I need help understanding a section of the proof of lemma 7.9 of the book "A course in Set Theory" by Ernest Schimmerling, more precisely the step where the author says
[...] Easily, we see that $\alpha = \sup{(C \cap \alpha)}$ and $\alpha = \sup{(D \cap \alpha)}$. [...]
Notice that I have read the other proofs of the same fact, for example in "Set Theory: The third millennium edition" by Thomas Jech, I simply cannot understand how the proof of lemma 7.9 follows from definition 7.8 in the book "A course in Set Theory" by Ernest Schimmerling.
Definition 7.8 Let $\kappa$ be a limit ordinal and $C$ be a set. Then:
- $C$ is unbounded in $\kappa$ iff $\sup{(C \cap \kappa)} = \kappa$
- $C$ is closed in $\kappa$ iff for every $\alpha < \kappa$, if $C \cap \alpha \neq \emptyset$, then $\sup{(C \cap \alpha)} \in C$.
- $C$ is club in $\kappa$ iff $C$ is closed and unbounded in $\kappa$.
Lemma 7.9 Assume that $\text{cf}(\kappa) > \omega$. Let $$\mathcal{F} = \{A \subseteq \kappa \mid \text{there exists a club}~C~\text{in}~\kappa~\text{such that}~C\subseteq A\}.$$ Then $\mathcal{F}$ is a filter over $\kappa$.
The part of the proof that confuses me goes as follows:
It enough to show that if $C$ and $D$ are club in $\kappa$, then so is $C \cap D$. First we show that $C \cap D$ is closed in $\kappa$. Let $\beta < \kappa$ and assume that $C \cap D \cap \beta \neq \emptyset$. Let $$\alpha = \sup{(C \cap D \cap \beta)}.$$ We must show that $\alpha \in C\cap D$. Easily, we see that $\alpha = \sup{(C \cap \alpha)}$ and $\alpha = \sup{(D \cap \alpha)}$.
Why do we have $\alpha = \sup{(C \cap \alpha)}$? How does it follow from the definition?
My investigation (I will only deal with $C$): Since $C \cap D \cap \beta \neq \emptyset$ we know that $C \cap \beta \neq \emptyset$ and $\alpha \neq \emptyset$. Assume that $\alpha$ is a successor ordinal (there exists an ordinal $\gamma$ such that $\alpha = \gamma + 1$), then $\sup{(\alpha \cap C)} \leq \gamma < \alpha$.
Where am I going wrong?
Thank you for your time.
First, let's assume that $\alpha$ is a limit ordinal.
If $\alpha=\sup(C\cap D\cap\beta)$, then first and foremost, $\alpha\leq\beta$ and very very clearly, $\alpha=\sup(C\cap D\cap\alpha)$.
So, since $C\cap D\subseteq C$, we have that $C\cap D\cap\alpha\subseteq C\cap\alpha$. Since $\sup(C\cap D\cap\alpha)=\alpha$ it must be that $\sup(C\cap\alpha)\geq\alpha$. But in what way can it not be equal?
Now, what happens if it wasn't a limit ordinal? In that case we really don't care. The problem with closure of sets of ordinals can only arise in the case where $\alpha$ is a limit ordinal.