# Thinking in terms of filters

In dealing with topological vector spaces it is more convenient to define a topology by specifying what the neighbourhoods of each point are going to be. It is well-known that the two approaches are equivalent: an open set will be a set, which, whenever it contains a point, contains a neighbourhood of this point; one can also say that an open set is a set which is a neighbourhood of each one of its points; on the other hand, a neighbourhood of a point x of E is simply a set which contains some open set containing x.

In order to define a topology by the system of the neighbourhoods of the points, it is convenient to use the notion of a filter. This is a very primitive notion, and the student should find it easy to become familiar with it, and to learn how to use filters, just as he learned how to use sequences. The notion of filter is perfectly independent of topology. A filter is given on a set which need not be carry any other structure. Let E be the set. A filter $\mathcal{F}$ is a family of subsets of E, submitted to three conditions:

($F_{1}$) The empty set $\phi$ should not belong to the family $\mathcal{F}$.

($F_{2}$) The intersection of any two sets, belonging to the family, also belongs to the family $\mathcal{F}$.

($F_{3}$) Any set, which contains a set belonging to $\mathcal{F}$ should also belong to $\mathcal{F}$.

The simplest example of a filter on a set E is the family of all subsets of E which contain a given subset A, provided the latter is non empty. With every infinite sequence of points of E is associated a filter. Let $x_{1}, x_{2}, \ldots$ be the sequence under consideration. The associated filter is the family of all subsets of E which have the following property:

(AF) The subset of E contains all elements $x_{1}, x_{2, \ldots}$ except possibly a finite number of them.

A family $\mathcal{B}$ of subsets of E is a basis of a filter on E if the following two conditions are satisfied:

($BF_{1}$) $\mathcal{B} \subset \mathcal{F}$ that is, any subset which belongs to $\mathcal{B}$ must belong to $\mathcal{F}$.

($BF_{2}$) Every subset of E belonging to $\mathcal{F}$ contains some subset of E which belongs to $\mathcal{B}$.

A familiar example of a basis of filter on the straight line is given by the family of all intervals $(-a,a)$ with $a>0$: it is a basis of the filter of the neighbourhoods of zero in the usual topology on the real line. Another useful example is the following one: Let $\mathcal{F}$ be the filter associated with a sequence $S = \{ x_{1}, x_{2}, \ldots, x_{n}\}$ For each $n=1,2,\ldots$ let us set $S_{n}=\{ x_{n}, x_{n+1}, \ldots\}$

and view $S_{n}$ as a subset of E.

Then the sequence of subsets $S = S_{1} \supset S_{2} \supset \ldots S_{n} \supset \ldots$ is a basis of $\mathcal{F}$.

Let $\mathcal{A}$ be some family of subsets of our set E. We may ask the question: is there a filter $\mathcal{F}$ having $\mathcal{A}$ as a basis (note that a filter can have several different bases) ? In view of the filter axioms, ($F_{1}, F_{2}, F_{3}$), that filter $\mathcal{F}$, if it exists, is completely and uniquely determined: it is the family of subsets of E which contains some subset belonging to $\mathcal{A}$. Observe that the latter property defines perfectly well a certain family, which we have called $\mathcal{F}$ of subsets of E. Then our question can be rephrased as follows: is $\mathcal{F}$ a filter? Obviously, $\mathcal{F}$ satisfies ($F_{3}$); it also satisfies ($F_{1}$) if we take care of requiring that no set belonging to $\mathcal{A}$ be the empty set. As for ($F_{2}$) it is equivalent as we see easily with the following property of $\mathcal{A}$:

(BF) The intersection of any two sets, belonging to $\mathcal{A}$ contains a set which belongs to $\mathcal{A}$.

The difference with condition ($F_{2}$) is that the intersection of two subsets which belong to $\mathcal{A}$ is not requested to belong to $\mathcal{A}$. but only to the contain some set belonging to $\mathcal{A}$. Thus, we may state : a basis of fliter on E is a family of nonempty subsets of E satisfying Condition (BF). The filter generated by the basis is uniquely determined by Condition ($BF_{2}$).

Next step: comparison of fliters. We want to be able to say: this filter is finer than this other filter. Keep in mind that filters are are sets of sets, or rather of subsets. In other words, filters are subsets of the set of subsets of of E, usually denoted by $\ss(E)$. As filters are subsets, (of some sets, in this case $\ss{E}$, there is a natural order relation among them: the inclusion relation. We can write $\mathcal{F} \subset \mathcal{F}^{'}$ if $\mathcal{F}$ and $\mathcal{F}^{'}$ are two filters on the same set E. It means that every subset of E which belongs to $\mathcal{F}$ also belongs to $\mathcal{F}^{'}$ (the converse being, in general, false). Instead of saying that $\mathcal{F}$ is contained in $\mathcal{F}^{'}$, one usually says that $\mathcal{F}^{'}$ is finer than $\mathcal{F}$ or that $\mathcal{F}$ is less fine than $\mathcal{F}^{'}$. Let $\mathcal{F}$ (respectively $\mathcal{F}^{'}$) be the family of all subsets of E which contains a given subset A (respectively $A^{'}$) of E; $\mathcal{F}^{'}$ is finer than $\mathcal{\mathcal{F}}$ if and only if $A^{'} \subset A$.

A topology on the set E is the assignment, to each point x of E, of a filter $\mathcal{F}(x)$ on E, with the additional requirement that the following two conditions be satisfied:

($N_{1}$) If a set belongs to $\mathcal{F}(x)$, it contains the point x.

($N_{2}$) If a set U belongs to $\mathcal{F}(x)$, there is another set $V$ belonging also to $\mathcal{F}(x)$ such that given any point $y$ of V, U belongs to $\mathcal{F}(y)$.

When these conditions are satisfied, we say that we have a topology on E and we call $\mathcal{F}(x)$ the filter of neighbourhoods of the point x. At frst sight Condition $(N_{2})$ may seem involved. It expresses, however, a very intuitive fact. Roughly speaking, it says that given any point z near x (that is, z is generic element of U), if a third point y lies sufficiently near to x (the sufficiently near is made precise by the neighbourhood of x, V, of which y is an element), then z lies near to y (that is, $z \in U \in \mathcal{F}(y)$). In the language of open sets $(N_{2})$ becomes evident: since U is a neighbourhood of x, U contains an open set containing x; let V be such an open set. Since V is open, and $V \subset U$; U is obviously a neighbourhood of each point of V. A basis of the filter $\mathcal{F}(x)$ is called a basis of neighbourhoods of x. This simple notion will play an important role in the forthcoming definitions.

Once we have the notion of filter of neighbourhoods of a point, hence of neighbourhood of a point (any subset of E belonging to the filter of neighbourhoods), we can quickly review the concepts that are used to describe a topology. As we have already said, an open set is a set which is a neighbourhood of each one of its points. A subset of E is closed if its complement is open. The closure of a set $A \subset E$ is the smallest closed set containing A. It will be denoted by $\overline{A}$. The following is easy to check: a point belongs to $\overline{A}$ if and only if everyone of its neighbourhoods meets $A$ (that is, has a nonempty intersection with A). The interior of a set is the largest open set contained in it; if A is the set, its interior will be denoted by $\AA$.

A very important notion is the one of a set A dense in another set B; both A and B are subsets of the same topological space E. Then, one says that A is dense in B if the closure $\overline{A}$ of A contains B. In particular, A is said to be dense in E (or everywhere dense) if $\overline{A}=E$. To say that A is dense in B means that given any neighbourhood of any point x of B, $U(x)$, there is a point y of A which belongs to $U(x)$, that is, $A \bigcap U(x) \neq 0$. A standard example of a set everywhere dense is the set of rational numbers Q, when regarded as a subset of the real line R (with the usual topology); note that the complement R-Q of Q is also dense in R. Examples of sets which are dense and open are given by the complement of a straight line in the plane or in space, by the complements of a plane in space, etc. Easy to check are the basic intersection and union properties about open or closed sets: that the intersection of a finite number of open sets is open (this follows immediately from the fact, itself obvious in virtue of Axiom $F_{2}$), that the intersection of a finite number of neighbourhoods of a point is again a neighbourhood of that point); that the union of any number of open sets, be that number finite or infinite, is open (this follows from the fact that the union of a neighbourhood of a point with an arbitrary set is a neighbourhood of the same point: Axiom $F_{3}$). Byy going to the complements, one concludes that finite unions of closed sets are closed, arbitrary intersections of closed sets are also closed, etc

Observe that a set E may very well carry severall different topologies. When dealing with topological vector spaces, we shall very often encounter this situation of a set, in fact a vector space, carrying several topologies (all compatible with the linear structure, in a sense that is going to be specified soon). For instance, any set may carry the following two topologies (which in practice are almost never used):

(1) the trivial topology: every point of E has only one neighbourhood, the set E itself; (note that in this case every point of E is a limit point and so there are simply too many points !)

(2) the discrete topology given any point x of E, every subset of E is a neighbourhood of x provided that it contains x; in particular, {x} is a neighbourhood of x, and constitutes in fact a basis of the filter of neighbourhoods of x. (note that in this case no point of E is a limit point of any subset of E.)

We may compare topologies, in analogy with the way we have compared filters. Let Let $\mathcal{T}, \mathcal{T}^{'}$ be two topologies on the same set E. We say that $\mathcal{T}$ is finer than $\mathcal{T}^{'}$ if every subset of E which is open for $\mathcal{T}^{'}$ is also open for $\mathcal{T}$, or equivalent, if every subset of E which is a neighbourhood of a point for $\mathcal{T}^{'}$ is also a neighbourhood of that same point for the topology $\mathcal{T}$. Let $\mathcal{F}(x)$ (respectively $\mathcal{F}^{'}(x)$) be the filter of neighbourhoods of an arbitrary point x of E in the topology $\mathcal{T}$ (respectively $\mathcal{T}^{'}$) : $\mathcal{T}$ is finer than $\mathcal{T}^{'}$, which we shorten into $\mathcal{T} \geq \mathcal{T}^{'}$, if, for every $x \in E$, $\mathcal{F}(x)$ is finer than $\mathcal{F}^{'}(x)$. Given two topologies on the same set, it may very well happen that none is finer than the other. If one is finer than the other, one says sometimes that they are comparable. The discrete topology is finer on a set E than any other topology on E; the trivial topology is less fine than all the others. Topologies on a set form thus a partially ordered set, having a maximal and a minimal element, respectively the discrete and the trivial topology.

The notion of a topology has been introduced in order to provide a solid ground for the notions of convergence and of continuity. Of course, the latter were correctly manipulated (or most of the time, at least) well before anybody thought of topology. We proceed now to give their general definition.

CONVERGENCE.

This concerns filters: filters are the “objects” which may (or may not) converge. When do we say that a filter $\mathcal{F}$ on a topological space E converges? We should recall that $\mathcal{F}$ is a family of subsets of E. If $\mathcal{F}$ is to converge to a point x of E, it means that elements of $\mathcal{F}$, which we repeat again, are subsets of E, get “smaller and smaller” about x, and that points of these subsets get “nearer and nearer” to x. This can be made precise in terms of neighbourhoods of x, which we have at our disposal, since E is a topological space: we must express the fact that, however small a neighbourhood of x is, it should contain some subset of E belonging to the filter $\mathcal{F}$ and consequently, all the elements of $\mathcal{F}$ which are contained in that particular one. But in view of Axiom ($F_{3}$), this means that the neighbourhood of x under consideration must itself belong to the filter $\mathcal{F}$, since it must contain some element of $\mathcal{F}$. The phrase “however small a neighbourhood of x is” has to be made mathematically meaningful: it simply means “whatever is the neighbourhood of x.” In brief, we see that the filter $\mathcal{F}$ converges to the point x if every neighbourhood of x belongs to $\mathcal{F}$, in other words, if $\mathcal{F}$ if finer than the filter of neighbourhoods of x, $\mathcal{F}(x)$. This is what the convergence to a point of a filter means.

We recall how the convergence of a sequence to a point is defined. Let $S = \{ x_{1}, x_{2}, \ldots \}$ be the sequence. We say that S converges to x if, given an arbitrary neighbourhood U of x, there is an integer n(U) such that $n \geq n(U)$ implies $x_{n} \in U$. Let $S = S_{1} \supset S_{2} \supset S_{3} \supset \ldots S_{n} \ldots$ be the subsequences introduced earlier: S converges to x if to every $U \in \mathcal{F}(x)$ there is an integer n(U) such that $S_{n(U)} \subset U$. As the subsets of $S_{n}$ of E form a basis of the filter associated with the sequence S, we see immediately that a sequence S converges to x if and only if the associated filter converges to x.

Note that a filter may converge to several different points. Suppose, for instance, that E carries the trivial topology: then every filter on E converges to every point of E. Note also that a filter may not converge: for instance, if it is the filter associated with some sequence and if this sequence does not converge. Another example is given by a filter on E which is not the filter of all subsets of E which contain a given point x — when E carries the discrete topology: in this topology, the only converging filters are the filters of neighbourhoods of the points. So much for convergence in general topological spaces.

CONTINUITY.

This concerns mappings. In point set topology, a map $f: E \rightarrow F$, this is to say a map from a topological space E into another topological space F, is said to be continuous if any one of the following two conditions is satisfied:

(a) given any point x of E and any neighbourhood V of the image $f(x) \in F$ of x, the preimage of V, that is to say the set

$f^{-1}(V) = \{ x \in E: f(x) \in V\}$

is a neighbourhood of x. In short,

$\forall {x} \in E$, $V \in \mathcal{F}(f(x))$ implies $f^{-1}(V) \in \mathcal{F}(x)$.

(b) the preimage of any open subset $\mathcal{O}$ of F, $f^{-1}(\mathcal{O}) = \{ x \in E: f(x) \in \mathcal{O}\}$ is an open subset of E.

The student can easily check the equivalence of (a) and (b). As for the intuitive meaning of these conditions, we may say the following. If the mapping f is to be continuous at the point x, it should mean that if $x^{'} \in E$ “converges to x”, then $f(x^{'})$ should converge to f(x). Note that “$f(x^{'}) converges to f(x)$” can be made precise in the following way: given an arbitrary neighbourhood of f(x), f(x) should eventually belong to it; and the “eventually” means here: provided that $x^{'}$ is sufficiently near to x. Thus, given an arbitrary neighbourhood V of f(x), if $x^{'}$ belongs to a sufficiently small neighbourhood of x, then $f(x^{'}) \in V$. The “sufficiently small” can only be determined by the existence of a certain neighbourhood U of x, such that, as soon as $x^{'} \in U$ then $f(x^{'}) \in V$. This is exactly property (a); to every neighbourhood V of f(x) there is a neighbourhood U of x such that

$x^{'} \in U$ implies $f(x^{'}) \in V$.

It is immediately seen that if a sequence $\{ x_{1}, x_{2}, \ldots\}$ converges in E to a point x, and if f is a continuous function from E into F, then the sequence $\{ f(x_{1}), f(x_{2}), \ldots\}$ converges to f(x) in F. Convergence of filters is also easily related to continuity of mappings. Let

$f: E \rightarrow F$

be a mapping from a set E into a set F. Let $\mathcal{F}$ be a filter on E. The image $f\mathcal{F}$ of $\mathcal{F}$ under f is defined as being the filter having the basis

$(f\mathcal{F})_{0} = \{ (f(U) \in F); U \in \mathcal{F}\}$

Observe that, in general, $(f\mathcal{F})_{0}$ is not itself a filter; it is always the basis of a filter (this is left as an exercise). Now, if the filter $\mathcal{F}$ converges to a point x in E and if f is a continuous function, then $f\mathcal{F}$ converges to f(x) in F. Indeed, the continuity of f implies that $f\mathcal{F}(x)$ is finer than $\mathcal{F}((x))$; this is simply a restatement of Property (a) above. If then $\mathcal{F}$ is finer than $\mathcal{F}(x)$ (which means that $\mathcal{F}$ converges to x), $f\mathcal{F}$ is finer than $f\mathcal{F}(x)$ and a fortiori finer than $\mathcal{F}(f(x))$.

We have only considered continuous funtions, which is to say functions defined everywhere and continuous everywhere. Of course, one may prefer to talk about functions continuous at a point. This is defined by the condition (where x is the point under consideration):

for every $V \in \mathcal{F}(f(x))$ then $f^{-1}(V)$ belongs to $\mathcal{F}(x)$,

or equivalently,

$f\mathcal{F}(x) \geq \mathcal{F}(f(x))$.

Let us insist on the fact that all the functions or mapping which will be considered in this book/blog (the series of topological vector spaces) are defined everywhere.

As a last remark, let us consider the case where F is identical with E as a set, but carries a different topology from the one given on E, and where f is the identity mapping of E onto F, I. The following two properties are obviously equivalent:

(i) $I: E \rightarrow F$ is continuous.

(ii) the topology of E is finer than the topology of F (these two topologies are defined on the same set).

Cheers,

Nalin Pithwa.

Reference: Topological Vector Spaces, Distributions and Kernels. Francois Treves. Dover Publications.

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