2
$\begingroup$

Let $s_n$ be the least $\Sigma_n-$admissible ordinal, so that $\mathrm{L}_{s_n}$ is a model of Kripke-Platek set theory with $\Sigma_n-$collection and $\Sigma_n-$separation.

Does $\mathrm{L}_{s_{n+1}}$, for $n>0$, contain a surjection from $\omega$ to $\mathrm{L}_{s_n}$?

Improve this question
$\endgroup$
3
  • $\begingroup$ Check your code. It does not render multiple statements. You have for example missed \ before Sigma at two places, AFAICT. $\endgroup$
    – Ben123
    Commented Jul 27 at 19:34
  • 1
    $\begingroup$ @Bent123 Yes, I already fixed it. $\endgroup$
    – Frode Alfson Bjørdal
    Commented Jul 27 at 19:36
  • $\begingroup$ Well, your $n > 0$ is not inside mathmode, but otherwise it looks better. $\endgroup$
    – Ben123
    Commented Jul 27 at 22:03

1 Answer 1

Reset to default
5
$\begingroup$

In fact much more is true: if $L_\alpha$ is the first level of the $L$-hierarchy satisfying some first-order theory $T$, then it will be pointwise-definable$^*$ and so $L_{\alpha+2}$ will contain an injection from $L_\alpha$ to $\omega$. (We need the "$+2$" here for the naive argument via the satisfaction relation, and in fact this is necessary in general as the example of $\alpha=\beta_0$ and $T=\mathsf{ZFC}^-$ shows.)

The result follows since $s_{n+1}$ is much larger than $s_n+2$. (EDIT: In fact we can do even better: since $s_n$ is smaller than the first gap ordinal $\beta_0$, we know that $L_{{s_n}+1}$ already contains a "counting" of $L_{s_n}$. But this takes more than the naive argument above, which applies to the problem as stated in the OP.)

$^*$To see this, note that the set $D$ of definable elements of $L_\alpha$ must be an elementary substructure of $L_\alpha$, and the Mostowski collapse map must be the identity by minimality of $\alpha$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ "We need the "+2" here for the naive argument via the satisfaction relation, and in fact this is necessary in general as the example of 𝛼=𝛽0 and 𝑇=𝖹𝖥𝖢− shows." As you remarked to that question that $\beta_0$ is the first ordinal with gap in $\mathrm{L}$, I thought that the situation would be different here. Why does the same result hold for gap-free ordinals smaller than $\beta_0$? $\endgroup$
    – Frode Alfson Bjørdal
    Commented Jul 27 at 22:14
  • 1
    $\begingroup$ @FrodeAlfsonBjørdal See my edit. I just wanted to focus on what we could get from the least amount of work. $\endgroup$
    – Noah Schweber
    Commented Jul 27 at 22:30
  • $\begingroup$ Philip Welch remarked "$𝐿_{\beta_{0}+1}$ sees that $L_{\beta_0}$ is countable, and so I suppose one may define a particular $\omega-$sequence in "$𝐿_{\beta_{0}+1}$ cofinal in it by just taking the $<𝐿$- least such." in his reply to my question mathoverflow.net/questions/138284/…. Does his supposition at that point perhaps conflict with the answer by Gabe Goldberg which we alluded to above? $\endgroup$
    – Frode Alfson Bjørdal
    Commented Jul 27 at 23:23
  • 1
    $\begingroup$ @FrodeAlfsonBjørdal I think that Philip means that such a bijection is definable over $L_{\beta_0+1}$, i.e. an element of $L_{\beta_0+2}$. But you'd have to ask him. $\endgroup$
    – Noah Schweber
    Commented Jul 27 at 23:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .