# a question on closed unbounded sets

For any limit ordinal $\alpha$, $C$ is a c.u.b. set (closed unbounded set) in $\alpha$ iff $C$ is closed and unbounded in $\alpha$.

In Kunen's book, the Lemma of II 6.8 says：If $cf(\alpha) > \omega,$ then the intersection of any family of less than $cf(\alpha)$ c.u.b. subsets of $\alpha$ is c.u.b.

I think the his proof is a little complex. I prove it by this.

Prove: Let $C_\beta$ be c.u.b. in $\alpha$ for $\beta< \lambda$, where $\lambda<cf(\alpha),$ and let $D=\bigcap_{\beta<\lambda} C_\beta$. Of course $D$ is closed in $\alpha$. We only need to show that $D$ is unbounded in $\alpha$. To see this, for any $\xi<\alpha$, we want to find a ordinal $\gamma$ in $D$ such that $\xi<\gamma.$ For the given $\xi$, for every $\beta<\lambda$, there is a $\gamma_\beta \in C_\beta$ such that $\xi<\gamma_\beta$ for every $C_\beta$ is unbounded in the $\alpha$. Let $\gamma=\bigcup\{\gamma_\beta: \beta<\lambda\}$. Since $\lambda<cf(\alpha),$ we have $\gamma<\alpha$. For every $C_\beta$ is unbounded in $\alpha$, there is a ordinal in $C_\beta$ which is bigger than $\gamma$, by the transitive, we have $\gamma \in C_\beta$ for every $\beta$. Therefore $\gamma \in D$. This completes the proof.

However, I proof the lemma without applying the condition $cf(\alpha) > \omega$. I don't know where I am wrong. Could someone give some suggestions to me on this question? Thanks:)

In Kunen's argument, why does the $g^\omega(\xi)$ is in every $C_\beta$? It is said" for each $\beta$, $C_\beta$ is unbounded in $g^\omega(\xi)$, so $g^\omega(\xi)\in C_\beta$." I can't reach this point. Any help will be appreciated:P

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If $\operatorname{cf}(\alpha)=\omega$, $D$ can be empty even if $\lambda=2$. Take $\alpha=\omega$: $\{2n:n\in\omega\}$ and $\{2n+1:n\in\omega\}$ are disjoint cub’s in $\omega$. If you want an uncountable example, take $\alpha=\omega_\omega$; $\{\omega_n:n\in\omega\}$ and $\{\omega_n+1:n\in\omega\}$ are disjoint cub’s. (The last sentence before Definition 6.7 even points this out.)

The specific error in your argument is the claim that $\gamma\in C_\beta$ for every $\beta$: the fact that $\gamma$ is less than some member of $C_\beta$ does not imply that $\gamma\in C_\beta$: $C_\beta$ is in general not a transitive set.

Added: In the textbook argument you have a strictly increasing sequence $\langle g^n(\xi):n\in\omega\rangle$ of ordinals less than $\alpha$; since $\operatorname{cf}(\alpha)>\omega$, $g^\omega(\xi)\triangleq\sup\{g^n(\xi):n\in\omega\}<\alpha$ as well. Moreover, $g^0(\xi)=\xi$, so $g^\omega(\xi)>\xi$. Now consider the cub $C_\beta$. Recall that $f_\beta(\xi)$ is the smallest member of $C_\beta$ larger than $\xi$ and that $g(\xi)=\sup\{f_\gamma(\xi):\gamma<\lambda\}\ge f_\beta(\xi)$, so at this point we have $$\xi<f_\beta(\xi)\le g(\xi)\;.\tag{1}$$ Then we repeat the process: $g^1(\xi)=g(g^0(\xi))=g(g(\xi))$, so $(1)$ becomes $$g(\xi)<f_\beta(g(\xi))\le g(g(\xi))=g^1(\xi)\;.\tag{2}$$ In general we have

$$g^n(\xi)<f_\beta(g^n(\xi))\le g(g^n(\xi))=g^{n+1}(\xi)$$ for each $n\in\omega$, so we have an infinite chain

$$\xi=g^0(\xi)<f_\beta(g^0(\xi))\le g^1(\xi)<f_\beta(g^1(\xi))\le g^2(\xi)<\dots\;,$$ where $f_\beta(g^n(\xi))\in C_\beta$ for each $n\in\omega$. Since the strictly increasing sequences $\langle f_\beta(g^n(\xi)):n\in\omega\rangle$ and $\langle g^n(\xi):n\in\omega\rangle$ are ‘interlocked’, they have the same supremum, and therefore $g^\omega(\xi)=\sup\{f_\beta(g^n(\xi)):n\in\omega\}$. Again, recall that $f_\beta(g^n(\xi))\in C_\beta$ for each $n\in\omega$, so $g^\omega$ is the limit from below of members of $C_\beta$: $C_\beta\cap g^\omega(\xi)$ is an unbounded subset of $g^\omega(\xi)$, and therefore, by the definition of cub, $g^\omega\in C_\beta$. And since $\beta<\lambda$ was arbitrary, $g^\omega\in D$.

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Oh..., however, in Kunen's argument, why does the $g^\omega(\xi)$ is in every $C_\beta$? It is said" for each $\beta$, $C_\beta$ is unbounded in $g^\omega(\xi)$, so $g^\omega(\xi)\in C_\beta$." I can't reach this point. –  Paul Jan 14 '12 at 5:15
@John: I’ve added an explanation of Ken’s argument to my original answer. –  Brian M. Scott Jan 14 '12 at 6:19
@Brain, your explanation is very helpful for me. The result $g^\omega \in C_\beta$ is obtained for $C_\beta$ is closed, since $f_\beta{(g^n(\xi))} \in C_\beta$ and $g^\omega$ is a limit. Am I right? However, why we need "$C_\beta\cap g^\omega(\xi)$ is an unbounded subset of $g^\omega(\xi)$"? –  Paul Jan 14 '12 at 7:45
@John: That’s just another way of saying that $g^\omega(\xi)$ is a limit point of $C_\beta\cap g^\omega(\xi)$, and therefore (since $C_\beta$ is closed) $g^\omega(\xi)\in C_\beta$. It also matches the way that cub was defined in Definition 6.6; that definition was chosen because it doesn’t require the reader to know any topology. –  Brian M. Scott Jan 14 '12 at 7:52
@Brain: O...I see:) Now the lemma of c.u.b is very clear to me. Thanks Brain for your much help. Very grateful:) –  Paul Jan 14 '12 at 8:00