Can we explicitly define two ordinals $\alpha$ and $\beta$ in the language of $\{\in\}$ such that the following hold?

  1. ZFC proves that $\alpha$ and $\beta$ exist.
  2. ZFC proves that $\beth_\beta \neq \aleph_\alpha$
  3. The following statements are independent of ZFC: $$\aleph_\alpha \leq \beth_\beta,\qquad\beth_\beta \leq \aleph_\alpha$$

Doesn't this happen if $\alpha=\omega$ and $\beta=1$?

Fact 2 follows from Konig's theorem and fact 3 can be proved by forcing arguments.

  • 3
    $\begingroup$ :-) I type too slow. $\endgroup$ Apr 3 '14 at 14:40
  • $\begingroup$ Sounds reasonable. Would it be fair to say that if $\alpha$ is a limit ordinal and $\beta$ is a non-limit ordinal, then $\aleph_\alpha$ and $\beth_\beta$ must be distinct? (Btw I do not see how to use Konig's theorem to show Fact 2). $\endgroup$ Apr 3 '14 at 14:44
  • $\begingroup$ @Andres: It's time you buy a good mechanical keyboard, then! $\endgroup$
    – Asaf Karagila
    Apr 3 '14 at 14:45
  • $\begingroup$ @user18921: Yes, for limit $\beta$, the cardinal $\beth_\beta$ is always a limit cardinal, so it cannot be $\aleph_\alpha$ for any successor cardinal $\alpha$; as for Koenig's theorem, it shows that $\aleph_\omega$ cannot be $\beth_1$. $\endgroup$
    – Asaf Karagila
    Apr 3 '14 at 14:48
  • 2
    $\begingroup$ To clarify: Konig's thoerem gives us that $cf(2^{\aleph_{0}})>cf(\aleph_{0})$. Since $\aleph_{\omega}$ has cofinality $\omega$, we get the result. $\endgroup$
    – UserB1234
    Apr 3 '14 at 14:57

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