Filters and the theory of convergence in topological spaces.

In order to study convergence in general topological spaces, conventional series (i.e. functions defined on the natural numbers) are too restrictive. There are today two generalizations, one is the concept of a filter, introduced by Henri Cartan, the other is the concept of a , introduced by Moore and Smith. The two theories are equivalent, but if one learns about filters I trust that anyone will agree that this is by far the most natural and elegant way to do things. 1. . A filter F on a X is a subset of P (X) satisfying the conditions,

(1a) ∅ 6∈ F and ∅= 6 F. (1b) If A ∈ F and B ∈ F, then A ∩ B ∈ F. (1c) If A ∈ F and A ⊆ B, then B ∈ F.

If F1 and F2 are filters on a set X we say that F2 is finer than F1 if F1 ⊆ F2. We can also say that F1 is coarser than F2. The set of all filters on a set X together with the fineness relation is a partially ordered set, and it is inductive. This means that any linearly ordered set of filters is dominated by a filter. S S In fact if L is a linearly ordered set of filters, then L is a filter, because for any sets X1 ∈ L S and X2 ∈ L there is a filter F ∈ L such that X1 ∈ F and X2 ∈ F . By Zorn’s Lemma every filter is dominated by a maximal filter. A maximal filter is called an ultrafilter. Here are some examples of filters. If X is a nonempty set and x ∈ X, the trivial filter generated by x, is Triv(x) = {A ⊆ X | x ∈ A}. Notice that there are no filter strictly finer than a trivial filter, so the trivial filters are examples of ultrafilters. If X is infinite there are ultrafilters that are not trivial. If X is a and x ∈ X is a point, the set of neighborhoods of x is a filter, which we call the neighborhood-filter of x, Nbh(x) = {A | A is a neighborhood of x}. We always have Nbh(x) ⊆ Triv(x). If (S, ) is a directed preordered set, there is a coarsest filter containing all sections or “tails” of S. A section is a subset of the form σ(a) = {x ∈ S | a  x}. This filter is called the Fr´echet filter on S. It can be shown that there are uncountably many ultrafilters dominating the Fr´echet filter on N. 2. Filter basis. A basis B for a filter on a set p˚a X is a subset of P (X), satisfying the conditions,

(2a) ∅ 6∈ B and ∅= 6 B. (2b) If A ∈ B and B ∈ B, there is a set C ∈ B such that C ⊆ A and C ⊆ B.

Condition 2b says that a filter basis with the relation ⊇ is a . If B is an arbitrary subset of P (X), and we define F(B) = {A | ∃(B ∈ B) B ⊆ A}.The reader may verify that B is a basis of a filter if and only if F(B) is a filter, and F(B) is the coarsest filter containing B. Note that if B1 and B2 are basis, then F(B1) ⊆ F(B2),if and only if for every B ∈ B1 there is a set A ∈ B2 such that A ⊆ B. In other words it is the “small” sets in a filter basis that determines the associated filter. We say that two filter basis are equivalent, if they generate the same filter.

1 3. Induced filters. Let f : X → Y be a function and let F be a filter on Y . If f(X) ∩ A 6= ∅ for all A ∈ F, f ∗(F) = {f −1(A) | A ∈ F} is a filter on X. Whenever f ∗ is defined it preserves the ordering of filters, and ∗ is functorial, meaning that (gf)∗ is defined for F, if and only if g∗ is defined for F, f ∗ is defined for g∗(F), and f ∗(g∗(F)) = (gf)∗(F). The filter f ∗(F) is called the pullback of the filter F by f. Much more important is the notion of push forward or direct image. If F is a filter on X, the set {f(A) | A ∈ F} is not necessarily a filter on Y . This defect however is very easy to repair because f(F) is a filter basis on Y . We leave the proof to the reader. More generally if B is a filter basis on X, then f(B) is a filter basis on Y . We define f∗(F) to be the filter generated by f(F). We leave to the reader to convince himself (herself) that f∗ preserves ordering, and that ∗ is functorial. The proof of the lemma below we leave for the reader. Lemma 3a If f : X → Y is a function and F is a filter on X, then A ∈ f∗(F) if and only if f −1(A) ∈ F. The following is a very important property of an ultrafilter. Definition 3b We say that a subset U ⊆ P (X) has the ultrafilter property, if for arbitrary sets A and B, A ∪ B ∈ U if and only if A ∈ U or B ∈ U. Lemma 3c A filter U is an ultrafilter if and only if it has the ultrafilter property. From the last two Lemmas it follows that the push forward of an ultrafilter is an ultrafilter. Theorem 3d If f : X → Y is a function, and U is an ultrafilter on X, then f∗(U) is an ultrafilter on Y . 4. Convergence. We now look at filters on topological spaces A point x is called an accumulation point for a filter F, if every neighborhood of x meets every set in F. In other words Nbh(x) ∪ F is a filter basis. We write F for the set of accumulation points of the filter F. A point x in a space X is called a point for a filter F, or a limit for the filter F, if every neighborhood of x is a member of F. This means that Nbh(x) ⊆ F. We writex = lim(F) when x is a limit for F. There is a weakness to this notation because a filter may have more than one limit point. If X is given the , every point is a limit point for every filter. Here are the formal definitions. Let X be a topological space, x a point in X, and F a filter on X. Definition 4a x ∈ F if and only if there is a filter G finer than both Nbh(x) and F. Definition 4b x = lim(F) means that Nbh(x) ⊆ F. A filter that has a limit point is called convergent, and otherwise divergent. Lemma 4c If F is an ultrafilter every accumulation point is a limit point. Lemma 4d A point x is an accumulation point for a filter F, if and only if there is an ultrafilter U that dominates F, and converges to x. Lemma 4e In a Hausdorff space every convergent filter has exactly one limit point. Let x be a point in a space X. Let Conv(x) denote the set of all filters converging to x. All the filters in Conv(x) are finer than the neighborhood filter of x, and since Nbh(x) ∈ Conv(x), we have Nbh(x) = T Conv(x). This means that the topology of a space is determined by the convergent filters. Lemma 4f A function f : X → Y is continuous if and only if for every filter F and every point x ∈ X which is a limit for F, f(x) is a limit for f∗(F). The proof is left to the reader. The next Lemma is crucial in the proof of Tychonov’s Theorem.

2 Q Lemma 4g A filter F on a product space Y = i∈I Xi converges to x, if and only if each filter Fi = πi ∗(F) converges to xi = πi(x). Proof. The condition is necessary by Lemma 4f. To prove that it is sufficient we show that Nbh(x) ⊆ F. Let W be a neighborhood of x. Then there is a finite index set K ⊆ I T −1 and for each k ∈ K a neighborhood Vk of xk such that k∈K πk (Vk) ⊆ W . The fact that −1 lim(Fk) = xk, means that πk (Vk) ∈ F, and by the filter properties it follows that W ∈ F. 5. Characterizing compactness using filters. Lemma 5a A space is compact if and only if every filter has an accumulation point. Proof. From Lemmas’ 4c and 4d, it follows that Lemma 5a is equivalent to Lemma 5b, so we prove that instead. Lemma 5b A space is compact if and only if every ultrafilter is convergent. Proof. Suppose X is compact and let U be an ultrafilter on X. Let C = {A | A ∈ U}. Every finite sub family of C has a nonempty intersection, and since X is compact there is a point x ∈ T C. If V is a neighborhood of x, X = V ∪ X \ V ∈ U, so by the ultrafilter property, Lemma 3b, and since x 6∈ X \ V , we have V ∈ U. This shows that x = lim(U). To prove the converse let C be a family of closed subsets of X with the property that every finite sub family of C has nonempty intersection. In that case there is an ultrafilter U, with C ⊆ U. Let x be a limit for U, and let F ∈ C. Every neighborhood of x is in U and hence meets F . Since F is closed it follows that x ∈ F , and we have proved that x ∈ T C. Q Theorem 5c The products pace Y = i∈I Xi is compact if and only if all the Xi’s are compact. Proof. The only if part follows because the image of a compact by a continuous map is compact, and the projections are all continuous. The if part is a consequence of Lemmas’ 3d, 4g and 5b.

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