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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 maximal filter: given some \(x \in X\), the
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
Similarly, a filter \(F\) on \(X\) is called normal, if, for each function \(f: X \to \bigcup X\), 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 no fine filter is principal, every principal filter is normal: assume \(F\) is principal, witnessed by \(x \in X\), and \(f: X \to \bigcup X\). Then, if \(\{s \in X: f(s) \in s\} \in F\), we have \(f(x) \in x\), 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 fine!
If \(F\) is a fine filter on \(X\), and \(X\) is a [[well-ordered set]], then all elements of \(F\) have the same cardinality as \(X\), and therefore fine (or even normal fine) ultrafilters may more closely approximate the notion of largeness - assume \(Y \in F\) and \(|Y| < |X|\). Let \(\leq\) be the well-order on \(X\), and let \(x\) be the \(\leq\)-least element of \(X\) which is greater than all elements of \(Y\). By fineness, we have \(\{s \in X: x \in s\} \in F\), and therefore \(\{s \in X: x \in s\} \cap X \in F\) since \(F\) is a filter. However, \(\{s \in X: x \in s\} \cap X = \emptyset\), contradicting \(\emptyset \notin F\).
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