# club filter

If $\kappa $ is a regular^{} uncountable cardinal then $\mathrm{club}(\kappa )$, the filter of all sets containing a club subset of $\kappa $, is a $\kappa $-complete^{} filter closed under diagonal intersection called the *club filter*.

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

It is a $\kappa $ complete filter because the intersection^{} of fewer than $\kappa $ club sets is a club set. To see this, suppose $$ is a sequence^{} of club sets where $$. 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 $$. Let $\u27e8{\beta}_{1,i}\u27e9$ be an increasing sequence with ${\beta}_{1,1}>\beta $ and ${\beta}_{1,i}\in {C}_{i}$ for every $$. Such a sequence can be constructed, since every ${C}_{i}$ is unbounded. Since $$ and $\kappa $ is regular, the limit of this sequence is less than $\kappa $. We call it ${\beta}_{2}$, and define a new sequence $\u27e8{\beta}_{2,i}\u27e9$ similar to the previous sequence. We can repeat this process, getting a sequence of sequences $\u27e8{\beta}_{j,i}\u27e9$ where each element of a sequence is greater than every member of the previous sequences. Then for each $$, $\u27e8{\beta}_{j,i}\u27e9$ is an increasing sequence contained in ${C}_{i}$, and all these sequences have the same limit (the limit of $\u27e8{\beta}_{j,i}\u27e9$). This limit is then contained in every ${C}_{i}$, and therefore $C$, and is greater than $\beta $.

To see that $\mathrm{club}(\kappa )$ is closed under diagonal intersection, let $\u27e8{C}_{i}\u27e9$, $$ be a sequence, and let $$. Since the diagonal intersection contains the intersection, obviously $C$ is unbounded. Then suppose $S\subseteq C$ and $sup(S\cap \alpha )=\alpha $. Then $S\subseteq {C}_{\beta}$ for every $\beta \ge \alpha $, and since each ${C}_{\beta}$ is closed, $\alpha \in {C}_{\beta}$, so $\alpha \in C$.

Title | club filter |
---|---|

Canonical name | ClubFilter |

Date of creation | 2013-03-22 12:53:11 |

Last modified on | 2013-03-22 12:53:11 |

Owner | Henry (455) |

Last modified by | Henry (455) |

Numerical id | 5 |

Author | Henry (455) |

Entry type | Definition |

Classification | msc 03E10 |

Defines | club filter |