**Question 1.** Suppose that $0^{\#}$ exists. Are there set generic filters $g,h$ over $L$ ($g,h \in V$) such that for any $f$ ($\in V$) which is generic over $L$ we have that $L[g] \cup L[h] \not \subseteq L[f]$?


In other words: Does $0^{\#}$ imply the failure of the upward directedness of the set generic universe over $L$?


I have the following partial result:


**Corollary 2.** Suppose $0^{\#}$ exists. Let $\kappa$ be the least innaccessible
 cardinal of $L$. Then there are $g,h$ which are $\mathbb{C}$-generic
 over $L$ ($\mathbb{C}$ is Cohen forcing) such
 that for any $f$ which is $\mathbb{P}$-generic over $L$ for some
 $\mathbb{P} \in J\_{\kappa}$ we have
 $L[g] \cup L[h] \not \subseteq L[f]$.


That observation was inspired by:


**Proposition 3.** (Mostowski?) Let $M$ be a countable transitive model of $\mathrm{ZFC}$ and let
 $\mathbb{C}$ be Cohen forcing. Then there are $\mathbb{C}$-generic
 filters $g,h$ over $M$ such that for any transitive $\mathrm{ZFC}$ model $N$
 with the same ordinal height of $M$ we have
 $M[g] \cup M[h] \not \subseteq N$.


In particular there is no set generic filter $f$ over $M$ such that
 $M[g] \cup M[h] \subseteq M[f]$.


Proof. Let $\alpha = M \cap \mathrm{Ord} < \omega\_{1}$ and fix a bijection
 $f \colon \omega \to \alpha$. Define
 $E \subseteq \omega \times \omega$ via
 $$
 (m,n) \in E \iff f(m) \in f(n).
 $$
 Let $z \in ^{\omega}2$ code $E$ in an absolute fashion. E.g. we may let
 $$
 z(k) = 1 \iff k = 2^{m}\cdot3^{n} \wedge (m,n) \in E.
 $$
 Clearly any transitive model having $z$ as an element has $E$ as an
 element as well and, since $E$ is a well-order of order type
 $\alpha$, must also have $\alpha$ as an element.


It now suffices to construct Cohen reals $c,d$ over $M$ such that
 they combined code the real $z$. We
 construct $c,d \in ^{\omega}2$ as follows:


Since $M$ is countable, we may fix an enumeration
 $(D\_{k} \mid k < \omega)$ of all dense subsets of $\mathbb{C}$ that
 are elements of $M$. Let $c\_{0} \in D\_{0}$ and let
 $d\_{0} = (0^{\mathrm{length}(c\_{0})}) ^{\frown} (1) ^{\frown} z(0) ^{\frown}
 y\_{0}$ for some $y\_{0}$ such that $d\_{0} \in D\_{0}$. Given
 $c\_{k}, d\_{k}$ for some $k < \omega$ we let
 $$
 c\_{k+1} = c\_{k} ^{\frown} (0^{\mathrm{length}(d\_{k})- \mathrm{length}(c\_{k})}) ^{\frown} (1) ^{\frown} x\_{k+1}
 $$
 for some $x\_{k+1}$ such that $c\_{k+1} \in D\_{k+1}$ and
 $$
 d\_{k+1} = d\_{k} ^{\frown} (0^{\mathrm{length}(c\_{k+1}) - \mathrm{length}(d\_{k})}) ^{\frown} (1) ^{\frown} y\_{k+1}
 $$
 for some $y\_{k+1}$ such that $d\_{k+1} \in D\_{k+1}$. We then let
 $c = \bigcup\_{k < \omega} c\_{k}$ and
 $d = \bigcup\_{k < \omega} d\_{k}$. The initial segments of $c$ and $d$ look as follows
 $$
 \begin{array}{cc|c|c|c|c}
 c = & c\_{0} & 0 \ldots 0 & 1 \ x\_{1} & 0 \ldots 0 & \ldots \\
 d = & 0 \ldots 0 & 1 \ z(0) \ y\_{0} & 0 \ldots 0 & 1 \ z(1) \ y\_{1} & \ldots
 \end{array}
 $$
 and the blocks of $0$'s in $c$ and $d$ now allow us to reconstruct $z$ from
 $c,d$ via a recursive function. (Q.E.D.)


Proof of Corollary 2. Since $0^{\#}$ exists, $\kappa$ exists and is countable in
 $V$. Since all dense subsets of $\mathbb{C}$ that $L$ can see are in
 $J\_{\omega\_{1}^{L}} \subseteq J\_{\kappa}$, there are only countably
 many such dense sets. Just like before we fix a real $z$ that codes
 the countable ordinal $\kappa$ and construct $g,h$ which are
 $\mathbb{C}$-generic over $L$ such that they combined code
 $z$. Suppose there were some $\mathbb{P} \in J\_{\kappa}$ and some
 $f$ which is $\mathbb{P}$-generic over $L$ with
 $L[g] \cup L[h] \subseteq L[f]$. $g,h,f$ are, of course,
 $\mathbb{C}$(respectively $\mathbb{P}$)-generic over $J\_{\kappa}$
 and we would have $g,h \in J\_{\kappa}[f]$. But now $J\_{\kappa}[f]$
 would have to contain $\kappa$ as an element. Contradiction! (Q.E.D.)




---


Side note: The proof above only uses that there is some $\kappa$ which is worldly in $L$ but countable in $V$ -- the assumption that $0^{\#}$ exists is overkill. It's just the first setting that came to mind when I thought about an inner model that has a lot of generic filters in $V$.



Does $0^{\#}$ imply the failure of upward directedness in the set generic universe over $L$?
Note that we can use $g,h$ to code any real of $V$ in a recursive fashion (recursive in the Cohen reals added by $g$ and $h$). In particular we can use them to code $0^{\#}$ -- a real that cannot be added to $L$ via set forcing. This shows that $0^{\#}$ does indeed imply the failure of upward directedness of the set generic universe over $L$.


On the other hand, if $\mathrm{HOD}^V = L$, any set is generic over $L$ (via the Vopenka algebra) and hence the set generic universe over $L$ is upward directed.


