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?



Adding a closed unbounded set containing of only limit ordinals with a special property
Maybe the following idea works: 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)$.


