# Adding a closed unbounded set containing of only limit ordinals with a special property

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?

• 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 $p$. Commented Oct 15, 2021 at 11:11
• We can add only limit ordinals Commented Oct 15, 2021 at 11:18

(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 $$f(\beta_i)$$.