6
$\begingroup$

I was wondering whether it is consistent to have $\frak{c} = \aleph_{\frak{c}}$ where $\frak{c} = 2^{\aleph_0}$ is the cardinality of the reals (over ZFC). If so, what interesting consequences of this statement are known (besides ¬CH)? I was curious about this because in some sense $\frak{c}$ is the largest possible number of cardinals below $\frak{c}$, and this is partly motivated by the idea that $\frak{c}$ may be so large as to be 'unreachable' via approximation by fewer smaller cardinals, which seems similar in nature to an opinion of Cohen on CH.

From what I have read, I think that it is consistent (relative to ZFC) for $\frak{c}$ to be the $ω_1$-th fixed-point of $\aleph$, which would be one possibility satisfying $\frak{c} = \aleph_{\frak{c}}$. But can $\frak{c}$ be the $\frak{c}$-th fixed-point of $\aleph$, and does this yield even more interesting consequences?

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ All the things you ask for are consistent by Easton's theorem. For the last part, continuum can be the $\omega_1$-th cardinal $\alpha$ such that $\aleph_\alpha=\alpha$. $\endgroup$
    – Wojowu
    Commented Jun 24, 2019 at 10:14
  • 3
    $\begingroup$ @Wojowu: Why Easton? Might as well argue that these things are consistent because starting with a supercompact cardinal, if you add that many Cohen reals, you get the wanted result. (The point I am trying to make is that Cohen and Solovay already prove this result, there's no reason involving class forcing over all regular cardinals here.) $\endgroup$
    – Asaf Karagila ♦
    Commented Jun 24, 2019 at 10:17
  • 2
    $\begingroup$ @AsafKaragila That's a good point. I appeal to Easton because that's one theorem of this sort which I know by the name :P $\endgroup$
    – Wojowu
    Commented Jun 24, 2019 at 12:49
  • $\begingroup$ @Wojowu: Letting $k$ be the $ω_1$-th aleph-fixed-point, $k$ cannot be the $k$-th aleph-fixed-point, so it doesn't answer my last question, but Asaf said my comment works. $\endgroup$
    – user21820
    Commented Jul 1, 2019 at 14:20
  • $\begingroup$ Robert Solovay, "$2^{\aleph_0}$ can be anything it ought to be", The theory of models. Proceedings of the 1963 International Symposium at Berkeley. Zbl 0202.30701 $\endgroup$
    – Goldstern
    Commented Aug 28, 2019 at 9:51

1 Answer 1

Reset to default
8
$\begingroup$

Yes. Start with a model of $\sf CH$, then take the least fixed point with uncountable cofinality. Call that $\kappa$. Now add $\kappa$ Cohen reals.

Since fixed points form a club of ordinals, you can iterate the fixed points enumeration. Repeat that $\omega_1$ times, then take the least one of cofinality $\omega_1$ in that club. Now call that $\kappa$, and add that many Cohen reals.

Assuming no large cardinals get involved, that means that $\frak c$ is singular. This by itself implies that Martin's Axiom fails, and that Cichon's diagram is not trivial, since some of the cardinal characteristics are provably regular.

Other than that, I don't believe we can say a lot more without adding more assumptions on the universe.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks, but what about my last question? And I was hoping ZFC alone can prove some interesting things given that $\frak{c}$ is an $\aleph$-fixed-point. Anyway what can you say under extra assumptions? $\endgroup$
    – user21820
    Commented Jun 24, 2019 at 11:57
  • $\begingroup$ Am I right that since we can define (over ZFC) the class function $F$ that maps $k$ to the $k$-th aleph-fixed-point, we can define an ordinal $m$ that is a fixed-point of $F$, so that $m$ would be the $m$-th aleph-fixed-point, and then by Easton's theorem gives consistency of $\frak{c}$ being the $\frak{c}$-th aleph-fixed-point relative to ZFC? $\endgroup$
    – user21820
    Commented Jul 1, 2019 at 8:07
  • $\begingroup$ Yeah, that's about right. $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 1, 2019 at 8:16
  • $\begingroup$ Okay then thanks, my question is answered! =) $\endgroup$
    – user21820
    Commented Jul 1, 2019 at 8:29

Not the answer you're looking for? Browse other questions tagged or ask your own question.