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The following theorem and proof are in Applications of the proper forcing axiom, the Baumgartner's paper in the book Handbook of Set-theoretic topology.

$3.6$ THEOREM. Assume PFA. Suppose that for each $\alpha < \omega_1$ a set $S_\alpha \subseteq \omega_1$ is given such that, for every limit ordinal $\beta < \omega_1$, $S_{\alpha} \cap \beta$ has ordertype $< \beta$. Then there is a closed unbounded set $C$ such that $ \forall \alpha < \omega_1\ C \cap S_{\alpha}$ is finite.

PROOF. Let $P$ consist of all $p$ for which there is a closed unbounded set $C \subseteq \omega_1$ containing only limit ordinals so that $p$ is a finite subset of the enumerating function of $C$. Let $Q$ be the set of all pairs $(p, x)$, where $p \in P$ and $x \in [\omega_1]^{<\omega}$. Let $(p_1, x_1) \leq (p_2, x_2)$ iff $ p_1 \supseteq p_2, \ x_1 \supseteq x_2$ and $\forall \alpha \in x_2 \ \text{range}(p_1 - p_2) \cap S_\alpha =0$. Now force with $Q$.

As usual we have to start by proving that $\forall \alpha < \omega_1$ the set $D_\alpha =\{ (p, x) \in Q : \alpha \in dom(p)\}$ is dense in $Q$.

For $D_0$ let we choose $(p, x) \in Q$ such that $\alpha_0 \in x,\ \omega \in S_{\alpha_0},\ p=\{(1, \omega . 2)\}$. For each $p' \supseteq p$ such that $0 \in dom(p')$ and every $x'\supseteq x,\ \omega \in \text{range}(p' - p) \cap S_{\alpha_0}$. Thus $(p', x') \nleq (p,x)$.

This counterexample shows $D_0$ isn't dense. Do we not need density? Or should we add something to the terms and definitions?

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    $\begingroup$ Why should $\omega \in \operatorname{range}(p' \setminus p)$? $S_{\alpha_0}$ does not have ordertype $\omega \cdot 2$ below $\omega \cdot 2$, so there is some $\gamma < \omega \cdot 2$, $\gamma \notin S_{\alpha_0}$. Just extend by adding $(0,\gamma)$ to $p$. $\endgroup$ Commented Oct 15, 2021 at 11:11
  • $\begingroup$ We can add only limit ordinals $\endgroup$ Commented Oct 15, 2021 at 11:18

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Given a condition as above, also require the following:

(1) for each $\alpha, f(\alpha)$ is indecomposable,

(2) suppose $dom(p)=\{\beta_0 < \beta_1 < \cdots < \beta_n\}$. Then there exists a closed subset $C$ of $f(\beta_n)$ of order type $\beta_n$ such that $C \cap \bigcup_{\gamma \in x}S_\gamma \subseteq range(p)$, for each $i$, $C \cap f(\beta_i)$ has order type $\beta_i.$ Furthermore, if $\beta_i$ is limit, then $C \cap f(\beta_i)$ is unbounded in $ f(\beta_i)$.

This answer is given by @Mohammad-Golshani here

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