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It is possible to impose further conditions, other than the four in the definition of the filter and nonprincipality. This includes \(\gamma\)-completeness, for a cardinal \(\gamma\), which asserts that the filter is closed not just intersection of two sets, but of \(< \gamma\)-many sets. Note that any filter is \(\omega\)-complete. A cardinal \(\kappa\) with a \(\kappa\)-complete ultrafilter on \(\kappa\) is precisely a [[measurable]] cardinal, and thus the existence of such a cardinal is unprovable in [[ZFC]], assuming its consistency.
For a set \(X\), a filter \(F\) on \(X\) is called fine if, for each \(x \in \bigcup X\), \(\{s \in X: x \in s\} \in F\). In particular, an ultrafilter on a cardinal \(\kappa\) is fine (in this case, also known as uniform) if, for each \(\alpha < \kappa\), \(\{\sigma: \alpha < \sigma < \kappa\} \in F\). NoteThis thatalso makes sense, ifsince "almost all" ordinals below \(X \subseteq kappa\bigcup) Xwill be in this interval. If \(F\), thenis everya fine filter on \(X\), and \(X\) is nonprincipal:a assume[[well-ordered set]], then all elements of \(F\) ishave principalthe same cardinality as \(X\), witnessedand bytherefore fine (or even normal fine) ultrafilters may more closely approximate the notion of largeness - assume \(xY \in F\) and \(|Y| < |X|\). ThenLet \(x\leq\) be the well-order on \in(X\), and let \bigcup(x\) be the \(\leq\)-least element of \(X\) so,which ifis greater than all elements of \(FY\). wereBy finefineness, thenwe have \(\{s \in X: x \in s\} \in F\), thusand therefore \(x\{s \in X: x \in s\} \cap X \in F\) -since \(F\) is a contradiction!filter. However, \(\{s \in X: x \in s\} \cap X = \emptyset\), contradicting \(\emptyset \notin F\).
Similarly,Note a filter \(F\) on \(X\) is called normalthat, if, for each function \(f: X \tosubseteq \bigcup X\), if \(\{s \in X: f(s) \in s\} \in F\), then there is some \(x \in \bigcup X\) so that \(\{s \in X: f(s) = x\} \in F\). This definition is inspired by the [[Fodor's lemma|pressing-down lemma]]. Dually to the fact that noevery fine filter is principal, every principal filter is normalnonprincipal: assume \(F\) is principal, witnessed by \(x \in X\),. andThen \(f: Xx \toin \bigcup X\). Thenso, if \(F\) were fine, then \(\{s \in X: f(s)x \in s\} \in F\), we havethus \(f(x) \in x\), and so, letting \(x' = f(x)\), we have \(x' \in \bigcup X\), and \(\{s \in X: f(s) = x'\} \in F\). Note, however, that there can be nonprincipal filters which are normal - it is even possible for a filter to be both normal and finecontradiction!
If \(F\) isSimilarly, a fine filter on \(XF\), andon \(X\) is acalled [[well-orderednormal, set]]if, thenfor alleach elements offunction \(F\)f: haveX the\to same\bigcup cardinality as \(X\), and therefore fineif \(or\{s even\in normalX: finef(s) ultrafilters\in mays\} more\in closelyF\), approximatethen thethere notionis of largeness - assumesome \(Yx \in F\bigcup X\) andso that \(|Y|\{s <\in |X|\: f(s). Let= x\(} \leqin F\). beThis thedefinition well-orderis oninspired \(X\),by andthe let[[Fodor's \(x\)lemma|pressing-down belemma]]. Dually to the \(\leq\)-leastfact elementthat ofno \(X\)fine whichfilter is greaterprincipal, thanevery allprincipal elementsfilter ofis normal: assume \(YF\). Byis finenessprincipal, wewitnessed haveby \(\{sx \in X:\), xand \in(f: X s\}to \inbigcup FX\),. andThen, thereforeif \(\{s \in X: xf(s) \in s\} \cap X \in F\), sincewe have \(F\f(x) is\in ax\), filter.and Howeverso, letting \(\{sx' = f(x)\in), X:we have \(x' \in s\} \capbigcup X = \emptyset\), contradictingand \(\emptyset{s \notinin X: f(s) = x'\} \in F\). Note, however, that there can be nonprincipal filters which are normal - it is even possible for a filter to be both normal and fine!
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