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Notes on Ultrafilters

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Notes on Ultrafilters Notes on Ultrafilters Alex Kruckman Toolbox Seminar 11/7/12 1 Basic theory Let X be a set. An (ultra)filter on X is a consistent choice of which subsets of X are \large". Definition. A filter on X is F ⊆ P(X) such that 1. X 2 F (the whole set is large). 2. ; 62 F (the empty set is not large). 3. If A 2 F and A ⊆ B, then B 2 F (any set containing a large set is large). 4. If A; B 2 F, then A \ B 2 F (large sets have large intersection). Examples. • The trivial filter: F = fXg • The principal filter generated by x 2 X: F = fA ⊆ X j x 2 Ag • The cofinite filter (X infinite): F = fA ⊆ X j X n A is finiteg Note that filters have the finite intersection property (FIP): \ If A1;:::;An 2 F, then Ai 6= ;: i Lemma (FIP). Any subset S ⊆ P(X) with the FIP has a minimal filter containing it, the filter generated by S. Proof. Close S downward under finite intersections, then upward under supersets. Definition. A filter F on X is an ultrafilter if 5. For any A ⊆ X, A 2 F or X n A 2 F (every set is either large or co-large). Sn Exercise 1. Given a filter F, show that F is an ultrafilter iff i=0 Ai 2 F implies that Ai 2 F for some i (a large set cannot be a finite union of small sets). Note that principal filters are ultrafilters, but the trivial filter and the cofinite filter are not. Are there any nonprincipal ultrafilters? The answer is yes, but not constructively! The following lemma is a weak form of the Axiom of Choice, i.e. it is not provable in ZF. 1 Lemma (Ultrafilter lemma). Every filter is contained in an ultrafilter. Proof. Zorn's Lemma on the poset of filters on X containing F. We need to check the following: • The union of a chain of filters is a filter. (Easy.) • A maximal filter is an ultrafilter. (Let F be maximal. If F is not an ultrafilter, take A with A2 = F and X n A2 = F. Then F [ fAg has the FIP, and F is properly contained in the filter generated by F [ fAg, contradicting maximality.) Corollary. Any subset S ⊆ P(X) with the FIP is contained in an ultrafilter. Exercise 2. Every nonprincipal ultrafilter contains the cofinite filter. 2 Generalized limits Consider a function f : N ! [0; 1], determining a sequence a0; a1; a2;::: We say that lim an = L if for every open set U containing L, all but finitely many natural numbers n!1 are mapped into U by f, i.e. f −1(U) is in the cofinite filter on N. Now given distinct points a and b in [0; 1], the sequence a; b; a; b; : : : does not converge. If we take disjoint neighborhoods a 2 U, b 2 V , f −1(U) and f −1(V ) are both infinite, coinfinite subsets of N. An ultrafilter on N would give preference to one of these two sets and decide whether a or b should be the limit of the sequence. Definitions We'll take a more general perspective by transporting (ultra)filters from the index set to the target space (from N to [0; 1] in the example above). Exercise 3. Let F be an (ultra)filter on the set X, and let f : X ! Y be a map of sets. −1 Then f∗F = fA ⊆ Y j f (A) 2 Fg is an (ultra)filter on Y . Definition. A filter F on a topological space Y converges to a point y 2 Y if for all open sets U containing y, U 2 F. Now we can define generalized limit points of maps. With this definition, the usual notion of convergence of a sequence is F-convergence, where F is the cofinite filter on N. Definition. Let X be a set, F an filter on X, Y a topological space, and f : X ! Y a map of sets. Then y 2 Y is an F-limit point of f if f∗F converges to Y . Filter convergence works best with ultrafilters on compact Hausdorff spaces, as illustrated by the following theorem. 2 Theorem (Ultrafilter convergence theorem). Let Y be a topological space. 1. Y is compact if and only if every ultrafilter F on Y converges to at least one point. 2. Y is Hausdorff if and only if every ultrafilter F on Y converges to at most one point. Proof. 1. Suppose for contradiction that Y is compact, but F has no limit points. Then S for all y 2 Y , there is an open set Uy containing y such that Uy 2= F. So Y = y Uy, Sn and by compactness, Y = i=1 Uyi . But Y 2 F, so some Uyi 2 F, contradiction. S Conversely, suppose that Y is not compact. Then there is an open cover Y = Ui T i with no finite subcover. So i(Y n Ui) = ;, but no finite subintersection is empty. Then f(Y n Ui)gi has the FIP, so we can take an ultrafilter F containing it. Now for any point y 2 Y , y is contained in some Ui, and Ui 2= F, since (Y n Ui) 2 F. So y is not a limit point of F. 2. Suppose for contradiction that Y is Hausdorff, but y 6= y0 are both limit points of F. Take disjoint open sets y 2 U, y0 2 U 0. Now U; U 0 2 F, but U \ U 0 = ;, contradition. Conversely, suppose that Y is not Hausdorff. Then there are points y 6= y0 such that every open neighborhood of y intersects every open neighborhood of y0. This means that fU j y 2 U openg [ fU 0 j y0 2 U 0 openg has the FIP. Let F be an ultrafilter containing it. Then y and y0 are both limit points of F. So if f : X ! Y is a function from a set to a compact Hausdorff space, f has a unique F-limit for every ultrafilter F on X. If F is the principal filter generated by x 2 X, then the F-limit of f is f(x). Application: Stone-Cech compactification The Stone-Cech compactification of a set X is a compact Hausdorff space βX along with a map of sets η : X ! βX satisfying the following universal property: Given a compact Hausdorff space Y and a map of sets f : X ! Y , there is a unique continuous map φ : βX ! Y such that φ ◦ η = f. η X / βX C CC CC 9! φ f CC C! Y For those initiated into category theory, the Stone-Cech compactification is a functor β : Set ! CHaus which is left-adjoint to the forgetful functor CHaus ! Set, and βX can be seen as the free compact Hausdorff space on X. As usual, βX is unique up to unique isomorphism. The construction is as follows. Underlying set: βX = fF j F is an ultrafilter on Xg. Topology: Given A ⊆ X, let UA = fF j A 2 Fg. The sets UA are a basis (of clopen sets). The universal map η: x 7! the principal ultrafilter generated by fxg. Given f : X ! Y , define φ : βX ! Y by F 7! the F-limit of f. 3 Exercise 4. Check the details. • βX is a compact Hausdorff space for all X. In fact, it is a Stone space: compact, Hausdorff, and totally disconnected. • β is a functor (given a map g : X ! X0, there's only one reasonable choice for βg : βX ! βX0 taking ultrafilters on X to ultrafilters on X0) • The diagram commutes. • φ is continuous. • φ is unique. Note that since every ultrafilter on a finite set is principal, if X is finite, βX = X with the discrete topology. Application: Tychenoff's theorem The convergence theorem allows us to give a snappy proof of Tychenoff's theorem. Theorem (Tychenoff's theorem). A product of compact spaces is compact. Q Proof. Let hXiii be a collection of compact spaces. To show that X = i Xi is compact, it suffices to show that every ultrafilter F on X has a limit point. th Let πi : X ! Xi be the i projection map. Then (πi)∗F is an ultrafilter on Xi, and since Xi is compact, it has a limit point xi 2 Xi. I claim that x = (xi) 2 X is a limit point of F. −1 The topology on X is generated by sets of the form V = πi (U), where U is an open set in Xi. So any open set containing x contains a finite intersection of sets of this form which contain x. Since F is closed under supersets and finite intersections, it suffices to show that −1 if x 2 V = πi (U), then V 2 F. Suppose x 2 V . Then xi 2 U, so U 2 (πi)∗F, since xi is a limit point of (πi)∗F. But then by definition of (πi)∗, V 2 F. This completes the proof. 3 Ultraproducts Application: Prime ideals in products of rings Qn Exercise 5. In a finite product of rings i=0 Ai, all prime ideals have the following form: f(a0; : : : ; ai; : : : ; an) j ai 2 pg, where p is a prime ideal in Ai for some i. Hence a prime ideal in the product is determined by a choice of a prime idea in one of the factors, and Qn ∼ `n Spec( i=0 Ai) = i=0 Spec(Ai) (you can think of this as an isomorphism of sets, topological spaces, or schemes).
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