Club filter

In mathematics, particularly in set theory, if κ is a regular uncountable cardinal then $\operatorname{club}(\kappa)$, the filter of all sets containing a club subset of κ, is a κ-complete filter closed under diagonal intersection called the club filter.

To see that this is a filter, note that $\kappa\in\operatorname{club}(\kappa)$ since it is thus both closed and unbounded (see club set). If $x\in\operatorname{club}(\kappa)$ then any subset of κ containing x is also in $\operatorname{club}(\kappa)$, since x, and therefore anything containing it, contains a club set.

It is a κ-complete filter because the intersection of fewer than κ club sets is a club set. To see this, suppose $\langle C_i\rangle_{i<\alpha}$ is a sequence of club sets where α < κ. Obviously $C=\bigcap C_i$ is closed, since any sequence which appears in C appears in every C_i, and therefore its limit is also in every C_i. To show that it is unbounded, take some β < κ. Let $\langle \beta_{1,i}\rangle$ be an increasing sequence with β_1,1 > β and $\beta_{1,i}\in C_i$ for every i < α. Such a sequence can be constructed, since every C_i is unbounded. Since α < κ and κ is regular, the limit of this sequence is less than κ. We call it β₂, and define a new sequence $\langle\beta_{2,i}\rangle$ similar to the previous sequence. We can repeat this process, getting a sequence of sequences $\langle\beta_{j,i}\rangle$ where each element of a sequence is greater than every member of the previous sequences. Then for each i < α, $\langle\beta_{j,i}\rangle$ is an increasing sequence contained in C_i, and all these sequences have the same limit (the limit of $\langle\beta_{j,i}\rangle$). This limit is then contained in every C_i, and therefore C, and is greater than β.

To see that $\operatorname{club}(\kappa)$ is closed under diagonal intersection, let $\langle C_i\rangle$, i < κ be a sequence of club sets, and let C = Δ_{i < κ}C_i. To show C is closed, suppose $S\subseteq \alpha<\kappa$ and $\bigcup S=\alpha$. Then for each $\gamma\in S$, $\gamma\in C_\beta$ for all β < γ. Since each C_β is closed, $\alpha\in C_\beta$ for all β < α, so $\alpha\in C$. To show C is unbounded, let α < κ, and define a sequence ξ_i, i < ω as follows: ξ₀ = α, and ξ_{i + 1} is the minimal element of $\bigcap_{\gamma<\xi_i}C_\gamma$ such that ξ_{i + 1} > ξ_i. Such an element exists since by the above, the intersection of ξ_i club sets is club. Then $\xi=\bigcup_{i<\omega}\xi_i>\alpha$ and $\xi\in C$, since it is in each C_i with i < ξ.

References

• Jech, Thomas, 2003. Set Theory: The Third Millennium Edition, Revised and Expanded. Springer. ISBN 3-540-44085-2.

---

This article incorporates material from club filter on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.