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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. (Woodin?) 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 $V$.

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 $V$.

Question 1. Suppose that $0^{\#}$ exists. Are there set generic filter $g,h$ 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. (Woodin?) 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 $V$.

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. (Woodin?) 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 $V$.

Question 1. Suppose that $0^{\#}$ exists. Are there set generic filter $g,h$ 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. (Woodin) 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 $V$.

Question 1. Suppose that $0^{\#}$ exists. Are there set generic filter $g,h$ 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. (Woodin?) 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 $V$.