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Let $S$ be a stationary costationary subset of $\omega_1$ and let $\mathbb{P}$ be the usual poset that shoots a club through $S$ (with closed initial segments, ordered by end-extension). Does $\mathbb{P}$ force CH?

Of course, if CH already holds, then it continues to hold in the extension, since $\mathbb{P}$ doesn't add reals. Also, if we do not assume that $S$ is costationary, the answer is yes, since the intersection of the generic club with the club contained in $S$ codes a Cohen subset of $\omega_1$.

I am mostly interested in this question because I would like to know more about what the generic club looks like. It will obviously be a fresh subset of $\omega_1$ (in the sense that all of its initial segments are in the ground model), but I wonder whether it also codes the ground model reals in some way.

I would also welcome comments about the situation at larger cardinals $\kappa$ (with $S$ appropriately fat).

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Unless I'm misunderstanding the forcing, I think the answer is yes:

Let $G$ be such a generic club. For $\alpha\in G$ and $n\in\omega$, let $\alpha_n$ be the $n$th element of $G$ after $\alpha$. Now say that an ordinal $\beta\in S$ is an $S$-successor if there is a $\gamma\in S$ with $\gamma<\beta$ such that $S\cap (\gamma,\beta)=\emptyset$, and $\beta\in S$ is an $S$-limit otherwise. Note that $S$ contains $\omega_1$-many $S$-limits and $S$-successors.

Now consider the real $$r(\alpha)=\{n: \alpha_n\mbox{ is an $S$-successor}\}.$$ I believe that every ground model real will generically appear as $r(\alpha)$ for some $\alpha$; and this gives a surjection in $V[G]$ from $\omega_1$ to $\mathbb{R}$.

I think a similar trick works for arbitrary $\kappa$. Define $\alpha_\eta$ as above for $\eta<\kappa$, with the caveat that if $\eta$ is a limit then $\alpha_\eta$ is the $(\eta+1)$th element of $G$ after $\alpha$ (since limit elements of $G$ are forced to be $S$-limits). Then, defining $r(\alpha)$ as above, I believe every ground subset of $\kappa$ is represented as $r(\alpha)$ for some $\alpha$, so the generic extension satisfies $2^\kappa=\kappa^+$.

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    I think this is right. A tiny bit of care is needed to prevent the coding points from accumulating towards a point not in $S$, but that's fine since $S$ contains many closed copies of $\omega^2+1$. – 2017-01-11
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    @MihaHabič That's a good point. Alternatively, we can let $r(n)=1$ if there is an element of $S$ between $\alpha_n$ and $\alpha_{n+1}$, and $0$ otherwise. This seems obstacle-free. – 2017-01-11
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After reading Noah's answer, I realized that, in fact, forcing with $\mathbb{P}$ will add a Cohen subset of $\omega_1$ regardless of whether $S$ is costationary or not. This then also answers the question (and gives other information as well). $\newcommand{\P}{\mathbb{P}}$

There is a projection from $\P$ to $\operatorname{Add}(\omega_1,1)$ that works as follows: given a condition $p\in\P$ let $p_\xi$ be its $\xi$th element. We can interpret $p_{\xi+1}$ as giving the $\xi$th bit of a condition in $\operatorname{Add}(\omega_1,1)$, based on whether $p_{\xi+1}$ is an isolated point of $S$ or not. Let $p^*\in\operatorname{Add}(\omega_1,1)$ be the condition derived from $p$ in this way.

The key to seeing that the map $p\mapsto p^*$ is a projection is the fact that any stationary subset of $\omega_1$ is fat, meaning that it contains closed copies of any $\alpha<\omega_1$. So, given a Cohen condition $q\leq p^*$, we can find in $S$ a closed copy of a very large countable successor ordinal above $\max(p)$ that gives us enough space to code the extra information from $q$ into $\bar{p}\leq p$ with $\bar{p}^*\leq q$.

The same argument should work to show that shooting a club through a fat stationary subset of $\kappa$ adds a Cohen subset of $\kappa$.

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    Yes, that seems right. – 2017-01-11