Club filter

In mathematics, particularly in set theory, 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 thus both closed and unbounded (see club set). 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 $⟨C_i{⟩}_{i<\alpha }$ is a sequence of club sets where $\alpha <\kappa .$ 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 $\beta <\kappa .$ Let $⟨{\beta }_{1,i}⟩$ be an increasing sequence with ${\beta }_{1,1}>\beta$ and ${\beta }_{1,i}\in C_i$ for every $i<\alpha .$ Such a sequence can be constructed, since every $C_i$ is unbounded. Since $\alpha <\kappa$ and $\kappa$ is regular, the limit of this sequence is less than $\kappa .$ We call it ${\beta }_2,$ and define a new sequence $⟨{\beta }_{2,i}⟩$ similar to the previous sequence. We can repeat this process, getting a sequence of sequences $⟨{\beta }_{j,i}⟩$ where each element of a sequence is greater than every member of the previous sequences. Then for each $i<\alpha ,$ $⟨{\beta }_{j,i}⟩$ is an increasing sequence contained in $C_i,$ and all these sequences have the same limit (the limit of $⟨{\beta }_{j,i}⟩$). 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 $⟨C_i⟩,$ $i<\kappa$ be a sequence of club sets, and let $C={\mathrm{\Delta }}_{i<\kappa }{C}_i.$ 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 $\beta <\gamma .$ Since each $C_{\beta }$ is closed, $\alpha \in C_{\beta }$ for all $\beta <\alpha ,$ so $\alpha \in C.$ To show $C$ is unbounded, let $\alpha <\kappa ,$ and define a sequence ${\xi }_i,$ $i<\omega$ as follows: ${\xi }_0=\alpha ,$ and ${\xi }_{i+1}$ is the minimal element of $\bigcap _{\gamma <{\xi }_i}{C}_{\gamma }$ such that ${\xi }_{i+1}>{\xi }_i.$ Such an element exists since by the above, the intersection of ${\xi }_i$ club sets is club. Then $\xi =\bigcup _{i<\omega }{\xi }_i>\alpha$ and $\xi \in C,$ since it is in each $C_i$ with $i<\xi .$

See also

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References

• Jech, Thomas, 2003. Set Theory: The Third Millennium Edition, Revised and Expanded. Springer. Template:Isbn.