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Let $\kappa$ be a regular, uncountable cardinal. Let $A$ be an unbounded set, i.e. $\operatorname{sup}A=\kappa$. Let $C$ denote the set of limit points $< \kappa$ of $A$, i.e. the non-zero limit ordinals $\alpha < \kappa$ such that $\operatorname{sup}(X \cap \alpha) = \alpha$. How can I show that $C$ is unbounded? I cannot even show that $C$ has any points let alone that it's unbounded. (Jech page 92)

Thanks for any help.

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Do you mean unbounded? – Asaf Karagila Feb 14 '12 at 15:22
I do yes, sorry. Edited my post. – Paul Slevin Feb 14 '12 at 15:42
up vote 4 down vote accepted

Fix $\xi\in \kappa$, since $A$ is unbounded there is a $\alpha_0\in A$ so that $\xi<\alpha_0$. Now, construct recursively a strictly increasing sequence $\langle \alpha_n: n\in \omega\rangle$. Let $\alpha=\sup\{\alpha_n: n\in \omega\}.$ Since $\kappa$ is regular and uncountable, we have $\alpha<\kappa.$ It is also easy to see that $\sup(A\cap\alpha)=\alpha$.

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Thank you. Is there a quick way to show that $C$ is also closed? I proved that any sequence $\langle \alpha_{\xi} \mid \xi < \gamma \rangle (\gamma < \kappa)$ has its limit in $C$, but I was wondering if there was a clever trick? – Paul Slevin Feb 14 '12 at 16:07
@PaulSlevin: Suppose $c_\alpha\in C$ then you can choose a member from each sequence approaching $c_\alpha$ in $A$, and have a sequence which approaches $\sup c_\alpha$, so $C$ is closed. – Asaf Karagila Feb 14 '12 at 16:23
@Asaf Why do we need regularity and uncountability here? If we do, the following must be incorrect. We have $\{\alpha_n\}=A\cap\{\alpha_n\}\subseteq A\cap\alpha$, and so $\alpha=\sup\{\alpha_n\}\leq\sup(A\cap\alpha).$ Also, $A\cap\alpha\subseteq\alpha,$ ando so $\sup(A\cap\alpha)\leq\sup\alpha=\alpha.$ What's wrong with this reasoning? – Bartek May 14 '13 at 14:47
Are $\alpha_n$ elements of $A$? – Bartek May 14 '13 at 14:49
@Bartek: Yes, $\alpha_n$ are elements of $A$. We want to show that the set of limit points of the club $A$ is unbounded in $\kappa$. So we construct a sequence in $A$, and we take its limit, which has to be $\alpha$ as long as $\alpha\neq\kappa$ itself. But since $\kappa$ is regular and uncountable, every countable sequence is bounded and therefore $\alpha<\kappa$ and we're done. It seems as if you read the answer, but forgot what the question was. – Asaf Karagila May 14 '13 at 15:02

Since $\kappa$ is regular this means that the order type of $A$ is $\kappa$, and for every $\delta<\kappa$ we have that the cofinality of $\delta<\kappa$ as well.

Now suppose that $A$ is bounded in all limit points above $\beta$, without loss of generality $A\cap\beta=\varnothing$. Define a regressive function on limit ordinals:


This is indeed well defined, since $A$ is bounded below $\delta$. Since $\mathrm{Lim}$ is a club set, therefore stationary, this function is constant on a stationary subset.

In turn this means that unboundedly many times $\delta\cap A$ is reaching the same maximum, in particular this means that $A$ is bounded below $\kappa$, in contradiction!

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