The question is motivated by Toni's question "Approximation of infinite set in generic extension" (see Approximation of infinite set in generic extension).

Before I state the question, let me add some remarks. In what follows, it is always assumed that $V$ is our ground model:

**Remark 1.** If we force to add $\lambda-$many Cohen reals by $Add(\omega, \lambda),$ then we get a cardinal preserving extension $W$ in which there is a set $C \subset \lambda$ of size $\lambda,$ such that $C$ contains no countable set in $V$. But note that in $W, 2^{\aleph_0} \geq \lambda,$ and so $GCH$ may fail in it (if $\lambda\geq \aleph_2$).

**Remark 2.** If we allow collapsing cardinals (by collapsing $\aleph_1$ to $\aleph_0$ or forcing with Namba forcing), then for any regular cardinal $\lambda,$ we can find a $GCH$ preserving extension $W$ of $V$ such that in $W$ there is a club $C\subset \lambda$ which avoids points of countable $V-$cofinality. This $C$ contains no countable set in $V$ and has finite intersection with any countable set in $V$.

**Remark 3.** If there are $\lambda-$many measurable cardinals, then there is a cardinal and $GCH$ preserving extension $W$ of $V$ with the same reals as $V$ such that $W$ contains a set $C$ of ordinals of size $\lambda$ which contains no countable set in $V$ and has finite intersection with any countable set in $V$.

Also note that if we require such a set $C$ in a cardinal preserving and not adding new reals extension, then some large cardinals are needed.

Now my questions are as follows:

**Question 1.** Suppose $V$ satisfies $GCH$ and contains no inner models with measurable cardinals. Is there a $GCH$ and cardinal preserving extension $W$ of $V$ such that in $W$ there is a set $C\subset \lambda$ of size $\lambda,$ for $\lambda\geq \aleph_3,$ such that $C$ has finite intersection with any countable ground model set?

**Question 2.** Suppose $V$ satisfies $GCH$ and contains no inner models with measurable cardinals. Is there a $GCH$ and cardinal preserving extension $W$ of $V$ such that in $W$ there is a set $C\subset \lambda$ of size $\lambda,$ for $\lambda\geq \aleph_3,$ such that $C$ contains no countable set from $V$.

**Update.**

Regarding Prof. Hamkins answer, I would like to add a few more comments (both of them are joint work with M. Gitik).

**A.** Assuming the existence of enough measurable cardinals, there is a pair $(V_1, V_2)$ of models of $ZFC$ with the same cardinals and reals, such that if $\kappa$ is the first fixed point of the $\aleph-$function in them, then in $V_2$, then there is a splitting $(S_\sigma: \sigma<\kappa)$ of $\kappa$ into sets of size $\kappa,$ such that any $S_\sigma$ has finite intersection with any countable set in $V_1$. This shows that Hamkins argument does not extend to the first fixed point of the $\aleph-$function.

**B.** Suppose $V \subset V_1$ have the same cardinals and reals and $\delta$ is less than the first fixed point of the $\aleph-$function. if $X \subset \aleph_\delta, X\in V_1$ and $|X|\geq \delta^+$ then $X$ has a countable subset which is in $V$.

Our proof of **B** is essentially the same as Hamkins argument and is by induction on $\delta$. Now Hamkins argument suggests that if we require $X$ has finite intersection with any countable set in $V$, then we do not require $V$ and $V_1$ to have the same reals (of course if $V$ and $V_1$ have the same reals then the statements "$X$ does not contain a countable set in $V$" and "$X$ has finite intersection with any countable set in $V$" are equivalent).