the negation of the phenomenon in the quote you wrote at the beginning. The quote discusses a partition of a set which has greater cardinality than the set itself, rather than a partition whose cardinality is not less than or equal to the cardinality of the set itself. In the absence of Choice, these are not the same (indeed, Choice is equivalent to the statement that if $|A|\not\leq |B|$, then $|A|>|B|$). – Eric Wofsey Feb 23 '17 at 23:03
• Similarly, the statement you are asking about in your questions ("...more parts than elements") is not the same as your $\neg PP$. – Eric Wofsey Feb 23 '17 at 23:06
• @Eric Wofsey: "The quote discusses a partition of a set which has greater cardinality than the set itself, rather than a partition whose cardinality is not less than or equal to the cardinality of the set itself. In the absence of Choice, these are not the same [...]" -- Wow!, thanks! (OMG, I had no idea ...) In formulating my "suggested proposition $\lnot$PP" and my questions I had (merely) considered the variants with "the smoothest wording" without intending to express any particular "finer points" that you raise. At least: I tried to phrase my questions along the lines of the quote ... – user12262 Feb 23 '17 at 23:26

No, to both questions. But for different reasons.

The Axiom of Choice proves that if $A$ is any set, then any partition of $A$ has size of at most $A$. This is because there is a surjection from $A$ onto a partition of $A$, which using choice means there is an injection from the partition into $A$.

So assuming $\sf ZFC$, every set can be partitioned into at most its-cardinality-many parts.

The second question is open, because what you are asking is called The Partition Principle (which is what usually called $\sf PP$, by the way), and the question whether or not it implies the axiom of choice over $\sf ZF$ is the oldest open question in set theory, as of 2017. So we simply don't know the answer to this one.