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For (2), what happens if we are in $L[r]$ with $r$ a Sacks real? Since this has minimal degree, all non-constructible $x\subseteq\omega$ have the same degree of constructibility, and if we split a non-constructible set into two disjoint pieces, at least one must be non-constructible. This would be a consistent "no" answer for (2), as we may split any infinite subset of $\omega$ into two infinite pieces.

Edit:

Denis' comment on Mathias forcing got me to thinking, and a look back at Mathias' original "Happy Families" paper has what we need for a consistent "yes" answer to (2).

In particular, Theorem 8.2 of the paper tells us exactly what we want: if $F$ is a Ramsey ultrafilter in $L$ and $X$ is Mathias generic over $L$ with respect to $F$, then $Z\subseteq X$ and $X\leq_L Z$ implies $X\setminus Z$ is finite.

So (2) is actually independent of ZFC.

For (2), what happens if we are in $L[r]$ with $r$ a Sacks real? Since this has minimal degree, all non-constructible $x\subseteq\omega$ have the same degree of constructibility, and if we split a non-constructible set into two disjoint pieces, at least one must be non-constructible. This would be a consistent "no" answer for (2), as we may split any infinite subset of $\omega$ into two infinite pieces.

Edit:

Denis' comment on Mathias forcing got me to thinking, and a look back at Mathias' original "Happy Families" paper has what we need for a consistent "yes" answer to (2).

In particular, Theorem 8.2 of the paper tells us exactly what we want: if $F$ is a Ramsey ultrafilter in $L$ and $X$ is Mathias generic over $L$ with respect to $F$, then $Z\subseteq X$ and $X\leq_L Z$ implies $X\setminus Z$ is finite.

So (2) is actually independent of ZFC.

Note as well that (2) has a positive answer if the set of constructible reals is countable, as any infinite pseudo-intersection for the filter generated by $F$ in $V$ is Mathias generic over $L$ for $F$.

For (2), what happens if we are in $L[r]$ with $r$ a Sacks real? Since this has minimal degree, all non-constructible $x\subseteq\omega$ have the same degree of constructibility, and if we split a non-constructible set into two disjoint pieces, at least one must be non-constructible. This would be a consistent "no" answer for (2), as we may split any infinite subset of $\omega$ into two infinite pieces.

For (2), what happens if we are in $L[r]$ with $r$ a Sacks real? Since this has minimal degree, all non-constructible $x\subseteq\omega$ have the same degree of constructibility, and if we split a non-constructible set into two disjoint pieces, at least one must be non-constructible. This would be a consistent "no" answer for (2), as we may split any infinite subset of $\omega$ into two infinite pieces.

Edit:

Denis' comment on Mathias forcing got me to thinking, and a look back at Mathias' original "Happy Families" paper has what we need for a consistent "yes" answer to (2).

In particular, Theorem 8.2 of the paper tells us exactly what we want: if $F$ is a Ramsey ultrafilter in $L$ and $X$ is Mathias generic over $L$ with respect to $F$, then $Z\subseteq X$ and $X\leq_L Z$ implies $X\setminus Z$ is finite.

So (2) is actually independent of ZFC.

For (2), what happens if we are in $L[r]$ with $r$ a Sacks real? Since this has minimal degree, all non-constructible $x\subseteq\omega$ have the same degree of constructibility, and if we split a non-constructible set into two disjoint pieces, at least one must be non-constructible. This would be a consistent "no" answer for (2).

For (2), what happens if we are in $L[r]$ with $r$ a Sacks real? Since this has minimal degree, all non-constructible $x\subseteq\omega$ have the same degree of constructibility, and if we split a non-constructible set into two disjoint pieces, at least one must be non-constructible. This would be a consistent "no" answer for (2), as we may split any infinite subset of $\omega$ into two infinite pieces.