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Note: This is a question and thus doesn't need the usual type of context.

According to this:

It is a surprising fact that the statement $$A > B \implies \mathcal{P}(A) > \mathcal{P}(B)$$ is undecidable in $\mathrm{ZFC}$.

. . . and . . .

we define $A > B$ to mean: there exists an injection $B\to A$ but no bijection between these two sets.

Please may I have a reference for this?

Context:

It is indeed surprising. I can't say that I would understand a proof of it, but having a reference would go a long way to convincing whoever that it's true; it's the kind of thing I would share with fellow students. If the reference includes some prerequisites for the proof, that would be ideal.

Shaun
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    The forcing construction used in a first proof that CH is not provable where $\aleph_2$ Cohen reals are added to a model of GCH satisfies $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$. I don’t know of a better reference than Kunen’s book or any intro reference on forcing, but maybe someone will find something where it’s laid out more explicitly. – spaceisdarkgreen Dec 21 '23 at 14:46
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    Also, it’s a more difficult forcing construction, but if you’re willing to take the consistency of MA +$\lnot$CH as a black box, one can show $2^\kappa=2^{\aleph_0}$ for all infinite $\kappa<2^{\aleph_0}$. This is also in Kunen (and Jech, and most anywhere forcing is sold). – spaceisdarkgreen Dec 21 '23 at 14:54
  • I don't see the link between this and the Continuum Hypothesis, @spaceisdarkgreen, but thank you; I believe you. – Shaun Dec 21 '23 at 15:04
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    If CH is true, then $2^{\aleph_0}=\aleph_1< 2^{\aleph_1}$, so $B=\aleph_0$ and $A=\aleph_1$ is not a counterexample. Similarly, GCH implies there are no counterexamples. So any counterexample is going to need to come from a universe where GCH is false, and violating CH is one way to do it. (And as in my first comment, it turns out the simplest way of violating CH happens to work for producing a counterexample.) – spaceisdarkgreen Dec 21 '23 at 15:36
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    There's also good discussion here, here, here of the relevant constructions and theorems - "Easton's theorem" is hugely overkill but it gives some broader context/an extension of this fact. In some sense there are only two basic facts that ZFC proves about cardinalities of powersets. – Izaak van Dongen Dec 21 '23 at 16:43
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    @IzaakvanDongen (For regular cardinals ) – spaceisdarkgreen Dec 21 '23 at 18:30
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    By the way, the hypothesis that $2^{\aleph_0}=2^{\aleph_1}$ is sometimes known as Luzin's hypothesis, who used it to prove various results in general topology. – Jason Zesheng Chen Dec 21 '23 at 23:25

1 Answers1

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Cohen's original work would be enough.

Cohen, Paul Joseph, The independence of the continuum hypothesis. I, II, Proc. Natl. Acad. Sci. USA 50, 1143-1148 (1963); 51, 105-110 (1964). ZBL0192.04401.

The second paper, Lemma 21 and Lemma 22 give us exactly that if we are adding $\kappa$ many Cohen reals, and $\kappa^{\omega_1}=\kappa$ in the ground model, then $2^{\aleph_1}=\kappa$ in the extension, and therefore $2^{\aleph_0}=2^{\aleph_1}$ in that case.

Asaf Karagila
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