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An ultrafilter is a maximal filter: for every subset \(Y\) of \(X\), either \(Y\) is large or its complement (\(X \setminus Y\)) is large. The reason these are maximal is, because if \(F' \supset F\), then there is some \(Y \in F'\) so that \(Y \notin F\). Therefore \(X \setminus Y \in F\), and so \(X \setminus Y \in F'\), therefore \((X \setminus Y) \cap Y = \emptyset \in F'\), so \(F'\) can't be a filter.
One of the most natural examples of a filter is a principal filter: given some \(x \in X\), the principal filter with respect to \(x\) has \(Y\) large iff \(x \in Y\). It is easy to verify this is an ultrafilter, however this doesn't exactly match with one's intuition of largeness, since even singletons can be large in this interpretation. One therefore typically studies nonprincipal filters instead - another easy example is the Fréchet filter, the filter on a cardinal \(\kappa\) defined by \(X\) being large iff \(|\kappa \setminus X| < \kappa\). ItAll sets in the Fréchet filter are intuitively large since they must have cardinality \(\kappa\) - it is also easy to see this is in fact a filter, and not a misnomer; however, it is not an ultrafilter.
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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