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I've seen this written several places without proof, so I assume it's not difficult, but I am not getting it.

Let $\mathbb P$ be a $\kappa$-c.c. notion of forcing, and let $C\in M[G]$ be club in $\kappa$. I want to show there exists $D\in M$ such that $D\subset C$ is club in $\kappa$. Kunen suggests: let $f\in M[G], f:\kappa\rightarrow\kappa$ such that $\forall\alpha<\kappa(\alpha<f(\alpha)\in C$. Then because the forcing is $\kappa$-c.c., there exists $F\in M, F:\kappa\rightarrow P(\kappa)$ such that $\forall\alpha<\kappa(\alpha<f(\alpha)\in F(\alpha)\wedge(|F(\alpha)|<\kappa)^{M})$. But now I'm stuck. The union of the image of $F$ is a set cofinal in $\kappa$, which contains a cofinal subset of $C$ that is not necessarily closed. Any ideas? Thank you!

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This is how I think about it. Suppose that $\dot C$ is a $P$-name and some condition $p_0$ forces that $\dot C$ is club in $\kappa$. Consider the set $D$ of ordinal $\alpha\lt\kappa$ such that $p_0$ forces $\check\alpha\in\dot C$. This set is certainly closed, since if $p_0$ forces ordinals into $\dot C$, then by closure it will also force their supremum into $\dot C$. But $D$ is also unbounded. To see this, fix any ordinal $\alpha_0$. The club $C$ will have a next element after $\alpha_0$, and by the mixing lemma, we can find a name $\dot\beta_0$ such that $p_0$ forces that $\dot\beta_0$ is the next element of $\dot C$ after $\check\alpha$. (This is essentially the function $f$ you mention.) There is an antichain of possible values for $\dot\beta_0$---this is the function $F$ you mention---and so by the $\kappa$-c.c. of $P$, we can find an ordinal $\alpha_1$ such that $p_0$ forces $\dot\beta_0\lt\check\alpha_1$. Now continue with $\dot\beta_1$ and $\alpha_2$, $\dot\beta_2$ and so on in the same way. Thus, the condition $p_0$ forces that each $\dot\beta_n$ is in $\dot C$, so it forces that the supremum of these $\omega$ many ordinals is in $\dot C$. But this supremum is the same as the supremum of the $\alpha_n$, and so $p_0$ forces that the ordinal $\alpha_\omega=\text{sup}\alpha_n$ is in $\dot C$. Thus, we have found an ordinal in $D$ above $\alpha_0$, and so $D$ is unbounded, as desired.

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  • $\begingroup$ Joel, what is the "mixing lemma"? $\endgroup$ – Asaf Karagila Mar 6 '11 at 17:26
  • $\begingroup$ The mixing lemma is the assertion that if $A$ is an antichain and for each $p\in A$ we have a name $\tau_p$, then there is a single name $\tau$ such that $p$ forces $\tau=\tau_p$ for each $p\in A$. Thus, the name $\tau$ mixes together the names $\tau_p$. This is useful when you have an antichain of possible values with a certain property, to find a single name that is forced to have that property. This lemma can be used to prove the Fullness Lemma, which asserts that if $p$ forces $\exists x\varphi(x)$, then there is a name $\tau$ such that $p$ forces $\varphi(\tau)$. $\endgroup$ – JDH Mar 6 '11 at 19:35
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    $\begingroup$ So I suppose I could have appealed merely to the Fullness lemma in my answer, rather than mixing. $\endgroup$ – JDH Mar 6 '11 at 19:36
  • $\begingroup$ I studied forcing from Jech, I do recall the Fullness lemma, however through Boolean-Valued models it's simple enough just to present and prove it - as I do not recall the mixing lemma at all being present in that chapter. Seems relatively obvious though. $\endgroup$ – Asaf Karagila Mar 6 '11 at 22:42
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    $\begingroup$ Oh, I got it. There are less than $\kappa$ possibilities for the ordinal, therefore their sup is less than $\kappa$. $\endgroup$ – Vinicius Rodrigues Jan 2 '14 at 20:38

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