Well-pruned $\kappa$-Suslin trees have the $\kappa$-Baire property

Question

I'm having troubles seeing that a well-pruned $\kappa$-Suslin tree, with $\kappa$ uncountable and regular, has the $\kappa$-Baire property.

I saw in Kunen's Set Theory that it can be seen by proving that $T$ doesn't add $\kappa$-sequences, but anyway I want to prove it by a direct way. Namely, taking $\{D_\alpha\}_{\alpha<\lambda}$ open dense subsets with $\lambda<\kappa$ then $\bigcap_{\alpha<\lambda}D_\alpha$ is dense.

To provide some definitions:

• Given a cardinal $\kappa$ a forcing poset $\mathbb{P}$ has the $\kappa$-Baire property if the intersection of less than $\kappa$-many open dense subsets of $\mathbb{P}$ is dense.

  If $(T,\sqsupseteq)$ is a tree we can see it as a forcing poset $\mathbb{P}$ with the ordering $\sqsupseteq$, so we can talk about the $\kappa$-Baire property for $T$.

• A $\kappa$-Suslin tree is a tree of size $\kappa$ with no chains or antichains of size $\kappa$.

• A $\kappa$-tree $T$ is well-pruned if for every $x\in T$ above it there are $\kappa$-many nodes.

Answer 1

I'll just take a moment to set up some notations. Let $T$ be an arbitrary tree, with ordering $\sqsubseteq$.

• Given $t \in T$, $\operatorname{ht}_T (t) = \operatorname{ot} \{ s \in T : s \sqsubset t \}$ is the height of $t$ in $T$.
• For an ordinal $\alpha$, $T_\alpha = \{ t \in T : \operatorname{ht}_T (t) = \alpha \}$ is the $\alpha$^th level of $T$.
• For $t \in T$, $T^{(t)} = \{ u \in T : t \sqsubseteq u \}$.

Now to prove a couple little Lemmas.

• Lemma 1. If $S$ is a well-pruned $\kappa$-Souslin tree ($\kappa > \omega$ regular), then for each $s \in S$, $S^{(s)} \cap S_\alpha \neq \varnothing$ for all $\operatorname{ht}_S(s) \leq \alpha < \kappa$.

  Proof. Since the levels of $S$ are antichains, $| S_\alpha | < \kappa$ for all $\alpha < \kappa$.

  Let $s \in S$, and suppose $\operatorname{ht}_S(s) < \beta < \kappa$ is such that $S^{(s)} \cap S_\beta = \varnothing$. Then $S^{(s)} \subseteq \bigcup_{\alpha < \beta} S_\alpha$, however by well-prunedness of $S$ and regularity of $\kappa$ $$\kappa = | S^{(s)} | \leq | {\textstyle \bigcup_{\alpha < \beta}} S_\alpha | < \kappa,$$ a contradiction. ⬜

• Lemma 2. Let $S$ be a $\kappa$-Souslin tree ($\kappa > \omega$ regular), and let $D \subseteq S$ be open-dense. Then there is an $\alpha < \kappa$ such that $S_\xi \subseteq D$ for all $\alpha \leq \xi < \kappa$ (equivalently, $S_\alpha \subseteq D$).

  Proof. Consider the family $$A = \{ t \in D : S^{(t)} \subseteq D \wedge ( \exists s \sqsubset t ) ( s \notin D ) \}.$$ Since $D$ is open-dense in $S$, $A$ is a maximal antichain in $S$, and so has cardinality $< \kappa$. Taking $\alpha = \sup \{ \operatorname{ht}_S (t) : t \in D \}$ by the regularity of $\kappa$ it follows that $\alpha < \kappa$. It is straightforward to show that $\alpha$ is as desired. ⬜

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To finally prove that a well-pruned $\kappa$-Souslin tree $S$ is $\kappa$-Baire for $\kappa > \omega$ regular, take a family $\{ D_\eta \}_{\eta < \lambda}$ ($\lambda < \kappa$) of dense-open subsets of $S$. For each $\eta < \lambda$ by Lemma 2 there is an $\alpha_\eta < \kappa$ such that $S_{\alpha_\eta} \subseteq D_\eta$. If $\alpha = \sup_{\eta < \lambda} \alpha_\eta$, then by regularity or $\kappa$ we have that $\alpha < \kappa$. Note, too, that $S_\alpha \subseteq \bigcap_{\eta < \lambda} D_\eta$ (and so $S_\xi \subseteq \bigcap_{\eta < \lambda} D_\eta$ for all $\alpha \leq \xi < \kappa$). This combined with the conclusion of Lemma 1 implies that $\bigcap_{\eta < \lambda} D_\eta$ is dense in $S$.