Assume $\mathsf{ZF} + \mathsf{DC}$. Must there exist an injection from $\aleph_2$ to $\mathcal{P}(\mathbb{R})$? If not, what is the consistency strength of the nonexistence of such an injection?

I hope I'm not missing something obvious. Here are my thoughts so far:

  1. There is always a surjection from $\mathcal{P}(\mathbb{R})$ to $\aleph_2$.
  2. $\mathsf{ZFC}$, or more generally the existence of a wellordering of $\mathcal{P}(\mathbb{R})$, implies there is such an injection.
  3. $\mathsf{ZF} + \mathsf{AD}$ implies the existence of a surjection from $\mathbb{R}$ to $\aleph_2$, which implies there is such an injection.
  4. If there is an inner model of $\mathsf{ZFC}$ satisfying $(\aleph_1^V)^+ = \aleph_2^V$, for example in $V(\mathbb{R}^{V[G]})$ where $\kappa$ is inaccessible and $G \subset \operatorname{Col}(\omega,\mathord{<}\kappa)$ is a $V$-generic filter, then there is such an injection.
  • 2
    $\begingroup$ The problem with you people of determinacy and inner models is that you always think in terms of large cardinals and $L(\Bbb R)$ style models... But the answer is often simpler than that! :) $\endgroup$
    – Asaf Karagila
    Jan 27 '18 at 7:01
  • $\begingroup$ Why is there always a surjection from $\mathcal P(\mathbb R)$ to $ \aleph_2?$ Is that supposed to be obvious? $\endgroup$
    – bof
    Jan 27 '18 at 7:25
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    $\begingroup$ @bof There's a surjection from $\mathbb{R}$ to $\aleph_1$, so there's a surjection from $\mathcal{P}(\mathbb{R})$ to $\mathcal{P}(\aleph_1)$, which we can compose with a surjection from $\mathcal{P}(\aleph_1)$ to $\aleph_2$. $\endgroup$ Jan 27 '18 at 7:45

The consistency strength is just that of ZFC.

Start with GCH, and do a symmetric collapse of $\aleph_{\omega_1}$ to be $\aleph_2$. Since everything is $\sigma$-closed, you get DC for free, and the reals can even be well ordered.

But any injection from $\aleph_2$ into $\cal P(\Bbb R)$ would have had to be adjoined by some bounded part of the iteration (by homogeneity), so none exist, as with the usual arguments for $L(\Bbb R)$.

  • $\begingroup$ Ah, right. I should have seen the analogy with getting $\aleph_1 \not\le \mathbb{R}$. $\endgroup$ Jan 27 '18 at 7:15
  • 2
    $\begingroup$ Indeed. Of course, if you want $\aleph_2$ to also be regular, an inaccessible is needed. $\endgroup$
    – Asaf Karagila
    Jan 27 '18 at 10:46

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