# Perfect set property implies $\omega_1$ is a limit cardinal in $L$

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)?

• Kanamori's book, section 11. – Andrés E. Caicedo Mar 7 '14 at 14:40
• I checked that, but it seemed as if Kanamori discusses the result, but doesn't really present a proof. – Asaf Karagila Mar 7 '14 at 14:43
• (See 11.3 to 11.5, pp. 133-135.) – Andrés E. Caicedo Mar 7 '14 at 22:06
• 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! :) – Asaf Karagila Mar 7 '14 at 22:09
• I think it was a missed oppprtunity on his part, not to separate the result from regularity. Truss's paper complements section 11 nicely. – Andrés E. Caicedo Mar 7 '14 at 22:44

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).

• 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). – Asaf Karagila Mar 7 '14 at 21:39
• Yes, you are right. – Andrés E. Caicedo Mar 7 '14 at 21:52