On page 14 of the introduction to Vol. II of Gödel's collected works:

By a slightly more difficult argument one can show that GCH continues to hold if $V=L[a]$ and $a\subseteq\aleph_1$.

Does anyone know this argument?

  • $\begingroup$ This is also an exercise from Jech's Set Theory, 13.26 (p.198) $\endgroup$
    – Asaf Karagila
    Mar 11, 2013 at 8:24
  • $\begingroup$ (It is a natural question. It appears as an exercise in a few books.) $\endgroup$ Mar 11, 2013 at 22:44

1 Answer 1


Suppose $A\subseteq\omega_1$ and $V=L[A]$.

Trying to imitate the usual argument (which I assume you have seen), we should start with an appropriate version of the condensation lemma. What we get is that if $X\prec L_\alpha[A]$ for some infinite limit $\alpha$, then $X\cong L_\beta[B]$ for some $B$ and, in fact, if $\pi:X\to L_\beta[B]$ is the transitive collapse, then $B$ is simply $\pi''(X\cap A)$.

Suppose in particular that $X$ is countable, $X\prec L_{\eta}[A]$, $\eta$ large enough. We use the standard fact that $X\cap\omega_1$ is an ordinal $\gamma$. (This is because for any ordinal $\tau\in X\cap\omega_1$, $X\models$"$\tau$ is countable", so there is in $X$ a surjection $f:\omega\to\tau$. But $\omega\subset X$, so $\tau=f''\omega\subset X\cap\omega_1$.)

We have then that $\pi''(X\cap A)=\pi''(A\cap\gamma)=A\cap\gamma$. Any "real" $r\subset\omega$ belongs to some such $X$, and so $\pi(r)=r$, which means that $r\in L_\beta[A\cap\gamma]$ for some countable ordinals $\beta$ and $\gamma$.

We conclude by noting that any such structure $L_\beta[A\cap\gamma]$ belongs to $L_{\omega_1}[A]$, which has size $\omega_1$, and therefore there are only $\omega_1$ many reals, that is, $\mathsf{CH}$ holds.

For $\kappa>\omega$, we can just implement the usual argument directly, assuming $X$ has size $\kappa$ which is at least $\omega_1$, and $\kappa\subset X$, and $A\in X$. Now the collapse of $X$ has the form $L_\beta[A]$ for some $\beta<\kappa^+$. Given $Y\subset\kappa$, we may assume $Y\in X$, which gives us $\pi(Y)=Y\in L_\beta[A]$, so each subset of $\kappa$ belongs to $L_{\kappa^+}[A]$, which has size $\kappa^+$. The conclusion is that $2^\kappa=\kappa^+$ as well.

  • $\begingroup$ No harassment, :-) But the argument is right. $\endgroup$ Jun 1, 2020 at 20:44

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