I'm self-studying set theory and got stuck on this exercise:

Let $\kappa$ be a regular cardinal. Give an example of a sequence $\langle C_\alpha\mid\alpha<\kappa\rangle$ such that $C_\alpha$ is club in $\kappa$ for every $\alpha<\kappa$, but $$\bigcap\{C_\alpha\mid\alpha<\kappa\}=\emptyset.$$

To make some progress at all, I've tried specific cases. For $\kappa=\omega$, I define $f:\omega\to\omega$ as taking $n$ to the $n$'th prime, and then define

$$C_n:=\{f(n)k\mid k<\omega\},$$

which is clearly unbounded ($|C_n|=\aleph_0$) and closed (every $n\cap C_n$ will be finite, so $\sup(n\cap C_n)=\max(n\cap C_n)\in C_n$). Furthermore due to the $f(n)$'s being primes, we have

$$\bigcap\{C_n\mid n<\omega\}=\emptyset.$$

I'm trying to do the case with $\kappa=2^{\aleph_0}$ now, where I'm trying to exploit that $|\mathbb{R}|=2^{\aleph_0}$ and $|[0,1]|=2^{\aleph_0}$, so if we set $C_n:=[n,n+1]$ then the $C_n$'s are clubs (unbounded due to cardinality and closed due to closed intervals preserve limits) and disjoint. But I can only construct countably many such $C_n$'s, so it doesn't satisfy the example of needing $2^{\aleph_0}$ many disjoint clubs.

If you guys would be so kind as to hint me in the right direction, whether it be about the $2^{\aleph_0}$ case, the general case, or about how I might generalize my methods from my special cases to the general, I would appreciate it a great deal.



First of all, when talking about clubs we generally require that $\kappa$ has uncountable cofinality. There is a problem with clubs for cardinals whose cofinality is $\omega$, because one unbounded sequence without limit points is not really a club (but still closed and unbounded). It's true this question is true for the countable case, but its solution doesn't necessarily generalize. It's better to start with the uncountable case, and deduce from it the countable case (and seeing how you solved the countable case, you only need to solve the uncountable case anyway).

Secondly, you can't use $\Bbb R$. It's not a well-order. The idea is to use the order topology of the cardinal, not of an ordered set of the same cardinality. If you note, the real numbers is an ordered set whose cofinality is $\omega$, as the natural numbers is a cofinal sequence in that order. Furthermore, $[0,1]$ is not unbounded in $\Bbb R$ as it is very much bounded by $0$ and by $1$.

Thirdly, $2^{\aleph_0}$ can be singular. It is consistent, for example that $2^{\aleph_0}=\aleph_{\omega_1}$.

Fourthly, clubs cannot be disjoint. The whole idea is that clubs represent "almost everything", and when you intersection two "almost everything", you should get "almost everything" again. The exercise asks for you to show that it is possible to have a sequence of $\kappa$ clubs, that the intersection of all of them is empty.

Finally, to hint you for the solution, think about end-segments of $\kappa$. (with its order as an ordinal!)

  • $\begingroup$ For the reason you give, one doesn't usually talk about clubs in $\omega$ (or in uncountable cardinals of cofinality $\omega$) but this particular exercise still has a solution for $\kappa = \omega$. $\endgroup$ – Trevor Wilson May 30 '13 at 15:34
  • $\begingroup$ Trevor, true. But if one try to think of the $\omega$ case at first, one comes up with "even vs. odd" sort of split and one can get the wrong impression that two clubs can be disjoint. Or one can do the same as the OP and think that this is somehow an arithmetical thing... $\endgroup$ – Asaf Karagila May 30 '13 at 15:35
  • $\begingroup$ True, the exercise is misleading for $\kappa = \omega$ because there is a way to solve it that doesn't generalize. But it is still a reasonable exercise to solve for all infinite cardinals $\kappa$. $\endgroup$ – Trevor Wilson May 30 '13 at 15:48
  • $\begingroup$ Trevor, well. I suppose you're right. $\endgroup$ – Asaf Karagila May 30 '13 at 15:50
  • $\begingroup$ Ah yes, I can see why one would distinguish between clubs of countable and uncountable regular cardinals. Thank you for pointing out the thing about disjointness, made me realize I had been thinking about this problem completely wrong! If I were to define the $C_\alpha$'s like this: $C_0:=\kappa$, $C_{\alpha+1}:=\{\beta<\kappa\mid\beta>\alpha\}$ and $C_\delta:=\bigcap\{C_\alpha\mid\alpha<\delta\}$ for $\delta$ limit ordinal, I can see that $sup(C_\alpha\cap\kappa)=\kappa$ and closed due to $\kappa$ being regular, with no common element in all of the $C_\alpha$'s. Or am I missing something? $\endgroup$ – Dan Saattrup Nielsen May 30 '13 at 15:52

For $\alpha < \kappa$ let $C_\alpha = \kappa - \alpha$. And you're done.

  • $\begingroup$ why was this voted down? $\endgroup$ – DanielWainfleet Aug 26 '15 at 19:05

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