I am aware that $MA(\kappa)$ when $\kappa = 2^{\omega_0}$ is false, so please excuse me if this question is absolutely trivial, however I am an absolute beginner when it comes to these topics.
My question is : Where exactly the proof of Rasiowa-Sikorski Lemma fails to be extended to any $\kappa > \omega_0$ ?
To be more precise, I'll state the argument of such proof : Let $D=\lbrace D_n\rbrace_{n\in\omega}$ be a collection of dense subsets of ($\mathbb{P}$,$\leq$), a partially ordered set. Choose $d_1\in D_1$ and since $D_2$ is dense, there is some $d_2\in D_2$ such that $d_2\leq d_1$. By simple induction, we get a totally ordered subset of $\mathbb{P}$ - $\lbrace d_n\rbrace_{n\in\omega}$ - such that each $d_n\in D_n$. Then, take $\mathcal{F}$ to be the filter in $\mathbb{P}$ generated by $\mathcal{F}$, i.e. $\mathcal{F}=\lbrace q\in\mathbb{P}:\exists n\in\omega : d_n\leq q\rbrace$. Then, $\mathcal{F}\cap D_n\neq\emptyset$ for each $n\in\omega$.
My question is : using transfinite induction, can't we choose one element $d_\alpha\in D_\alpha$ such that $d_\beta\leq d_\alpha$ for any $\beta\leq\alpha$, for any collection $\lbrace D_{\alpha}\rbrace_{\alpha\leq\kappa}$ of dense subsets of $\mathbb{P}$ ? If so, why can't we just take the filter generated by $\lbrace d_{\alpha}\rbrace_{\alpha\leq\kappa}$?
Thanks for all clarifications and excuse me, once again, if this is absolutely trivial.
Consider the Cohen forcing. Or simply the infinite binary tree. Now suppose the you take $D_n$ to be the tail segment of the tree of all levels above the $n$th one (the rest of the enumeration doesn't matter). Start making these choices of $d_n\in D_n$.
What would be $d_\omega$? It would be an element in the full binary tree, such that for every $n$, there is some $d_n\in D_n$ such that $d_\omega$ extends $d_n$. But this is impossible, since the intersection of the $D_n$'s is empty. You cannot just argue like this.
In simpler, forcing terms, you're assuming that $\Bbb P$ is $\leq\kappa$-closed. Which certainly doesn't have to be the case.