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Suppose $M$ is a c.t.m and suppose $P$ is $Fn(I,2)$ where $I$ is infinite. Now suppose $G$ is $P$-generic, and $A \in M[G]$ is infinite set.

Is it guaranteed that the exist $B \in M$ such that $B \subset A$, and $B$ is infinite ? If not, is there any 'reasonable' condition that we can add to $M$ or $P$ to get the desired result ? (Like CH is $M$, or large enough $I$)

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If $P$ is a forcing notion in $M$ that adds no new countable sequences, then (trivially) every new infinite set $A$ of ordinals (or: $A \subseteq M$) contains a countably infinite set $B$, and $B$ must be in the ground model. (For example, $P$ could be $\sigma$-closed. There are many interesting forcing notion that have this property, but they are not interesting in the context of your question.)

If $P$ is any forcing notion in $M$ that adds a new countable sequence, then the answer to your question is "no". Indeed, let $x \in \kappa^\omega$ be a new sequence. Let $A$ be the set $\{ x \restriction n: n\in \omega\}$. This is a countable set with the property that for any infinite subset $B$ we have $x=\bigcup B$; so no such $B$ can be in the ground model.

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  • $\begingroup$ This illustrates that forcing is irrelevant here. In fact, choice is not needed either: If $N$ is an outer model of $M$ (with the same ordinals) and $N$ has a new countable set, then there is such a set with no infinite subset in $M$. If $N$ has no new countable sets, and any uncountable set contains a countable set, then any infinite set in $N$ has an infinite subset in $M$. So far, this is just as in Martin's answer. The last option is that $N$ has sets $X$ without countable subsets (infinite Dedekind finite sets). (Cont.) $\endgroup$ – Andrés E. Caicedo Oct 19 '13 at 2:09
  • $\begingroup$ (And $N$ has no new countable sets.) In this case, either $\mathcal P(X)$ or its power set $\mathcal P^2(X)$ is Dedekind infinite. Say it is the first option. Any countably infinite subset of $\mathcal P(X)$ is then in $M$, and its union is an infinite subset of $X$ in $M$. If $\mathcal P(X)$ is Dedekind finite, then the same argument gives an infinite subset of it in $M$, and its union is an infinite subset of $X$, again in $M$. (On the other hand, I doubt it is possible to have such pair $M,N$.) $\endgroup$ – Andrés E. Caicedo Oct 19 '13 at 2:15
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Theorem: Let $\kappa$ be a regular cardinal in $M$. Let $\mathbb{P} \in M$ be a separative partial order of size $\kappa$, and let $G$ be $\mathbb{P}$-generic over $M$. Then the following are equivalent:

(1) There is some $p \in G$ such that in $M$, $\mathbb{P} \restriction p$ has a dense subset of size $< \kappa$.

(2) For all sets of ordinals $A \in M[G]$ of size $\kappa$, there is $B \subseteq A$ such that $B \in M$ and $|B| = \kappa$.


Per Mohammad's request, I will give a proof. I think this is a nice theorem because it characterizes a certain combinatorial property of partial orders in terms of model comparison with generic extensions, like the distributivity properties. I have found some use of it in comparing submodels of generic extensions.

For a partial order $P$, let $d(P)$ be the least size of a dense subset of $P$. For a given $P$, and $q \leq p$ in $P$, $d(P \restriction q) \leq d(P \restriction p)$. Call an element of $P$ "$d$-stable" if $(\forall q \leq p) d(P \restriction q) = d(P \restriction p)$. By well-foundedness, the $d$-stable elements are dense.

Let $G$ be $P$-generic over $M$. Suppose there is $p \in G$ with $d(P \restriction p) < \kappa$, and let $D \subseteq P \restriction p$ witness this. Then $p \Vdash \kappa$ is regular. Let $A \in M[G]$ be a set of ordinals of size $\kappa$, and let $\dot{A}$ be a name for it. $A = \{ \alpha : (\exists q \in D \cap G) q \Vdash \check{\alpha} \in \dot{A} \}$, so for some $q \in D \cap G$, $B = \{ \alpha : q \Vdash \check{\alpha} \in \dot{A} \}$ has size $\kappa$.

Suppose now that $p \in G$ is $d$-stable, and $d(P \restriction p) = \kappa$. We will find a set of ordinals $A \in M[G]$ unbounded in $\kappa$ that contains no unbounded set from $M$. This suffices because, although $\kappa$ may no longer be regular, the function $f : \kappa \to A$ defined by $f(\alpha) =$ the least $\beta \in A$ above $\alpha$, is an object of size $|\kappa|$, and if it had a size-$\kappa$ subset $g \in M$, then the range of $g$ would be an unbounded subset of $A$ from $M$.

Let $D \subseteq P \restriction p$ be dense, and recursively construct $D' \subseteq D$ also dense, and with the property that for all $q \in D'$, $| \{ r \in D' : r \geq q \}| < \kappa$ (exercise). Let $D' = \langle p_\alpha : \alpha < \kappa \rangle$, and in $M[G]$ consider $A = \{ \alpha : p_\alpha \in D' \cap G \}$. $A$ is unbounded in $\kappa$ (exercise). If we had $q \in D'$ and an unbounded $B \in M$ such that $q \Vdash \check{B} \subseteq \dot{A}$, then $\{ p_\alpha : \alpha \in B \} \subseteq D'$, and by separativity, $q \leq p_\alpha$ for all $\alpha \in B$, which contradicts the property of $D'$.

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  • $\begingroup$ Dear Monroe, would you please give a sketch of the result $\endgroup$ – Mohammad Golshani Oct 26 '13 at 5:13
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The answer is no, and there is no condition on $M$ or $I$ that can ensure the desired property. The reason is that the generic filter $G$ itself can contain no infinite ground model set. To see this, note that if $p$ is any condition, then since it is finite, there are only finitely many conditions that weaken $p$, and so for any infinite set $B\subset P$ in $M$, there must be some $q\in B$ such that is not a weakening of $p$, and so we may strengthen $p$ in such a way that is incompatible with $q$. It follows that no condition can force that $B$ is contained in the generic filter, and so $G$ contains no infinite ground model set.

Another way to think about it is like this: the generic filter is adding a new generic subset $A\subset I$, namely, the set of which the generic function is the characteristic function. But this set $A$ can contain no infinite ground model set, since for any infinite set $B\subset I$ and any condition $p$, we may extend $p$ to a stronger condition that forces some element of $B$ out of $A$. So it is dense that any particular $B$ is not contained in $A$.

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