4
$\begingroup$

Specker proved in 1957 that if in $V$ every set of real numbers has the perfect set property, than in $L$, $\omega_1^V$ is actually a limit cardinal.

The original proof is in German, and I've been looking for an English account of the proof. I couldn't find it in all the usual places (usually just a reference). I do remember it not being that complicated when it was given in a course I attended a couple of years ago, so it seems strange that I couldn't find it in the books.

Can anyone help me find such English version of the proof (or give it here if it is short enough)?

$\endgroup$
  • $\begingroup$ Kanamori's book, section 11. $\endgroup$ – Andrés E. Caicedo Mar 7 '14 at 14:40
  • $\begingroup$ I checked that, but it seemed as if Kanamori discusses the result, but doesn't really present a proof. $\endgroup$ – Asaf Karagila Mar 7 '14 at 14:43
  • $\begingroup$ (See 11.3 to 11.5, pp. 133-135.) $\endgroup$ – Andrés E. Caicedo Mar 7 '14 at 22:06
  • $\begingroup$ Thanks. I think that it's the fact that he discusses inaccessibility, rather than being a limit cardinal that had me confused there. I should know better! :) $\endgroup$ – Asaf Karagila Mar 7 '14 at 22:09
  • $\begingroup$ I think it was a missed oppprtunity on his part, not to separate the result from regularity. Truss's paper complements section 11 nicely. $\endgroup$ – Andrés E. Caicedo Mar 7 '14 at 22:44
6
$\begingroup$

The usual proof (as in Kanamori's book, section 11) is as follows: Work in $\mathsf{ZF}$. Note first, with Bernstein, that if $\omega_1\le\mathfrak c$, then there is a set of reals without the perfect set property: Either $\omega_1=\mathfrak c$, so $\mathbb R$ can be well-ordered, and we can build Bernstein sets using the usual transfinite recursion, or else $\omega_1<\mathfrak c$, and any set of reals of size $\omega_1$ lacks the perfect set property (that there are exactly continuum many perfect sets, and that each perfect set contains a copy of the Cantor set and therefore has size $\mathfrak c$ are provable in $\mathsf{ZF}$).

Now, under the assumption that all sets of reals have the perfect set property, we argue that $\omega_1$ is a limit cardinal in $L[r]$ for all reals $r$: Suppose otherwise, so for some real $r$ and some $\kappa$, $\omega_1=\kappa^+$ in $L[r]$. Let $s$ be a real coding $r$ and a well-ordering of $\omega$ in type $\kappa$. In $L[s]$, we have that $\omega_1$ is computed correctly. But now we see that $\omega_1\le\mathfrak c$, as witnessed by $\mathbb R^{L[s]}$.

As Asaf points out in the comments, if $\omega_1$ is regular in $V$, this gives us that it is inaccessible in $L[r]$ for all reals $r$, but it is equiconsistent with $\mathsf{ZF}$ that $\omega_1$ is singular and yet the perfect set property holds, see

John Truss. Models of set theory containing many perfect sets. Ann. Math. Logic, 7, (1974), 197–219. MR0369068 (51 #5304).

$\endgroup$
  • $\begingroup$ See, now we only argue that $\omega_1$ is a limit cardinal. You need to assume regularity to get inaccessibility (and of course this is needed, as shown by Truss in his models where $\omega_1$ is singular and the perfect set property holds). $\endgroup$ – Asaf Karagila Mar 7 '14 at 21:39
  • $\begingroup$ Yes, you are right. $\endgroup$ – Andrés E. Caicedo Mar 7 '14 at 21:52

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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