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This is essentially exercise H8 (p.248) of Kunen's Set Theory: An Introduction to Independence Proofs (old edition), or exercise IV.7.58 (p.307) of Kunen's Set Theory (new edition).

Suppose $P$ is a notion of forcing in $M$ such that $\left | P \right | \leq \omega_{1}$ and $P$ is ccc. Suppose further $\Diamond$ holds in $M$. How does one show that $\Diamond$ also holds $M[G]$?

Here

  • $M$ is a countable transitive model (of $\mathsf{ZFC}$), and $M[G]$ is a generic extension of $M$ by the forcing $P$.
  • $P$ being ccc (countable chain condition) means that all antichains (sets of pairwise incompatible conditions) in $P$ are countable.
  • $\diamondsuit$ is the usual diamond principle:

    There is sequence $\langle A_\alpha : \alpha < \omega_1 \rangle$ such that $A_\alpha \subseteq \alpha$ for each $\alpha < \omega_1$, and for each $A \subseteq \omega_1$ the set $$\{ \alpha < \omega_1 : A \cap \alpha = A_\alpha \}$$ is stationary.

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    $\begingroup$ Please clarify your question. And try to see if it would be more appropriate here or at math.stackexchange. $\endgroup$ Jul 28, 2014 at 20:24
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    $\begingroup$ @JoonasIlmavirta He wants to know if $\Diamond$ is preserved by ccc forcing of size at most $\omega_1$. And the answer is yes: one uses the ground-model $\Diamond$-sequence to anticipate names for the desired subset of $\omega_1$, and this idea builds a $\Diamond$-sequence in the forcing extension. $\endgroup$ Jul 28, 2014 at 20:30
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    $\begingroup$ I don't really understand why the question was closed, and so I have voted to re-open. This is on-topic graduate-level material in set theory. $\endgroup$ Aug 19, 2014 at 7:16
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    $\begingroup$ The question uses completely standard terminology and notation---I think it is perfectly understandable to any set theorist without change. The title could be improved. I'd suggest, "Is $\Diamond$ preserved by c.c.c. forcing of size $\omega_1$?"; and the forcing tag could be added. $\endgroup$ Aug 28, 2015 at 13:08
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    $\begingroup$ While I voted to reopen this version of the question, I want to record that the version that was closed was mathoverflow.net/revisions/177302/2 which reads in full: "In $M$: $\left | P \right | \leq \omega_{1}$ and $P$ es c.c.c and $\Diamond$ is hold M. Shows that $\Diamond$ is hold $M[G]$. give one suggestion please" $\endgroup$
    – user9072
    Aug 28, 2015 at 15:48

1 Answer 1

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Let $\vec A=\langle A_\alpha\mid\alpha<\omega_1\rangle$ witness $\Diamond$ in $M$. Since $|P|\leq\omega_1$, we might as well assume that the underlying set of $P$ is a subset of $\omega_1$. By standard coding techniques, we may view a subset of $\omega_1$ as coding a $P$-name for a subset of $\omega_1$. So in $M[G]$, define $B_\alpha=\dot (A_\alpha)_G\cap\alpha$, if $A_\alpha$ codes the $P$-name $\dot A_\alpha$.

I claim that $\langle B_\alpha\mid\alpha<\omega_1\rangle$ is a $\Diamond$ sequence in $M[G]$. To see this, fix any $A\subset \omega_1$ in $M[G]$. So there is a $P$-name $\dot A$ such that $A=\dot A_G$. Let $A^*\subset\omega_1$ be a code of $\dot A$. By a closure argument, using the fact that $P$ is c.c.c., there is a club $C\subset\omega_1$ such that $A^*\cap\alpha$ codes $\dot A\upharpoonright \alpha$ for $\alpha\in C$. Since $A_\alpha=A^*\cap\alpha$ on a stationary set of $\alpha$ in $M$, it follows that $B_\alpha=(\dot A_\alpha)_G=A\cap \alpha$ for stationary many $\alpha$. So $\vec B$ witnesses $\Diamond$ in $M[G]$.

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