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It is well-known that the continuum hypothesis (CH) holds in the constructible universe $L$. Now suppose we add a nonconstructible subset $a_1$ of $\omega$, and consider the universe $L[a_1]$. Will CH be automatically true or automatically false in this new model, or is there an independence result?

If $CH$ still holds in $L[a_1]$, we may still add a subset $a_2$ of $\omega$ with $a_2 \not\in L[a_1]$. As long as CH is false, we may add a new subset of $\omega$ (in fact we can add nonconstructible elements as we please, regardless of whether CH holds or not). So for each ordinal $\alpha$, we have an axiom ${\sf Ax}_{\alpha}$ stating the existence of a function $a:\alpha \to {\cal P}(\omega)$ such that $a(\beta) \not\in L[(a(\gamma)_{\gamma < \beta})]$ for all $\beta < \alpha$.

The question can then be stated as, is there an $\alpha$ such that $ZFC+{\sf Ax}_{\alpha}$ implies that CH is false ? What is the smallest such $\alpha$ ?

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up vote 6 down vote accepted

On one hand, if we start with $V=L$ then of course that CH holds, let $V[G]$ be a generic extension which just added one new real (Cohen, random, etc.), of course it is not $V=L$ anymore however $2^\omega=\omega_1$ since we added relatively few subsets.

On the other hand, if we add $\omega_2$ many reals then of course the continuum hypothesis will not hold. If we then collapse $\omega_2$ to $\omega_1$ then CH is true again, but there are $\omega_1$ many (mutually) non-constructible reals.

This is enough to conclude that you need to have $\omega_2$ many (non-constructible) reals to deduce that CH is false.

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On the other hand, there are generic extensions of $L$ obtained by adding a single real, such that CH fails in the extension. This can be done, for example, by the tricks of iterative almost-disjoint coding, so that the real codes a subset of $\omega_1$, which codes a subset of $\omega_2$, which lists many new reals. – JDH Jan 19 '12 at 11:23
@JDH: Yeah, I heard of an interesting use of this trick in generating a class forcing which gives a non-ground model strictly between $L$ and $L[a]$ when $a$ is a single generic real (you just add enough about these codes each step) – Asaf Karagila Jan 19 '12 at 11:34
@JDH: Although this is sort of cheating... :-) – Asaf Karagila Jan 19 '12 at 11:36
Yes, this method is known as "coding the universe", due to Jensen. Every model $V$ of ZFC has a class forcing generic extension $V[G]$, which has a real $a$ such that $V[G]=L[a]$. – JDH Jan 19 '12 at 11:46
@JDH: Yeah, so I heard. I do not remember the exact details, I believe that in my first comment the resulting model does not satisfy the axiom of choice. I cannot recall at the moment, though. – Asaf Karagila Jan 19 '12 at 11:47

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