19
$\begingroup$

Recently I was talking to my friend and I have mentioned to him that it was proven that CH is not provably (over ZFC) equivalent to any statement in second-order arithmetic. However, today I found out that the result I was thinking about is the one mentioned in this answer, which says that GCH is independent of all sentences in analytical hierarchy. This now made me think:

Is there a sentence in the analytical hierarchy which is provably equivalent to continuum hypothesis?

I am expecting the answer to this question to be no, because CH concerns arbitrary sets of reals, but this is only an intuitive reason.

Thanks in advance.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
24
$\begingroup$

The collapsing forcing by countable partial functions from $\omega_1$ to $2^\omega$ is $\omega_1$-closed, hence it preserves $H_{\omega_1}$, and a fortiori the truth of all formulas in the analytical hierarchy; it also makes CH hold. Thus CH is not equivalent to any statement in the analytical hierarchy (assuming ZFC is consistent).

In fact, the argument implies that ZFC+CH is $\Pi^2_1$-conservative over ZFC, hence CH is not equivalent to a $\Pi^2_1$ statement. On the other hand, it is easy to see that it can be formulated as a $\Sigma^2_1$ statement.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I suppose it's standard terminology, but what does "$\omega_1$-closed" mean? It turned out to be unsuccessful to search for this on Wikipedia. $\endgroup$
    – Wojowu
    Commented Sep 18, 2015 at 17:37
  • 6
    $\begingroup$ It means that every countable chain of forcing conditions has a lower bound (in the present case, this amounts to the observation that the union of countably many compatible countable partial functions is again a countable partial function). This condition is also called “countably closed” or “$\sigma$-closed”. $\endgroup$
    – Emil Jeřábek
    Commented Sep 18, 2015 at 17:43
  • 6
    $\begingroup$ @Wojowu: It means $\kappa$-closed, for $\kappa=\omega_1$. It sounds like a snide comment, but it's not. The term $\kappa$-closed forcing is a lot easier to find. What Emil says is correct, for $\kappa=\omega_1$ it is often called $\sigma$-closed instead. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 18, 2015 at 18:13
  • 14
    $\begingroup$ It may be worth adding that, in a sense, $\mathsf{CH}$ is the $\Sigma^2_1$ statement. Woodin's $\Sigma^2_1$-absoluteness theorem can be stated as follows: Assuming enough large cardinals (a proper class of Woodin cardinals suffices) and that $\mathsf{CH}$ holds, if $\phi$ is a $\Sigma^2_1$ statement, and $\mathbb P$ is a poset forcing $\mathsf{CH}$, then $\mathbb P$ forces $\phi$ iff $\phi$ already holds. That is, informally, any $\Sigma^2_1$ statement compatible with $\mathsf{CH}$ already follows from $\mathsf{CH}$. $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 18, 2015 at 21:08

You must log in to answer this question.

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