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*of all the neighborhoods of_{x}*x*is a non-trivial filter: this is the*neighborhood filter*of*x*

- given a net (
*x*)_{i}_{i}_{ in }, the collection of all the subsets_{I, ⊑}*A*such that*x*is eventually in_{i}*A*is also a non-trivial filter: let us call this the*convergence filter*of the net.

Now remember that (*x _{i}*)

_{i}_{ in }

*converges to*

_{I, ⊑}*x*if and only if every open subset

*U*that contains

*x*is such that

*x*is eventually in

_{i}*U*. One obtains an equivalent definition by stating that every neighborhood

*A*of

*x*(i.e., in

*N*) is such that

_{x}*x*is eventually in

_{i}*A*. In other words, if and only if

*N*is included in the convergence filter of the net.

_{x}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 }

*of a net (*

_{J, ≤}*x*)

_{i}

_{i}_{ in }

*is a*

_{I, ⊑}*superset*of the convergence filter of (

*x*)

_{i}

_{i}_{ in }

*. Yes, you are right… filters are ordered the other way around: to simulate taking*

_{I, ⊑}*sub*nets 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*cover X, by construction, so by compactness one can find a finite set

_{x}*E*of such points such that the interiors of

*A*with

_{x}*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 }

*with no finite subcover. Consider the finite intersections of complements of sets*

_{I}*U*. None is empty. In particular, the collection of all supersets of such finite intersections is a non-trivial filter

_{i}*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*. (This is possible since (

_{i}*U*)

_{i}

_{i}_{ in }

*is a cover.) The complement of*

_{I}*U*is in

_{i}*F*, hence in

*UF*. Since

*U*is a neighborhood of

_{i}*x*and

*UF*converges to

*x*, i.e., contains all the neighborhoods of

*x*,

*U*also is in

_{i}*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 }

*. For each*

_{I}*i*in

*I*, let

*UF*be the collection of subsets

_{i}*A*of

*X*such that π

_{i}

_{i}^{-1}(

*A*) is in

*UF*. One easily checks that

*UF*is an ultrafilter, hence converges to a point

_{i}*x*. We choose one for each

_{i}*i*, using the Axiom of Choice, and build the tuple (

*x*)

_{i}

_{i}_{ in }

*. Finally, we check that*

_{I}*UF*converges to (

*x*)

_{i}

_{i}_{ in }

*: every neighborhood*

_{I}*B*of (

*x*)

_{i}

_{i}_{ in }

*contains an open rectangle ⋂*

_{I}

_{i}_{ in }

*π*

_{J}

_{i}^{-1}(

*A*

*) for some finite subset*

_{i}*J*of

*I*; as each

*A*

*is a neighborhood of*

_{i}*x*and

_{i}*UF*converges to

_{i}*x*,

_{i}*A*

*is in*

_{i}*UF*; by definition, π

_{i}

_{i}^{-1}(

*A*) is in

*UF*, so the finite intersection ⋂

_{i}_{ in }

*π*

_{J}

_{i}^{-1}(

*A*

*) is again in*

_{i}*UF*, and therefore also

*B*. QED.

In the above proof, *UF _{i}* is the image filter of

*UF*by the continuous map π

*. In general, the image of a filter*

_{i}*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)