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Let $X$ be an infinite set. Can one prove in ZF (without choice, and with finiteness defined as being equipotent to a finite ordinal) that the set $\mathfrak{P}_{< \omega}(X)$ of all finite subsets of $X$ cannot be mapped onto the set $\mathfrak{P}(X)$ of all subsets of $X$?

If not, does countable choice suffice?

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    An interesting near-miss: if $X$ is *amorphous*, then two copies of $\mathfrak{P}_{<\omega}(X)$ can map onto $\mathfrak{P}(X)$, since every subset of an amorphous set is either finite or cofinite. (Incidentally, the fact that even amorphous sets don't immediately provide a counterexample suggests to me that this may, surprisingly!, be provable in ZF alone.) – 2017-02-15
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    Noah Schweber: I already feel I am not nearly familiar with ZF without choice... – 2017-02-15
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    @Noah: Indeed, this is a theorem. Although truth be told, not everything that cannot be contradicted by an amorphous set can be provable. For example, every filter on an amorphous set can be extended to an ultrafilter. :) – 2017-02-15
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    @AsafKaragila True, but I find it's still a useful heuristic in many cases (like this one :P). – 2017-02-15
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    Oh, I definitely agree about that. I was just being nitpicky... I mean, a mathematician. :P – 2017-02-15
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    Noah, it turns out to be consistent that there is such a set. – 2017-02-15

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It is provable that $\mathcal P_\omega(X)$ is strictly smaller than $\mathcal P(X)$, but it is consistent that there is a surjection still.

Lorenz Halbeisen and Saharon Shelah, Consequences of arithmetic for set theory, J. Symbolic Logic 59 (1994), no. 1, 30--40.

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    You guys are quick! Thanks, I'll look it up right now. – 2017-02-15
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    Oops! I looked at the wrong paper, and I misread the result. It turns out to be consistent, although the injection-based order is always strict. I've amended my answer. – 2017-02-15
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    See the paper on Halbeisen's page [here](http://user.math.uzh.ch/halbeisen/publications/pdf/carla.pdf). – 2017-02-15
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    It's not we guys, it's @AsafKaragila ;-) I knew the answer (+1) would have been his before opening the post. – 2017-02-15
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    Ok, it's less surprising but still interesting. – 2017-02-15
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    @nombre: It's a partial answer, though. The role of countable choice is interesting here. The proof relies heavily on the fact that the construction is "a finite support construction". So doing the same with countable support (hence obtaining countable choice, and in fact DC) might give us that countable subsets can be mapped onto the power set, but not necessarily the finite subsets. So I'm not sure yet whether or not this is actually consistent with countable choice. So I wouldn't hurry to accept it just yet. – 2017-02-15
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    Having taken the time to understand the proof of the negative answer, I cannot say if it stands in a model of countable choice. I think I should keep the answer accepted since the main question as that of the provability in ZF. – 2017-02-20
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    @nombre: Yes, I agree that it is unclear. I'm thinking that perhaps one can show that if such a surjection exists, then the set is surely Dedekind-finite, and therefore countable choice would have to rule this out. But this remains to be seen. – 2017-02-20