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Suppose $M$ satisfies the $CH$ and that we force over $M$ with $\mathbb{P}=Fn(I,2)$ where $(\omega_{2} \leq |I|))^{M}$, that is, with the finite partial functions from $I$ to $2$. If $f \in M[G] \cap \omega^{\omega}$, then is there necessarily a function $g \in M \cap \omega^{\omega}$ for which {$n:f(n) \leq g(n)$} is infinite? Please give a suggestion to help me work this exercise from Kunen's book.

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Your question uses unexplained notation, mentions an exercise in a book but does not tell us what excercise nor what book, and it is quite not clear what you are trying to ask. – Mariano Suárez-Alvarez Jun 23 '13 at 20:23
No Joel, the question is not clear. There is in fact no question, and the sentences are hard to parse. We can sort of see what a possible question could perhaps be, given what is written, but that is a different story. – Andrés E. Caicedo Jun 23 '13 at 20:36
I tend to side with Joel. I think the appropriate reaction is to edit the question. (This is not done enough on MO!) – François G. Dorais Jun 23 '13 at 21:00
François, I have edited. – Joel David Hamkins Jun 23 '13 at 21:16
Well, Miguel, despite the closure of your question, I'd like to welcome you to MathOverflow. I'm truly very sorry to see a newcomer get such a rough reception. I know that this could be discouraging, but I'd like to encourage you not to become discouraged. Please try to look past it and simply try again with other questions! Indeed, you seem to be studying forcing, and I'd particularly encourage questions on forcing, which is definitely on-topic here at MO. You may find that you will get a more favorable reception with a more carefully worded question or fuller explanation. – Joel David Hamkins Jun 24 '13 at 2:07

Here is a sketch of a proof: Let $\dot f$ be a name for $f$. Wlog we may assume that $\dot f$ is a nice name and hence uses only countably many conditions. It follows that it is an $Fn(J,2)$-name for some countable set $J\subseteq I$. Let $(p_n)_{n\in\omega}$ be an enumeration of $Fn(J,2)$.

We construct a function $g\in M$ as follows. For each $n\in\omega$ choose $g(n)$ such that for all $k\leq n$ the following holds: if for some $m\in\omega$ we have $p_k\Vdash\dot f(n)\gt m$, then $g(n)\gt m$. It is possible to choose $g(n)$ in this way since for all $p\in Fn(J,2)$ there are only finitely many $m$ such that $p\Vdash\dot f(n)\gt m$.

We show that $g$ has the desired property. Suppose this is not the case. Let $p\in G$ and $n_0\in\omega$ be such that $$p\Vdash\forall n\geq n_0(\dot f(n)\gt g(n)).$$ For some $k\in\omega$, $p=p_k$. Now let $n\geq n_0,k$. Then $p_k\Vdash\dot f(n)\gt m$ for $m=g(n)$. By the choice of $g$, $g(n)\gt m=g(n)$, a contradiction.

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