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1 Contents Filter Bases and Nets Filter Bases and Ultrafilters: A Brief Overview Convergence of Filter Bases Ultrafilters Nets: A Brief Overview Translating From a Net to a Filter Base Limit Points of Sets and Filter Bases Images of Filter Bases Finer Filter Bases Tychonoff s Theorem A Point x is a Limit Point of E X iff Filter Base of Subsets of E Converging to x A Function is Continuous iff the Image of Every Convergent Filter Base is a Convergent Filter Base A Point x is a Cluster Point of a Filter Base B iff a Filter Base C Finer Than B With C x Compact iff Each Filter Base Has a Cluster Point iff Every Filter Base Has a Convergent Finer Filter Base Existence of Ultrafilters Containing Filter Bases Generalities on Ultrafilters Ultrafilters Converge to Their Cluster Points When They Have Them Every Ultrafilter on a Compact Space Converges A Maximality and Minimality Conditions on Filter Bases Necessary and Sufficient Maximality Condition For Being an Ultrafilter Surjective Maps Take Ultrafilters to Ultrafilters A Filter Base B on i I X i Converges B x iff π i [B] π i (x) For Each i I Tychonoff s Theorem Application of Filter Bases Index

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3 Filter Bases, Nets and Tychonoff s Theorem

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5 Filter Bases and Nets Filter Bases and Ultrafilters: A Brief Overview To motivate this somewhat, let us define a rudimentary notion of convergent sequences in a topological space. Definition (Convergent Sequences in Topological Spaces). In an arbitrary topological space, one says that a sequence (x n ) of points in the space X converges to the point x X provided that for every nbhd U of x, N U N, n N, x n U. In plain English: For every neighborhood U of x, there exists a positive integer N such that x n U for all n N. Definition (Filter Bases). Let X be a set. To say that B is a filter base on X means that Ø B P(X) \ {Ø} and for all B 1, B 2 B, there exists B 3 B such that B 3 B 1 B 2. In particular, B 1, B 2 B, B 1 B 2 Ø. Exercise (#347). Let (x n ) be a sequence in X. For each n N, let T n = {x n : n N}. Let T = {T N : N N}. Then T is a filter base on X, called the filter base of tails of (x n ). Convergence of Filter Bases Definition (Convergence of Filter Bases). Let X be a topological space, let B be a filter base on X and let x X. To say that B x means that for every nbhd U of x, there exists a B B such that B U. Exercise (#349). Let X be a topological space. Then TFAE: (a) X is Hausdorff. (b) For each filter base B on X and x, y X, if B x and B y, then x = y. Definition (Cauchy Filter Base on a Metric Space). Let (X, ρ) be a met. space and let B be a filter base on X. To say B is Cauchy (i.e., a Cauchy filter base) means that ε > 0, B B such that diam(b) ε. Ultrafilters Definition (Ultrafilter). Let X be a set. To say that U is an ultrafilter on X means that U is a filter base on X and for each filter base C on X, if U C then U = C. (These exist by a routine application of Zorn s lemma.) Theorem. If X is a topological space, then TFAE: (a) X is compact. (b) Every net/filter base in/on X has a cluster point. (c) Every net/filter base in/on X has a convergent subnet/is a subset of a convergent filter base. (d) Every ultrafilter on X converges. 5

6 6 Filter Bases and Nets Nets: A Brief Overview Now, let us introduce the concept of a net. Definition (Directed Set). A directed set is a set A equipped with a relation (which we may also write with the obvious meaning) such that (a) α A, α α; (b) if α β and β γ, then α γ; (c) for any α, β A, γ A, α, β γ. That is, a preorder with the condition that for any α, β A, γ A, α, β γ. Definition (Net). A net in a set X is a mapping from a directed set, say A, into X, say φ: A X. We shall usually denote such a mapping by x α α A Definition (Net Terminology). Let X a topological space, E X, and x α α A a net. (a) We say x α is eventually in E if α 0 A, α α 0, x α E. (b) We say x α is frequently in E if α A, β A, β α and x β E. (c) We say a point x X is a limit of x α (or x α converges to x, or x α x) if for every nbhd U of x, x α is eventually in U. (d) We say a point x X is a cluster point of x α if for every nbhd U of x, x α is frequently in U. Translating From a Net to a Filter Base Remark (Translating Between Filters Bases and Nets). Let x α α A be a net. Then x α α A induces the filter base of tails (or, the derived filter base) B = {{x α : α A, α α 0 }: α 0 A}. Conversely, given a filter base B, it is possible to construct an associated net, but that is not nearly as useful. Limit Points of Sets and Filter Bases Proposition. If X is a topological space, E X, x X, then x is a limit point of E iff there is a net in E \ {x} that converges to x (iff there is a filter base of subsets of E that converges to x), and x B iff there is a net in E that converges to x (iff there is a filter base B whose elements are all contained in E such that B x). Images of Filter Bases Proposition (Images of Filter Bases & Nets). Let X and Y be topological spaces. Let B be a filter base on X, f : X Y. Then the image of B is defined as f[b] def = {f[b]: B B}. The image of a filter base is also a filter base. The analogous statements hold for nets. Proposition. If X, Y are topological spaces, f : X Y, then f is continuous at x X iff for every net x α converging to x (resp., filter base B converging to x), f(x α ) converges to f(x) (resp., f[b] f(x)). Finer Filter Bases Definition (Finer Filter Bases). Let B and C be filter bases. To say that C is finer than B means that B B, C C, C B. Definition (Subnet). A subnet of a net x α α A is a net y β β B together with a map β α β from B to A such that

7 7 1. α 0 A, β 0 B such that if β β 0, then α β α y β = x αβ. Proposition. If x α α A is a net in (resp., B a filter base on) a topological space X, then x X is a cluster point of x α (resp., B) iff x α has a subnet that converges to x (resp., iff there is a filter base C on X such that C is finer than B and C x).

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9 Tychonoff s Theorem A Point x is a Limit Point of E X iff Filter Base of Subsets of E Converging to x Proposition. Let X be a topological space. x X is a limit point of E X iff there is a filter base of subsets of E that converges to x in X. Proof. ( = ) This means that every nbhd U of E in X contains a point in E other than x itself. Let B = {U X : U is a nbhd of x} and let B = {U E : U B }. Then since x is a limit point of E, Ø B (E) \ {Ø}. Moreover, U 1 U 2 Ø for any U 1, U 2 B, so (U 1 E) (U 2 E) = (U 1 U 2 ) E Ø as the intersection of two nbhds of x is still a nbhd of x and, hence, as x is a limit point of E, has nonempty intersection with E. Hence, clearly B is a filter base of subsets of E and B x in X. ( = ) Conversely, suppose B is a filter base of subsets of E such that B x in X. Then for every nbhd U of x in X, there exists B B such that B U. Hence clearly every nbhd of x in X intersects E. Corollary. Let X be a topological space and E X. Then x E iff there is a filter base B of subsets in E such that B x in X. Proof. Trivial from the above. Remark (Images of Filter Bases). Let X and Y be topological spaces. Let B be a filter base on X, f : X Y. Then the image of B is defined as f[b] def = {f[b]: B B}. The image of a filter base is also a filter base. A Function is Continuous iff the Image of Every Convergent Filter Base is a Convergent Filter Base Proposition. If X, Y are topological spaces, f : X Y, then f is continuous at x X iff for every filter base B converging to x, f[b] f(x). x is a Cluster Point of a Filter Base B iff a Filter Base C Finer Than B With C x Definition (Finer Filter Bases). Let B and C be filter bases. To say that C is finer than B means that B B, C C, C B. Definition (Cluster Point). A filter base B on a topological space X is said to have x as a cluster point if U B Ø for every nbhd U of x and B B. Proposition. If B a filter base on a topological space X, then x X is a cluster point of B iff there is a filter base C on X such that C is finer than B and C x. Proof. ( = ) Let C = {U B : B B and U is a nbhd of x}. Then for any nbhds U 1, U 2 of x, any B 1, B 2 B, (U 1 B 1 ) (U 2 B 2 ) = (U 1 U 2 ) (B 1 B 2 ) so that U 1 U 2 is a nbhd of x and B 3 B such that B 3 B 1 B 2 and hence, by assumption that x is a cluster point of B, that (U 1 U 2 ) (B 1 B 2 ) Ø. Obviously this is not empty. Hence, C is filter base and clearly C is finer than B. Moreover, it is clear that C x as every element is contained in some nbhd of x in X. ( = ) Conversely, if C is a finer filter base than B and C x in X. Then obviously x is a cluster point of B. 9

10 10 Tychonoff s Theorem Compact iff Each Filter Base Has a Cluster Point iff Every Filter Base Has a Convergent Finer Filter Base Theorem. Let X be a topological space. Then X is compact iff each filter base on X has a cluster point iff for every filter base on X, there is a finer filter base on X which converges in X. Proof. The last two statements are equivalent by an easy application of the preceding proposition, so we prove the equivalence of the first two. ( = ) Suppose X is compact and let B be a filter base on X. Suppose for the sake of a contradiction that B does not have a cluster point. Let U = { X \ B : B B }. Since B has no cluster point, x X, there exists an open nbhd U of x and a B B such that U B = Ø. Hence, since U is open, x / B since U c is closed, x / B U c and B is equal to the intersection of all closed sets containing B. By then U = { X \ B : B B } is an open cover of X and hence has a finite subcover of the form n j=1 X \ B j = X. But then taking complements again, we must have n j=1 B j = Ø. But then n j=1 B j = Ø which is impossible since B is a filter base and, hence, for any finite intersection of elements of B, there exists a nonempty element B of B contained in it. ( = ) Suppose for the sake of a contradiction that each filter base on X{ has a cluster point but X is not compact. Then there is an open cover U of X with no finite subcover. Let B = X \ } U 0 : U 0 U. Then B is a filter base on X as it is nonempty, does not contain the finite empty set and is downwards direct with respect to inclusion. Hence, B has a cluster point, say x. Then U 0 x X \ U 0 = X \ U 0 since U 0 is open. In particular, U U, x / U, so U does not cover X, a contradiction. finite U, Existence of Ultrafilters Containing Filter Bases Exercise (#718). Let B be a filter base on a set X. Then there exists an ultrafilter U on X such that B U. Proof. Let F = {C : C is a filter base on X and B C } ordered by inclusion. Then F Ø as B F. If Γ is a chain in F, then F F. Hence, F has a maximal element by Zorn s lemma, say U. Generalities on Ultrafilters Exercise (#720). If B and C are ultrafilters such that C B, then C is finer than B. Proof. Let B B, then C C such that C B since C B. Exercise (#721). Let X be a set. (a) Let B be a filter base on X and C a finer filter base on X. Then B C is a filter base on X. (b) Let U be an ultrafilter on X. If C is a filter base on X finer than U, then C U. Hence, as U is also finer than C, C and U are equivalent. Proof. (a) Put D = B C. Clearly Ø D (X) \ {Ø}. Let D, E D and WLOG suppose D B and E C. Since C is finer than B, D C, D D. Thus, F C, F E and F D D as C is a filter base on X. (b) By (a), U C is a filter base on X as C is finer than U ; of course, U U C, so by definition of an ultrafilter, U = U C. Hence, C U. Ultrafilters Converge to Their Cluster Points When They Have Them Exercise (#722). Let x X be a cluster point for an ultrafilter U on a space X. Then U x. Proof. Since x is a cluster point for the ultrafilter U, there is a filter base C on X such that C is finer than U and C x, as we saw before. But by #721(b), C U. Hence, if V is a nbhd of x in X, since C x, C C such that C V and since C U, we have C U. Thus, V a nbhd of x in X, C U, C V. Thus, U x.

11 11 Every Ultrafilter on a Compact Space Converges Exercise (#723). A space X is compact iff every ultrafilter on X converges in X. Proof. (a = b) By #718, each filter base on X has a cluster point. Let U be an ultrafilter on X. Then U has a cluster point, say x. By #722, U x. (b = a) Suppose each ultrafilter on X converges in X. Let B be any filter base on X. By #719, there exists 1 an ultrafilter U on X such that B U. Then, by assumption, x X, U x. By #720, U is finer than B and hence by #717, x is a cluster point for B. Thus, each filter base on X has a cluster point. Therefore, by #718, X is compact. Remark. We thus have the following equivalences X is compact Every ultrafilter on X converges in X Each filter base on X has a cluster point in X For every filter base on X, there is a finer filter base on X which converges in X. A Maximality and Minimality Conditions on Filter Bases Exercise (#724). Let B be a filter base on a set X. (a) For every E X, B B, either B E Ø or B (X \ E) = Ø. (b) Suppose E X, such that B B, B E Ø. Then C = {B E : B B} is a filter base finer than B. Proof. (a) Suppose not. Then there exists E X and B 1 B, B 1 E = Ø, B 2 B, B 2 (X \ E) = Ø. Put E 1 = E and E 2 = X \ E. Since B is a filter base, B B such that B B 1 B 2. Then Ø B = (B E 1 ) (B E 2 ) (B 1 E 1 ) (B 2 E 2 ) = Ø Ø = Ø. A contradiction. (b) Is obvious. Necessary and Sufficient Maximality Condition For Being an Ultrafilter Exercise (#725). A filter base U on a set X is an ultrafilter on X iff E X, either E U or X \ E U. Proof. ( = ) Suppose U is an ultrafilter on X. Let E X. Then by #724, E {E, X \ E}, U U, U E Ø. Let C = {U E : U U } By #724(b), C is a filter base on X which is finer than U. Since U is an ultrafilter on X, C U. In particular, E = X E U. ( = ) Suppose E X, either E U or X \ E U. Let C be a filter base on X with U C. Let C C. By assumption, C U or X \ C U. But this latter case is impossible as then X \ C C. Thus, C C, C U, so that C U. Hence, C = U so U is an ultrafilter on X. Surjective Maps Take Ultrafilters to Ultrafilters Exercise (#726). Let f : X Y be a surjective function and let U be an ultrafilter on X. Then V = {f[u]: U U } is an ultrafilter on Y. Proof. V is a filter base on Y as we know. Let F Y and put E = f 1 [F ]. By #725(b), either E U or X \ E U. If E U, then f[e] V, but f[e] = f [ f 1 [F ] ] = F Rng(f) = F Y = F. So F V. If X \ U U, then f[x \ E] V, but f[x \ E] = f [ f 1 [Y \ F ] ] = (Y \ F ) Rng(f) = Y \ F Y = Y \ F, so Y \ F V. Thus, V is an ultrafilter. 1 We have used the axiom of choice in this assertion as it is equivalent to Zorn s lemma.

12 12 Tychonoff s Theorem A Filter Base B on i I X i Converges B x iff π i [B] π i (x) For Each i I Lemma. Let X = X α be a product of nonempty topological spaces in the product topology. A filter base U on X converges in X iff the filter bases V α = π α [U ] on X α converge in X α for each α. In particular, U x if and only if π α [U ] π α (x) for each α. Proof. ( = ) Suppose U x X. Let V be a neighborhood of π α (x) in X α. Then πα 1 [V ] is a nbhd of x X by definition of the product topology. Hence, there exists U U such that U πα 1 [V ] and hence π α [U] V. But π α V α by definition. Hence, since the nbhd was arbitrary and α was arbitrary, V α π α (x) for each α. ( = ) Let U be a filter base on X and suppose the filter bases V α = π α [U ] on X α converge in X α for each α, say V α x α and put x = (x α ) X and let U be a nbhd of x in X. Then U contains a basis element U α where all but finitely many U α = X α, say α 1,..., α n. Since V αi x αi, for each α i, we may choose V αi U such that π[v αi ] U αi (this must exist, obviously). Since U is a filter base, there exists V U such that Ø V n i=1 V α i. But then clearly V U α. Hence, U x. Tychonoff s Theorem Tychonoff s theorem is listed as Exercise 727 in Professor Falkner s exercises. Theorem (Tychonoff s Theorem). An arbitrary product of nonempty compact topological spaces is compact in the product topology. Proof. Let (X α ) α A be a family of nonempty compact topological spaces and let X = α A X α. Let U be an ultrafilter on X. α A, let V α = {π α [U]: U U }. Then by #726, V α is an ultrafilter for each α A since π α : X X α is onto and each X α Ø. Hence, x α X α, V α x α by #723 since X α is compact and V α is an ultrafilter on X α. Let x = (x α ) α A. Then x X and U x since each V α x α by the lemma. Application of Filter Bases Theorem. Let {f α } α A be a subset of C 0 (X) and {c α } α A be a family of complex numbers. Suppose that for each finite set B A, there exists µ B M(X) (i.e., in the space of complex Radon measures) such that µ B 1 and such that for all α B, f α dµ B = c α. Then there exists µ M(X) such that µ 1 and f α dµ = c α for all α A. Proof. Let {µ B } B A and #B<. Then this family is naturally directed by inclusion (in fact, it is a net) and we pass to the derived filter base B = {{µ B : B B 0, #B < }: Ø B 0 A, #B 0 < }. This clearly is a filter base since B Ø, no member of B is empty, and for all B 1, B 2 B, there exists B 3 B such that B 3 B 1 B 2. By the Riez Representation theorem, ϕ: M(X) C 0 (X), isometrically. By the Banach-Alaoglu theorem, the unit ball in C 0 (X) is compact in the weak topology and thus the closed unit ball B is compact in M(X) (indeed, since each µ B 1, B is a filter base of subsets of this ball) we call the weak topology in M(X) the vague topology. Since every filter base in a compact set has a cluster point, B has a cluster point in B, say µ B notice that, then µ 1. In particular, we recall that this means that for every nbhd U of µ in the weak topology and every B B, U B Ø. But µ is a cluster point of B iff there is a finer filter base C on B such that C is finer than B and C µ. We assert that f α dµ = c α for each α. Invoking the Riesz representation theorem, this means that image of C under the isometric isomorphism from ϕ: M(X) C 0 (X) yields a convergence filter base with ϕ(c ) I µ. Since the weak topology is the topology of pointwise convergence, this means that for each f C 0 (X), ϕ(c )(f) I µ (f) and in particular for each f α. Thus, it suffices to show that the filter base ϕ(c )(f α ) c α in C. But for each nbhd U of c α in C, we may simply choose any element of C which is a superset of a finite subset of A containing α for then it will obviously belong to this nbhd. Hence, f α dµ = c α. This holds for all α A, proving our assertion. Proof. Here s another, slicker proof. Let V be the K-span of T = {f α } α A and define I : V C by I(f 1 + f 2 ) = f1 + f 2 dµ B where B A is any finite subset of A such that the (necessarily finite) decomposition of f 1 and f 2 into elements of T. It is clear this independent of our choice of element in T by linearity of the integral. Then, also clearly, I is a well-defined K-linear function. We assert that it is bounded on V. Let f V such that f u 1. Then I(f) = f dµ B for some finite B A. Since µ B 1, µ (X) 1. Hence, f dµ f d µ f u d µ f u 1 (this is pretty clear with µ the total variation of µ, but see Folland, Proposition 3.13(a), page 94 for the easy proof). Hence, in

13 particular, I 1 and thus, I(f) I f u f u on V. Therefore, I(f) is bounded above for each f V by the norm (we only really need it to be a seminorm) p(f) def = f u. Hence, by the Hahn Banach theorem, I extends to a continuous linear functional on all of C 0 (X) and satisfies I(f) f u for all f C 0 (X) thus, in particular, I 1. Then, by the Riez representation theorem, I = I µ for some µ M(X) with µ 1. Clearly µ does it. 13

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15 Index Cauchy, 5 cluster point, 6, 9 Convergence of Filter Bases, 5 converges, 5, 6 derived filter base, 6 directed set, 6 eventually, 6 filter base, 5 filter base of tails, 5, 6 finer, 6, 9 frequently, 6 iff, 6, 7, 9 12 limit, 6 net, 6 subnet, 6 Tychonoff s Theorem, 12 ultrafilter, 5 15