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(I am replacing prior nonsense with a completely different suggestion. I am also turning this into CW so details can be added by somebody with time (which, sadly, most likely won't be me). Comments prior to Feb. 9, 2011, refer to said prior nonsense.)

Start with $V=L$ and force to add a Matthias real $s$. Let $W$ be the resulting extension. Let $A$ be the set of reals $r$ that are Matthias generic over $L$ and such that $L[r]=W$. I strongly suspect that a real $r$ is in $A$ iff it differs from $s$ finitely often, and so $A$ is countable, ordinal definable, and lacks any ordinal definable members.

(I have briefly discussed this idea with other set theorists, but we did not elaborate any details.)

From Andreas Blass: The following began as a comment, but Andres suggested adding it to his answer, for improved visibility. As it stands, with ordinary Mathias forcing, this won't work, because if $r\subset\omega$ is a Mathias real then so is the result of shifting it to the right (or left) by 1, and it still generates the same model. Instead of a simple shift, you could apply any strictly monotone function from $L$. But suppose you did Mathias forcing with respect to the constructibly-first non-principal ultrafilter on $\omega$ in $L$. That would avoid this problem. (Note that Joel David Hamkins's comment also depends on the fact that Prikry forcing is with respect to an ultrafilter in the ground model.)

(I am replacing prior nonsense with a completely different suggestion. I am also turning this into CW so details can be added by somebody with time (which, sadly, most likely won't be me). Comments prior to Feb. 9, 2011, refer to said prior nonsense.)

Start with $V=L$ and force to add a Mathias real $s$. Let $W$ be the resulting extension. Let $A$ be the set of reals $r$ that are Mathias generic over $L$ and such that $L[r]=W$. I strongly suspect that a real $r$ is in $A$ iff it differs from $s$ finitely often, and so $A$ is countable, ordinal definable, and lacks any ordinal definable members.

(I have briefly discussed this idea with other set theorists, but we did not elaborate any details.)

From Andreas Blass: The following began as a comment, but Andres suggested adding it to his answer, for improved visibility. As it stands, with ordinary Mathias forcing, this won't work, because if $r\subset\omega$ is a Mathias real then so is the result of shifting it to the right (or left) by 1, and it still generates the same model. Instead of a simple shift, you could apply any strictly monotone function from $L$. But suppose you did Mathias forcing with respect to the constructibly-first non-principal ultrafilter on $\omega$ in $L$. That would avoid this problem. (Note that Joel David Hamkins's comment also depends on the fact that Prikry forcing is with respect to an ultrafilter in the ground model.)

(I am replacing prior nonsense with a completely different suggestion. I am also turning this into CW so details can be added by somebody with time (which, sadly, most likely won't be me). Comments prior to Feb. 9, 2011, refer to said prior nonsense.)

Start with $V=L$ and force to add a Matthias real $s$. Let $W$ be the resulting extension. Let $A$ be the set of reals $r$ that are Matthias generic over $L$ and such that $L[r]=W$. I strongly suspect that a real $r$ is in $A$ iff it differs from $s$ finitely often, and so $A$ is countable, ordinal definable, and lacks any ordinal definable members.

(I have briefly discussed this idea with other set theorists, but we did not elaborate any details.)

(I am replacing prior nonsense with a completely different suggestion. I am also turning this into CW so details can be added by somebody with time (which, sadly, most likely won't be me). Comments prior to Feb. 9, 2011, refer to said prior nonsense.)

Start with $V=L$ and force to add a Matthias real $s$. Let $W$ be the resulting extension. Let $A$ be the set of reals $r$ that are Matthias generic over $L$ and such that $L[r]=W$. I strongly suspect that a real $r$ is in $A$ iff it differs from $s$ finitely often, and so $A$ is countable, ordinal definable, and lacks any ordinal definable members.

(I have briefly discussed this idea with other set theorists, but we did not elaborate any details.)

From Andreas Blass: The following began as a comment, but Andres suggested adding it to his answer, for improved visibility. As it stands, with ordinary Mathias forcing, this won't work, because if $r\subset\omega$ is a Mathias real then so is the result of shifting it to the right (or left) by 1, and it still generates the same model. Instead of a simple shift, you could apply any strictly monotone function from $L$. But suppose you did Mathias forcing with respect to the constructibly-first non-principal ultrafilter on $\omega$ in $L$. That would avoid this problem. (Note that Joel David Hamkins's comment also depends on the fact that Prikry forcing is with respect to an ultrafilter in the ground model.)

[Mmm. Nonsense, isn't it? Deleted. Sorry.]

(I am replacing prior nonsense with a completely different suggestion. I am also turning this into CW so details can be added by somebody with time (which, sadly, most likely won't be me). Comments prior to Feb. 9, 2011, refer to said prior nonsense.)

Start with $V=L$ and force to add a Matthias real $s$. Let $W$ be the resulting extension. Let $A$ be the set of reals $r$ that are Matthias generic over $L$ and such that $L[r]=W$. I strongly suspect that a real $r$ is in $A$ iff it differs from $s$ finitely often, and so $A$ is countable, ordinal definable, and lacks any ordinal definable members.

(I have briefly discussed this idea with other set theorists, but we did not elaborate any details.)