# General Topology/Uniform spaces

**Definition (uniform structure)**:

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

- , being the
*diagonal*of - , where
- , 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.

- Since for all entourages (),
- Due to , is closed under finite intersections
- is closed under supersets by definition
- 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.

**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]

- 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 .
- 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 .