General Topology/Uniform spaces

From Wikibooks, open books for an open world
Jump to navigation Jump to search

Definition (uniform structure):

Let be a set. A uniform structure on is a filter of such that

  1. , being the diagonal of
  2. , where
  3. , where for general

Definition (uniform space):

A uniform space is a set together with a uniform structure on it.

In this definition, if are contained in a sufficiently small entourage, they are considered "close" to each other. That is, a uniform structure provides a means of determining when two arbitrary points are close. This is the intuition behind this definition. A very important special case of a uniform space are metric spaces, which we'll learn about in the next chapter. Uniform spaces are a generalisation of metric spaces, and many of the notions and theorems carry over from metric spaces to uniform spaces, and we'll immediately treat them in full generality.

Definition (entourage):

Let be a set with a uniform structure . An entourage is simply an element of .

Definition (entourage-induced neighbourhood):

Let be a uniform space and an entourage of . For , define


A uniform structure induces a topology on its space.

Proposition (uniform structure induces topology):

Let be a space with a uniform structure . Then there exists a unique topology on so that a basis for each neighbourhood filter of points is given by as ranges over all entourages.

Proof: Define to be the filter generated by the sets as ranges over all entourages, and observe that indeed these sets are a filter subbasis, since they all contain , so that the characterisation of filter subbases is applicable. Claim that satisfies 1.-4. of the characterisation of a topology by its neighbourhoods.

  1. Since for all entourages (),
  2. Due to , is closed under finite intersections
  3. is closed under supersets by definition
  4. Let . Pick so that and then so that . For all , we have , so that is in

Henceforth, we shall consider a uniform space as a topological space with this topology.

Definition (V-small):

Let be a uniform space and let be an entourage of . A subset is called -small iff for all , we have .

Proposition (every uniform space is regular):

Let be a uniform space (with uniformity ). Then is regular.

Proof: Let be closed and . Since is closed, is an open neighbourhood of . Hence, pick a symmetric entourage s.t. , and then another symmetric entourage s.t. . Then and

are disjoint, since otherwise, if , there is s.t. and also , so that , a contradiction to .

Definition (Cauchy filter):

Let be a uniform space. A Cauchy filter is a filter on such that for any entourage of , there exists such that , that is, is -small.

Definition (completeness):

Let be a uniform space. is called complete iff each Cauchy filter on converges to some point in .

Definition (total boundedness):

Let be a uniform space, and let be its entourage filter. is totally bounded if and only if for each , there exist finitely many points so that


The following is a generalisation of the Heine–Borel theorem.

Theorem (compact iff totally bounded and complete):

Let be a uniform space, and a subset. is compact if and only if it is totally bounded and complete.

(On the condition of the ultrafilter lemma.)

Proof: Suppose first that is compact, and let be an arbitrary entourage. Note that is an open cover of , so that we may choose a finite subcover in order to achieve total boundedness. Let then be a Cauchy filter of subsets of , and suppose that does not converge to any point of . For each , select a nonempty set of entourages sufficiently small so that for all , and then sufficiently small so that . Then choose by compactness a finite subcover (where for , and define

, .

Since is a Cauchy filter, it will contain a -small set . Then pick arbitrary, and so that . Then we will have for , that and , so that , that is, , and we conclude that and , a contradiction.

Suppose now that is not compact. By the characterisation of compactness by filter convergence, pick a filter on which does not admit a refinement that converges to a point of (note that this does not use the axiom of choice). By the ultrafilter lemma, pick a maximal filter that contains . Upon proving that is Cauchy, we obtain a contradiction, since Cauchy filters converge in as is complete. Let hence be any entourage of , and pick so that . Then is -small for all . By definition of the subspace topology and since is totally bounded, pick so that . Suppose that for all , there existed so that . Then set and observe that for all , and taking the union over all we get that , a contradiction to being a filter. Hence, pick so that for all and observe that , for otherwise we could properly extend by extending by . But is -small, so that is Cauchy.

Definition (uniform continuity):

Let be uniform spaces with uniform structures resp. . A function is said to be uniformly continuous if and only if


Proposition (uniform continuity implies continuity):

Let be a uniformly continuous function, where is the uniform structure of and is the uniform structure of . Then is continuous.

Proof: Let be a filter of that converges to a point , so that . Let be the filter on that is generated by , and let be a neighbourhood of . By definition of the topology on induced by the uniform structure, pick so that . By uniform continuity, pick so that . Then , so that , but for we have so that , and we get .

Definition (fundamental system of entourages):

Let be a uniform space with uniform structure . A fundamental system of entourages is a filter basis for .

Proposition (inverse image of uniform structure generates a uniform structure):

Let be a set, let be a uniform space (with uniform structure ) and let be a function. Then

is a filter basis for a uniform structure on

Proof: First note that whenever , then contains the diagonal, since we have for a certain , that is,


and clearly, for , we have . Therefore, every also contains the diagonal (as it contains a set of ). Further, is a filter base, since taking preimages commutes with intersections, and is closed under finite intersections, being a filter itself. Then let , and pick so that . Pick so that . Then pick so that . We claim that if we set , then (). Indeed, if and , then , so that . Finally, .

Proposition (least upper bound of uniform structures):

Let be a set and let be a family of uniform structures on . Then there exists a (unique) least upper bound uniform structure on of the 's, namely the filter generated by


and the topology induced by it coincides with the least upper bound topology of the topologies on that are induced by the 's.

Proof: First we claim that as given above is a uniform structure. Indeed, is closed under finite intersections and hence forms a filter base. Further, suppose that , and pick and in resp. resp. ... resp. . Then for each , observe for one that (so that ), and then pick so that . Then .

Definition (initial uniform structure):

Let be a set, let be a family of uniform spaces with uniform structures , and let be functions. The initial uniform structure on is defined to be the least upper bound uniform structure induced by the uniform structures .

Proposition (separation of compact subsets of open sets by an entourage):

Let be a uniform space with uniformity . Let be open, and let be compact. Then there exists a symmetric entourage such that , where


Proof: For each , pick a symmetric entourage such that (passing to the union so as to avoid the axiom of choice) and then a symmetric entourage such that .

Proposition (the topology induced by a final uniform structure coincides with the respective final topology):

Let be a set, and let be a collection of uniform spaces. Suppose further that we are given a function for each . Then the final topology on that is induced by the functions coincides with the topology that is induced by the final uniform structure induced by the functions .

Proof: It suffices to show that the neighbourhood filters of the two topologies coincide. Hence, we select an arbitrary point . Suppose first that is an enclosure of with respect to the final uniform structure induced by the s.

Exercises[edit | edit source]

  1. Let be a set, and let and be uniform structures on so that they generate the same topology and is compact with respect to . Prove that in fact .
  2. Let be a topological space whose topology is induced by both of the two uniform structures and . Suppose that is complete with respect to the uniform structure induced by . Show that is complete with respect to the uniform structure induced by .