1

Is the following statement consistent with ZF (obviously, this is not true in ZFC)?

There are cardinals $\kappa, \lambda$ such that $\kappa \neq \lambda$ and $2^\kappa = 2^\lambda$.

Asaf Karagila
  • 38,140
Sumac
  • 113
  • 1
    Surely this type of thing has been asked multiple times at MO? One instance is here https://mathoverflow.net/questions/82769/is-it-consistent-with-zfc-that-for-all-ordinals-alpha-beta-omega-it-hold/82782#82782 (which asks for a stronger statement). – Todd Trimble May 09 '17 at 11:03
  • I am sorry, I failed to find the answer through searching. The moderators can delete my question. – Sumac May 09 '17 at 11:10
  • It's okay to keep the question. This one should be easy for others to find in future searches, esp. since the title is in plain English. – Todd Trimble May 09 '17 at 12:19
  • 3
    According to MO users, that different cardinals should not have power sets of the same size is the most reasonable-sounding statement which is independent of ZFC (with 224 votes as of right now). See this answer. – Burak May 09 '17 at 12:21

1 Answers1

6

Yes, this is possible, even with ZFC. In Cohen's model, where the continuum hypothesis fails, he gets ZFC plus $2^\omega=2^{\omega_1}$. This statement is known as Luzin's hypothesis, an alternative to CH.

More generally, the situation for regular cardinals is described by Easton's theorem, which says that if ZFC plus the GCH holds, then for any function $E:\kappa\mapsto E(\kappa)$ on the regular cardinals satisfying the properties that $\kappa\leq\lambda\to E(\kappa)\leq E(\lambda)$ and $\kappa<\operatorname{cof}(E(\kappa))$, then there is a forcing extension in which $2^\kappa=E(\kappa)$.

In particular, one can easily make functions $E$ that will thereby ensure ZFC plus many instances of $2^\kappa=2^\lambda$.

Joel David Hamkins
  • 224,022