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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).

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  • $\begingroup$ I guess you intend to assume GCH in $V$? (Otherwise, no cardinal preserving extension can have GCH.) $\endgroup$ – Joel David Hamkins Oct 19 '13 at 10:46
  • $\begingroup$ Yes, of course. $\endgroup$ – Mohammad Golshani Oct 19 '13 at 11:26
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The answer to your original question (now question 1) is no, this is impossible, and the GCH and measurable cardinals are not involved.

Theorem. There is no cardinal-preserving forcing extension $V[G]$ with a set $C\subset\aleph_3$ having finite intersection with every countable set in $V$.

Proof. Suppose that there is such an extension $V[G]$ with such a set $C$. Let $\alpha$ be the supremum of the first $\omega$-many elements of $C$. So $\alpha\lt\aleph_3$ and $C\cap\alpha$ is a subset of $\alpha$ with finite intersection with every ground model countable set. Since $\alpha$ has size $\aleph_2$ in $V$, we may apply a bijection $\pi:\alpha\to\aleph_2$ in $V$ to get a set $B\subset\aleph_2$, namely $B=\pi[C]$, such that $B$ has finite intersection with every ground model set. So we have reduced to $\aleph_2$. Now, let $\beta$ be the supremum of the first $\omega$ many elements of $B$. So $\beta\lt\aleph_2$ and $B\cap\beta$ is a subset of $\beta$ having finite intersection with every countable ground model set. There is a bijection in $V$ of $\beta$ with $\aleph_1$, so $B\cap\beta$ is isomorphic to a set $A\subset\aleph_1$, by an isomorphism in the ground model, which has finite intersection with every ground model set. Let $\gamma$ be the supremum of the first $\omega$ many elements of $A$. So $\gamma$ is a countable ordinal in $V$, but $A\cap\gamma$ is infinite, a contradiction. QED

Clearly, the argument can be generalized beyond $\aleph_3$.

Your revised question (question 2) seems to be trivialized by the case of simply adding a Cohen real. This adds a set $C\subset\omega$, which is therefore also a subset of $\aleph_3$, but it contains no infinite ground model set as explained in the answers to Toni's question Approximmation of infinite set in generic extension. I suppose you intend to add a cofinal subset to $\aleph_3$?

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  • $\begingroup$ Your answer is no "having finite intersection with countable ground model sets". What about "not containing countable ground model sets". $\endgroup$ – Mohammad Golshani Oct 19 '13 at 12:10
  • $\begingroup$ I edited the question a little, to separate two cases mentioned in my previous comment. $\endgroup$ – Mohammad Golshani Oct 19 '13 at 12:13
  • $\begingroup$ Of course when talking about a subset of $\lambda,$ I require the set to have size $\lambda.$ I added it to the question to avoid triviality. $\endgroup$ – Mohammad Golshani Oct 20 '13 at 5:27

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