# the generalized Delta System Lemma

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.

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 $$\kappa.$$
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.