# Adding large sets not containing countable ground model sets

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).

• I guess you intend to assume GCH in $V$? (Otherwise, no cardinal preserving extension can have GCH.) – Joel David Hamkins Oct 19 '13 at 10:46
• Yes, of course. – Mohammad Golshani Oct 19 '13 at 11:26

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 $\aleph_3$?

• Your answer is no "having finite intersection with countable ground model sets". What about "not containing countable ground model sets". – Mohammad Golshani Oct 19 '13 at 12:10
• I edited the question a little, to separate two cases mentioned in my previous comment. – Mohammad Golshani Oct 19 '13 at 12:13
• 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. – Mohammad Golshani Oct 20 '13 at 5:27