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

enter image description here enter image description here enter image description here 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 $\kappa.$


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.