Suppose CH + "there are no Cohen reals over L". Can we force the negation of CH without adding any Cohen real over L?

3$\begingroup$ Use random reals. $\endgroup$ – Thomas Benjamin Jan 4 '16 at 15:15

$\begingroup$ Does adding \aleph_2 random reals work? Or \aleph_2 Sacks reals? And if so, how do one show that? $\endgroup$ – Ohad Drucker Jan 4 '16 at 15:35

1$\begingroup$ One key point here is that if there are no Cohen reals over L, then $\omega_1$ is computed correctly in $L$. $\endgroup$ – Asaf Karagila Jan 4 '16 at 16:28

$\begingroup$ Thanks! So should I understand from your comments that if CH + "no L  Cohen reals", then after adding $\aleph_2$ random reals, $\mathbb{V}[G] \models \neg CH$ + "no L  Cohen reals" ? $\endgroup$ – Ohad Drucker Jan 4 '16 at 18:34
Let $\mathbb{S}$ denote Sacks forcing. Let $\mathbb{S}_{\omega_2}$ denote the $\omega_2$length countable support iteration of Sacks forcing.
Let $G_{\omega_2} \subseteq \mathbb{S}_{\omega_2}$ be generic over $L$ for $\mathbb{S}_{\omega_2}$. By the usual results around proper forcing, you can show that $L[G_{\omega_2}] \models 2^{\aleph_0} = \aleph_2$.
You can show that no Cohen reals over $L$ are added by showing that every real of $L[G_{\omega_2}]$ is contained in a ground model coded closed meager set. Note that Cohen forcing is forcing equivalent to $\mathbb{P}_{I_\text{meager}}$, the forcing of nonmeager Borel sets. Hence a Cohen real is not contained in any ground model meager set.
For simplicity, let first consider one Sacks forcing $\mathbb{S}$. Let $\tau \in V^{\mathbb{S}}$ and $\tau$ be a name for a real not in the ground model. Then do a fusion argument to produce a condition $p$ so that at every split node $s$ of $p$, $p_{s0}$ and $p_{s1}$ determines a certain finite amount of $\tau$ and what $p_{s0}$ determines about $\tau$ and what $p_{s1}$ determines about $\tau$ differs in at least two places. If you let $T$ be the set of all finite strings $t$ so that $t$ is an initial segment of what $p_s$ can determined about $\tau$ for some $s \in {}^{<\omega}2$, then you get a tree so that at each split node the two sides differ two times before splitting again. By the definability of forcing, $T \in L$. You can show that body $[T]$ is meager closed set and $p \Vdash \tau \in [\check T]$.
If you understand how to do this for 1Sacks forcing, now do an iterated Sacks forcing version of this. This tends to be quite heavy in notation. See Geschke and Qickert $\textit{On Sacks Forcing and the Sacks property}$ for more information. Essentially the argument above is a modified version of their proof of the 2localization property. Also see their paper on how to do the iterated Sacks version of the 2localization property.
It seems that I understood the question as is it ever possible to force the continuum to be $\aleph_2$ without adding Cohen generic over $L$. It seems the far more interesting question is whether for any universe $V$ (perhaps not equal to $L$) which does not have Cohen generic over $L$, is there a forcing extension of $V$ which does not add Cohen generics over $L$.

$\begingroup$ Link to the paper you mention by Geschke and Qickert: math.unihamburg.de/home/geschke/papers/OnSacks_main.pdf $\endgroup$ – Asaf Karagila Jan 4 '16 at 22:21

3$\begingroup$ Does this work if the ground model is something other than L? Can Sacks forcing over a model of CH containing no LCohen reals add an LCohen real? $\endgroup$ – Todd Eisworth Jan 5 '16 at 21:32

$\begingroup$ Is it clear that $x$ would also be Sacksgeneric over $L$? What if, say, $V$ itself is a Sacksgeneric extension of $L$? I can't see why $x$ cannot code some extra information, enough to ensure that there are inner models of $L[x]$ which are not $L$. $\endgroup$ – tci Jan 6 '16 at 13:10

$\begingroup$ @tci You are right, a Sacks generic for $V$ may not be Sack's generic for $L$. $\endgroup$ – William Jan 7 '16 at 1:34