the generalized Delta System Lemma

Question

I'm trying to understand the proof of the generalized Delta System Lemma in Kunen's Set Theory, the 2013 edition.

This stationary set $T$ we've constructed using Fodor's Lemma (Pressing Down Lemma) has a nonempty intersection with $D$, but how can we know that there is more than one element in $T\cap D$? I don't understand what the role of $D$ is here.

I also don't understand this statement: Since $T$ is stationary, and hence of size $\kappa$, while $|[\nu]^{<\lambda}|<\kappa$, there is, by Lemma III.6.7, a stationary $W\subseteq T$ such that the $A_\alpha \cap \nu$, for $\alpha\in W$, are all the same set $R$.

Is this stationary $W$ a subset of $T\cap D$? To use Lemma III.6.7, are we setting $f:T\rightarrow [\nu]^{<\lambda}$, $f(\alpha)=A_\alpha \cap \nu$? And why is the fact that $T$ has size $\kappa$ relevant?

I don't really get what the overall picture is. I'd appreciate some more explanation. Thank you in advance.

Answer 1

Regarding $T$ and $D:$ Since $\kappa$ is regular and uncountable, the intersection of any 2 clubs of $\kappa$ is club in $\kappa.$ So $T\cap D$ is stationary in $\kappa.$ Because if $D'$ is any club in $\kappa,$ then so is $D\cap D',$ so $\emptyset \ne T\cap (D\cap D')=(T\cap D)\cap D'.$ So $T\cap D $ not only has more than one member; we have $|T\cap D|=\kappa.$

That is, if $\kappa$ is regular and uncountable, the intersection of a club $D$ of $\kappa$ with a stationary $T$ of $\kappa$ is stationary in $\kappa.$

Answer 2

I believe I can clear up this last paragraph:

This is not mentioned in the proof, but I think we do in fact need to use $f : T \cap D \to [\nu]^{< \lambda}$ by $\alpha \mapsto A_\alpha \cap \nu$ to apply III.6.7 and take $W \subseteq T \cap D$. As DanielWainfleet's answer mentions, we can do this since $T \cap D$ is in fact stationary in $\kappa$ (and of size $\kappa$); this is crucial for what follows.

The first result is that $A_\alpha \cap \nu = R$ for all $\alpha \in W$. But we still need to show that $\{A_\alpha \; : \; \alpha \in W\}$ actually forms a $\Delta$-system with root $R$: Indeed, if we arrange that $W \subseteq T \cap D$, then, taking $\alpha, \beta \in W$, the preceding argument applies to show that $A_\alpha \cap A_\beta \subseteq \nu$, so $$ A_\alpha \cap A_\beta = A_\alpha \cap A_\beta \cap \nu = (A_\alpha \cap \nu) \cap (A_\beta \cap \nu) = R $$ as desired.