3
$\begingroup$

There are two important numbers that in some meaningful sense describe "how well-orderable" the reals are:

  1. Hartogs' Number $H(\Bbb R)$, also notated as $\aleph(\Bbb R)$, the least ordinal/well-ordered cardinal that doesn't inject into $\Bbb R$
  2. The ordinal $\Theta$, also notated as $\aleph^*(\Bbb R)$, the least ordinal/well-ordered cardinal that $\Bbb R$ doesn't surject onto

The first number $H(\Bbb R)$ can be thought of as describing the supremum of the cardinalities of all well-orderable subsets of $\Bbb R$, whereas the second number $\Theta$ can be thought of as describing the supremum of the cardinalities of all well-orderable equivalence classes of $\Bbb R$.

With choice, these are all equal to $\mathfrak c^+$, the cardinal after $\mathfrak c$.

My question: without choice, do we have any results regarding:

  1. How large each of these numbers can be?
  2. How small each of these numbers can be?
  3. Which number is larger or if they can be equal?

I know that $AD$ determines some of these strongly enough to relate them to large cardinals (I believe Woodin cardinals), but I'm also interested in what the possibilities are without something that strong.

$\endgroup$
  • 1
    $\begingroup$ $\aleph_1\leq\aleph(\Bbb R)\leq\aleph^*(\Bbb R)$. And that's the only provable thing here. $\endgroup$ – Asaf Karagila Jan 30 '19 at 21:42
  • 1
    $\begingroup$ With choice, those aren't $\mathfrak c+1$, but rather $\mathfrak c^+$, the next initial ordinal after $\mathfrak c$. $\endgroup$ – Wojowu Jan 30 '19 at 21:49
  • 3
    $\begingroup$ You believe Woodin cardinals? What have they told you? $\endgroup$ – Asaf Karagila Jan 30 '19 at 22:32
  • 2
    $\begingroup$ Ah yes, Woodin Cardinals... They arose one day in a dream and spoke to me in a strange language, which I could only understand if I believed I could understand. But if I didn't truly believe, they would be inconsistent and vanish... $\endgroup$ – Mike Battaglia Jan 30 '19 at 22:51
  • 4
    $\begingroup$ @Mike: Is this a call of Cthulhu kind of situation? $\endgroup$ – Asaf Karagila Jan 30 '19 at 23:49
11
$\begingroup$

Other than $\aleph_1\leq\aleph(\Bbb R)\leq\aleph^*(\Bbb R)$ and at least one of them is sharp, not much is provable. And of course, both of these inequalities are easy to prove.

We can have $\aleph(\Bbb R)$ be any uncountable cardinal, even singular. And we can have $\aleph^*(\Bbb R)$ be any cardinal satisfying the above inequality, even singular.

The whole thing can be achieved with clever combination of collapsing cardinals and adding Cohen reals. The only thing to note, though, is that if $\aleph(\Bbb R)=\aleph_1$ then:

  1. $\aleph(\Bbb R)<\aleph^*(\Bbb R)$,
  2. If $\aleph_1$ is regular (e.g. in presence of countable choice, or determinacy), then $\omega_1$ is inaccessible to reals, and in particular inaccessible in $L$ (not to be confused with $\omega_1^L$ is inaccessible, of course).
  3. As a consequence, we had to collapse cardinals.

The point is that starting from CH, given two uncountable cardinals $\aleph_1<\kappa\leq\lambda$ we can arrange a model just by adding Cohen reals (so cofinalities and cardinals are not changed), where $\aleph(\Bbb R)=\kappa$ and $\aleph^*(\Bbb R)=\lambda$.

If we also want $\aleph(\Bbb R)=\aleph_1$, then we need to be more subtle about it and collapse some limit cardinal (e.g $\aleph_\omega$ or an inaccessible).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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