Given a forcing poset $\mathbb{P}$ it is said that $\mathbb{P}$ is countably closed if given $\{p_n\}_n$ a decreasing sequence of conditions there exists a condition $q\in \mathbb{P}$ below them. As well as, recall that a sequence $\{A_\alpha\}_{\alpha<\omega_1}$ such that $A_\alpha\subset \alpha$ is a $\diamondsuit$ sequence if for all $A\subset\omega_1$ then the set $\{\alpha\in\omega_1: \alpha\cap A=A_\alpha\}$ is stationary in $\omega_1$.
Is somebody willing to give a proof of countably closed forcing preserves $\diamondsuit$ sequences? That's, if $\{A_\alpha\}_{\alpha<\omega_1}$ is a $\diamondsuit$ sequence in $M$ it's still $\diamondsuit$ sequence in $M[G]$.
Thank you.
Cesare.
I've found a possible argument to prove this. First of all note that since $\mathbb{P}$ is countably closed then $\omega_1$ is preserved and thus also the notion of stationariety.
Suppouse that $\{A_\alpha\}_\alpha$ is a $\diamondsuit$ sequence in $M$ but not in $M[G]$. Then, there exists $A\in\mathcal{P}^{M[G]}(\omega_1)$ and a club $C\subset \omega_1$ such that for all $\alpha\in C$ then $A\cap \alpha\neq A_\alpha$. Thus, for every $\alpha\in C$ WLOG there exists a $\beta_\alpha\in A\cap \alpha\setminus A_\alpha$. If $f$ is a continious bijection between $\omega_1$ and $C$ then, by the forcing theorem, it can be translated to the following: for every $\alpha\in \omega_1$ there exists $p_\alpha\in G$ such that $p_\alpha\Vdash \beta_\alpha\in \dot{A}\cap \dot{f}(\alpha)\setminus A_{\dot{f}(\alpha)}$.
Now consider the set $A'=\{\beta\in\omega_1: \exists \alpha\in \omega_1\exists p\in\mathbb{P}\;\;p\Vdash \beta\in \dot{A}\cap \dot{f}(\alpha)\setminus A_{\dot{f}(\alpha)}\}\in M$. It's obvious that $A'\supset \{\beta_\alpha\}_{\alpha\in\dot{\omega_1}}$. Since $\{A_\alpha\}_\alpha$ is a $\diamondsuit$ sequence then $S=\{\alpha\in\omega_1:\,A'\cap \alpha=A_\alpha\}$ is stationary in $M$ and thus in $M[G]$. Therefor, there exists $\gamma\in\omega_1$ such that $A'\cap f(\gamma)=A_{f(\gamma)}$. This last equality implies that $\beta_\gamma\in A_{f(\gamma)}$ but $p_\gamma\Vdash \beta_\gamma\in \dot{A}\cap \dot{f}(\gamma)\setminus A_{\dot{f}(\gamma)}$ and $p_\gamma\in G$. Contradiction.