It’s been a long time, and I haven’t given any news from the Summer Topology Conference.  There I met Frédéric Mynard.  Frédéric stressed the importance of filters to me, and I should mention a few of the nice things one can do with them.

I’ve probably said so already, but in writing a book, you need to make choices.  One of the first decisions you have to make, beyond deciding what you want to put into the book, is to decide what you will not put into the book.  If you are not firm about that, then soon enough your book project will become a monster, something so big that you’ll never finish it, or which will become so expensive nobody will care.

One of the choices I had made early on was on the notion of convergence.  Well, I had in fact decided not to talk about convergence at all, thinking you could probabily do everything with opens only.  (That is true, but it is another story.)  But convergence is the starting point, if not the central point, of topology!

So I needed to talk about convergence.  Convergence of sequences is fine, but behaves well only on first-countable spaces.  Then we have two choices:

• Moore-Smith convergence (of nets)
• Convergence in terms of filters, due to Henri Cartan (well… everybody thinks H. Cartan invented them, but Frédéric tells me Leopold Vietoris has already used them, and before that, already, Giuseppe Peano).

In the book, I concentrated on the first.  Nets have a big advantage: it is easy to understand them, they look so much like sequences.  Sure, there are a few traps to avoid, such as the fact that you need subnets to be indexed by possibly different index sets, but globally, they’re nice and readable.  On the opposite, convergence in terms of filters looks arcane.  Of course, I first learned about filters from Bourbaki, and Bourbaki does not care to explain much of the intuition.  This may have helped grow myself a prejudice against filters.

By the way, I still did talk about filters, and they play a minor role throughout the book.  But I avoided them as convergence was concerned.

Let me repair this.

A filter (of subsets) on a space X is a collection F of subsets such that:

• if A is in F, and B is a larger subset, then B is in F as well
• if A and B are both in F, then their intersection is in F, too
• F is non-empty, or equivalently the whole space X is an element of F.

The only filters that matter are the non-trivial ones, and this additional requirement is sometimes included in the definition of filters.  A filter is non-trivial if and only if it does not contain all subsets; equivalently, if and only if the empty set is not in the filter.

There are two filters that one should keep in mind:

• given a point x in X, the collection N_x of all the neighborhoods of x is a non-trivial filter: this is the neighborhood filter of x
  
• given a net (x_i)_{i in I, ⊑}, the collection of all the subsets A such that x_i is eventually in A is also a non-trivial filter: let us call this the convergence filter of the net.

Now remember that (x_i)_{i in I, ⊑} converges to x if and only if every open subset U that contains x is such that x_i is eventually in U.  One obtains an equivalent definition by stating that every neighborhood A of x (i.e., in N_x) is such that x_i is eventually in A.  In other words, if and only if N_x is included in the convergence filter of the net.

This leads one to define convergence of filters, instead of nets, in a pretty natural (if abstruse) way:

A filter converges to a point x if and only if

it contains the neighborhood filter N_x of x.

And now, everything translates into the language of filters.  For example, the convergence filter of a subnet (x_{α (j)})_{j in J, ≤} of a net (x_i)_{i in I, ⊑} is a superset of the convergence filter of (x_i)_{i in I, ⊑}.  Yes, you are right… filters are ordered the other way around: to simulate taking subnets with filters, you have to take larger filters.

But what is the point?  I shall illustrate this through a few pearls.

Ultrafilters and why Kelley’s Theorem is complicated.

Look at the poset Filt(X) of all non-trivial filters of subsets of X.  We order filters by inclusion, and soon realize that Filt(X) is a dcpo.

So one can use Zorn’s Lemma: every non-trivial filter is contained in some maximal non-trivial filter.  Such maximal non-trivial filters are exactly the ultrafilters of opens of X.

Yes, those of Definition 4.7.34 of the book: the ultrafilters are those filters F such that for every subset A of X, either A or its complement is in F.  Note that every ultrafilter is non-trivial.  Conversely, no non-trivial filter can contain both a subset and its complement, since otherwise it would contain their (empty) intersection.  Every ultrafilter is maximal: if F is an ultrafilter, then we cannot add any new subset A to it: since A is new, it would not be in F, so its complement would be, and then adding A to F would produce a trivial filter.  Conversely, every maximal non-trivial filter is an ultrafilter.  Consider indeed any subset A of X that is not in F.  We must show that its complement is in F.  Build the collection F’ of all those subsets that contain some intersection A ∩ B of A with an element B of F.  This is a filter, and is strictly larger than F, because A is not already in F (take B=X).  Since F is a maximal non-trivial filter, F’ must be trivial, meaning that A ∩ B is empty for some element B of F: now the complement of A must be a superset of B, hence be in F.

That every non-trivial filter is contained in some ultrafilter is the filter-theoretic counterpart of Kelley’s Theorem 4.7.35, and is much easier to prove!

In fact, the reason why Kelley’s Theorem is hard to prove, as F. Mynard rightly notices, is that we would like to find an ultranet as some kind of maximal net, for the converse-of-subnet relation, but nets on a space form a proper class, not a set, so we cannot apply Zorn’s Lemma.

On the opposite, the collection of filters on a set is a set.  So the proof goes by converting a net intro a non-trivial filter, finding a maximal non-trivial filter containing it—this is an ultrafilter that is called a section filter of the net—, and finally converting back the ultrafilter into an ultranet.

Compactness.

One of the first successes in the history of filters was to give a short proof of Tychonoff’s Theorem.  As essentially every concept in point-set topology, compactness can be characterized through filters.

Theorem: A space X is compact if and only if every ultrafilter on X converges to a point.

Of course, this is the filter analogue of Exercise 4.7.36 on nets.

Proof. In one direction, assume X is compact, and let UF be an ultrafilter.  We must show that UF contains all the neighborhoods of some point.  If that were not true, then every point x of X would have a neighborhood A_x that is not in UF.  (We pick one for each x by using the Axiom of Choice, as usual.)  The interiors of A_x cover X, by construction, so by compactness one can find a finite set E of such points such that the interiors of A_x with x in E still cover X.  In particular, ⋃_{x in E} A_x=X, and by taking complements, the intersection of the complements of A_x, x in E, is empty.  Since UF is an ultrafilter, all these complements are in UF, so their intersection is again in UF, contradicting the fact that UF is non-trivial.

Conversely, assume that X is not compact.  We can find an open cover (U_i)_{i in I} with no finite subcover.  Consider the finite intersections of complements of sets U_i.  None is empty.  In particular, the collection of all supersets of such finite intersections is a non-trivial filter F.  By Zorn’s Lemma, one can find an even larger ultrafilter UF.  Assume UF converged to some point x.  Find i in I such that x is in U_i.  (This is possible since (U_i)_{i in I} is a cover.)  The complement of U_i is in F, hence in UF.  Since U_i is a neighborhood of x and UF converges to x, i.e., contains all the neighborhoods of x, U_i also is in UF: but its complement was also in UF, contradiction.  So UF does not converge.  QED.

We obtain directly:

Theorem (Tychonoff): every product of compact spaces is compact.

Proof.  Consider an ultrafilter UF on a product of compact spaces (X_i)_{i in I}.  For each i in I, let UF_i be the collection of subsets A of X_i such that π_i^-1 (A) is in UF.  One easily checks that UF_i is an ultrafilter, hence converges to a point x_i.  We choose one for each i, using the Axiom of Choice, and build the tuple (x_i)_{i in I}.  Finally, we check that UF converges to (x_i)_{i in I}: every neighborhood B of (x_i)_{i in I} contains an open rectangle ⋂_{i in J} π_i^-1 (A_i) for some finite subset J of I; as each A_i is a neighborhood of x_i and UF_i converges to x_i, A_i is in UF_i; by definition, π_i^-1 (A) is in UF, so the finite intersection ⋂_{i in J} π_i^-1 (A_i) is again in UF, and therefore also B.  QED.

In the above proof, UF_i is the image filter of UF by the continuous map π_i.  In general, the image of a filter F by a map f:X → Y is defined as f[F] = {B ⊆ Y | f^-1 (B) is in F}.  I’ll let you show that f is continuous at x if and only if for every filter F that converges to x, f[F] converges to f(x), hence that the continuous maps are those that preserve convergence of filters.

I’ll stop there for today, but I still have a few other things to say about filters and convergence.  This will be for another time.

— Jean Goubault-Larrecq (October 26th, 2013)