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we know that in this model, the relation $\supseteq_G$ is reflexive, transitive and directed. This was used in [13] to calculate MLG in that model to be S4.2; we are reusing parts of this proof in our main result.

DEFINITION 14. Let $\kappa$ be a regular cardinal. The forcing poset Add($\kappa$) which adds a Cohen subset of $\kappa$ is the following:

  1. $p \in \text{Add}(\kappa)$ if $p$ is a function and there is a $\gamma < \kappa$ such that $\text{dom}(p) = \gamma$ and $\text{ran}(p) \subseteq {0, 1}$.

  2. If $p, q \in \text{Add}(\kappa)$, then $p \le q$ if $p \supseteq q$.

DEFINITION 15. Let $\text{SuccR}^\mathcal{L}$ denote the class of successors of infinite regular cardinals in $\mathcal{L}$. Define in $\mathcal{L}$ the following (class-sized) partial order with Easton support:

P:=∏γ∈SuccRLAdd(γ). \mathbb{P} := \prod_{\gamma \in \text{SuccR}^\mathcal{L}} \text{Add}(\gamma).

That is, $p \in \mathbb{P}$ if

  1. $p$ is a function such that $\text{dom}(p) \subseteq \text{SuccR}^\mathcal{L}$;

  2. for each $\gamma \in \text{dom}(p)$, $p(\gamma) \in \text{Add}(\gamma)$;

  3. for each regular cardinal $\gamma$, $\lvert\text{dom}(p) \cap \gamma\rvert < \gamma$.

The ordering is defined by $p \le q$ if $p \supseteq q$.

Also, for each $p \in \mathbb{P}$ and each $\gamma \in \text{SuccR}^\mathcal{L}$, we can decompose $p$ into three parts:

p<γ=p∣[0,γ);pγ=p∣[γ,γ];p>γ=p∣(γ,∞). \begin{align*} p<\gamma &= p|[0, \gamma); \\ p_{\gamma} &= p|[\gamma, \gamma]; \\ p>\gamma &= p|(\gamma, \infty). \end{align*}

Using this decomposition, for each $\gamma \in \text{SuccR}^\mathcal{L}$, we can decompose $\mathbb{P}$ into three parts:

P<γ={p<γ∣p∈P};Pγ={pγ∣p∈P};P>γ={p>γ∣p∈P}. \begin{align*} \mathbb{P}_{<\gamma} &= \{p_{<\gamma} | p \in \mathbb{P}\}; \\ \mathbb{P}_{\gamma} &= \{p_{\gamma} | p \in \mathbb{P}\}; \\ \mathbb{P}_{>\gamma} &= \{p_{>\gamma} | p \in \mathbb{P}\}. \end{align*}

It is clear that $\mathbb{P} \cong \mathbb{P}_{<\gamma} \times \mathbb{P}_{\gamma} \times \mathbb{P}_{>\gamma}$.

THEOREM 16. Assume $\mathbf{V}=\mathbf{L}$. Let $\gamma$ be an infinite successor cardinal. Let $Q_\gamma := \mathbf{P}_{<\gamma} \times \mathbf{P}_{>\gamma}$. Then forcing with $Q_\gamma$ does not add a Cohen subset of $\gamma$.

PROOF. This result is essentially due to McAloon [21]. Since $\mathbb{P}{>\gamma}$ is clearly $\leq \gamma$-closed, we only need to show that forcing with $\mathbb{P}{<\gamma}$ does not add a Cohen subset of $\gamma$. But for this, notice that by GCH and the fact that $\gamma$ is the successor of a regular cardinal, it has the $\gamma$-c.c. and consequently does not add Cohen subsets of $\gamma$. ⊥

Fix any transitive set $M \models ZFC$, consider $\mathbb{P}$ to be defined in $\mathcal{L}^M$, and let $G$ be $\mathcal{L}^M$-generic for $\mathbb{P}$, and let $R := \mathcal{L}^M[G]$. If $\gamma$ is an infinite successor cardinal of $\mathcal{L}^M$, we write $G_{>\gamma} := G \cap \mathbb{P}_{>\gamma}$. Let

τ:=∀κ∈SuccRL∃G⊆κ(G is an L-Cohen subset of κ) and \tau := \forall \kappa \in \text{SuccR}^{\mathcal{L}} \exists G \subseteq \kappa (\mathcal{G} \text{ is an } \mathcal{L}\text{-Cohen subset of } \kappa) \text{ and}