Filter

A filter is a particular notion used to define ultraproducts/ultrapowers and various large cardinals above measurable cardinals, although they also have some relation to indescribable and greatly Mahlo cardinals. Formally, a filter on a set \(X\) is a collection \(F\) of subsets of \(X\) satisfying the following conditions:

• \(X \in F\).
• \(\emptyset \notin F\).
• If \(A \in F\) and \(B \in F\) then \(A \cap B \in F\).
• If \(A \in F\) and \(A \subseteq B\) then \(B \in F\).

Intuitively, a filter is a predicate which chooses out exactly which subsets of \(X\) are large; namely:

• \(X\) is large.
• The empty set is not large.
• The intersection of large sets is large.
• A superset of a large set is large.

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 \(|\kappa \setminus X| < \kappa\). It is also easy to see this is a filter, 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 \(\alpha < \kappa\), \(\{\sigma: \alpha < \sigma < \kappa\} \in F\). Note that, if \(X \subseteq \bigcup X\), then every fine filter is nonprincipal: assume \(F\) is principal, witnessed by \(x \in X\). Then \(x \in \bigcup X\) so, if \(F\) were fine, then \(\{s \in X: x \in s\} \in F\), thus \(x \in x\) - a contradiction!

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 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 \(\emptyset \notin F\).