club filter

If $\\kappa$ is a regular uncountable cardinal then $\\operatorname{club}(\\kappa)$ , 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 $\\kappa\\in\\operatorname{club}(\\kappa)$ since it is obviously both closed and unbounded. If $x\\in\\operatorname{club}(\\kappa)$ then any subset of $\\kappa$ containing $x$ is also in $\\operatorname{club}(\\kappa)$ , 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 $\\langle C_{i}\\rangle_{i<\\alpha}$ 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 $\\langle\\beta_{1,i}\\rangle$ 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 $\\langle\\beta_{2,i}\\rangle$ 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 $i<\\alpha$ , $\\langle\\beta_{j,i}\\rangle$ 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 $\\beta$ .

To see that $\\operatorname{club}(\\kappa)$ is closed under diagonal intersection, let $\\langle C_{i}\\rangle$ , $i<\\kappa$ be a sequence, and let $C=\\Delta_{i<\\kappa}C_{i}$ . 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 $\\beta\\geq\\alpha$ , and since each $C_{\\beta}$ is closed, $\\alpha\\in C_{\\beta}$ , so $\\alpha\\in C$ .

单词	ClubFilter
释义	club filter If $\\kappa$ is a regular uncountable cardinal then $\\operatorname{club}(\\kappa)$ , 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 $\\kappa\\in\\operatorname{club}(\\kappa)$ since it is obviously both closed and unbounded. If $x\\in\\operatorname{club}(\\kappa)$ then any subset of $\\kappa$ containing $x$ is also in $\\operatorname{club}(\\kappa)$ , 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 $\\langle C_{i}\\rangle_{i<\\alpha}$ 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 $\\langle\\beta_{1,i}\\rangle$ 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 $\\langle\\beta_{2,i}\\rangle$ 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 $i<\\alpha$ , $\\langle\\beta_{j,i}\\rangle$ 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 $\\beta$ . To see that $\\operatorname{club}(\\kappa)$ is closed under diagonal intersection, let $\\langle C_{i}\\rangle$ , $i<\\kappa$ be a sequence, and let $C=\\Delta_{i<\\kappa}C_{i}$ . 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 $\\beta\\geq\\alpha$ , and since each $C_{\\beta}$ is closed, $\\alpha\\in C_{\\beta}$ , so $\\alpha\\in C$ .
随便看	corollary of Cauchy integral theorem CorollaryOfEulerFermatTheorem corollary of Euler-Fermat theorem CorollaryOfKummersTheorem corollary of Kummer’s theorem CorollaryOfMorleysTheorem corollary of Morley’s theorem CorollaryOfSchurDecomposition corollary of Schur decomposition CorollaryToTheCompositumOfAGaloisExtensionAndAnotherExtensionIsGalois corollary to the compositum of a Galois extension and another extension is Galois CorrespondenceBetweenNormalSubgroupsAndHomomorphicImages correspondence between normal subgroups and homomorphic images CorrespondenceOfNormalSubgroupsAndGroupCongruences correspondence of normal subgroups and group congruences CorrespondingAnglesInTransversalCutting corresponding angles in transversal cutting Coset coset CosineAtMultiplesOfStraightAngle cosine at multiples of straight angle CosinesLaw cosines law CotangentBundle cotangent bundle