Ultrafilters

Definition: A filter on a set is an ultrafilter if for every , then either or it’s complement .

Definition: A filter on a set is maximal if there is no filter on such that .

A filter on a set is maximal if and only if is an ultrafilter.

Theorem: Every filter can be extended to an ultrafilter. Proof: Let be a filter on a set . Let be the set of all filters on such that and partially ordered set . If is a chain in , then from Forcing, is a filter and so an upper bound of on . By Zorn’s lemma there exists a maximal element . From the above, is an ultrafilter.

Ultraproducts

Let be an ultrafilter on a set , a vocabulary, and

An ultraproduct is defined as

Domain: For is an equivalence relation obtained from the ultrafilter.

, i.e., two functions and are equivalent if and only if the set of indices such that is in the ultrafilter

For a constant symbol , then , i.e., the ultraproducts interpretation of a constant symbol is defined as the equivalence class (for the relation ) of the function which maps and index to the interpretation of the constant symbol in

For a relation symbol of arity , then

For a function symbol of arity , then The atomic formula is true in the ultraproduct if it is true in ‘almost’ all factors This can also be applied to other filters (not just ultrafilters) although the product is a ‘reduced’ product and not an ultraproduct.

Łoś Theorem (ultraproducts): For a formula of vocabulary (in the language , which we are working in) and elements , then

Proof: Induction over the ‘shape’ of the formula . of the form for :

  • and (i.e.,

  • (i.e., or equivalently, )

  • there is w/

  • Special case: and

Corollary: Th() = Th()

Interesting case: Let be a non principal ultrafilter on , then the arithmetic structure has infinitely large members (larger than other integers)

Ultraproducts can be used to give an alternate proof of the compactness theorem.

References

Forcing [Ultrafilter](Ultrafilter - Wikipedia) jech_03