Les espaces métriques sont des « espaces uniformes » de nature par-ticulière; les espaces uniformes n’ont été définis de manière générale qu’en 1937, par A. Weil (XI). Auparavant on ne savait utiliser les notions et les résultats relatifs à « structure uniforme » que lor-squ’il s’agissait d’espaces métriques: ce qui explique le rôle import-ant joué dans beaucoup de travaux sur la topologie, par les espaces métriques ou métrisables (et en particulier par les espaces compact métrisables) dans des questions où la distance n’est d’aucune utilité véritable.
Nicolas Bourbaki [Bou06, TG II.43]
The basic reference on uniform spaces is due to Bourbaki [Bou06] and we follow their terminology. A recent account on this topic is provided by Page [Pag11].
Every metric space (𝒳, 𝑑)has an associated uniform structure generated by the entourages of the form {(𝑥, 𝑦) ∈ 𝒳 × 𝒳 ∣ 𝑑(𝑥, 𝑦) < 𝜀} for some 𝜀 > 0.
Very importantly, any compact Hausdorff topological space admits a unique uniform structure that agrees with its topology [Bou06, Theorem 1, II.27].
In particular, every Boolean space, i.e. compact Hausdorff zero-dimensional space, is unambiguously seen as a uniform space.
In this section, we focus on those uniform spaces which arise as a uniform subspace of some Boolean space. We prove a simple characterisation of uniform continuity between such uniform spaces, which is reminiscent of topological continuity. This characterisation is of crucial importance in the next section.
The following notion simplifies greatly the study of these uniform subspaces.
Definition 3.12. Let𝑆 be a subset of a Boolean space𝒳. A subset𝐵of 𝑆 is called a block of 𝑆 (relatively to 𝒳) if there exists a clopen 𝐶 of 𝒳 such that 𝐵 = 𝐶 ∩ 𝑆. We write Blocks(𝑆) for the Boolean subalgebra of 𝒫(𝑆)of blocks of 𝑆.
The uniform structure of a uniform subspace of a Boolean space 𝒳is essen-tially given by its blocks as we show in Lemma 3.17 below (see also [Bou06,
Exercice 12, II.38]). As a consequence, uniform continuity between such spaces admits of the following simple characterisation.
Proposition 3.13. Let 𝒳 and 𝒴 be two Boolean spaces, and let 𝑆 ⊆ 𝒳 and 𝑇 ⊆ 𝒴 be endowed with the induced uniform structure. Then a function 𝑓 ∶ 𝑆 → 𝑇 is uniformly continuous if and only if for all 𝐵 ∈ Blocks(𝑇 ) we have 𝑓−1(𝐵) ∈ Blocks(𝑆).
When a subset𝑆 of a Boolean space𝒳is endowed with the induced uniform structure, then the completion of the uniform space𝑆 coincide with the closure 𝑆 of 𝑆 in 𝒳1. Therefore a function 𝑓 ∶ 𝑆 → 𝑇 as in Proposition 3.13 is uniformly continuous if and only if there exists a unique continuous map 𝑓 ∶ 𝑆 → 𝑇 such that𝑓↾𝑆 = 𝑓.
Although Proposition 3.13 is folklore, we did not find any reference for this very statement (see however the work of Gehrke, Grigorieff, and Pin [GGP10]
in relation with automata theory and the one by Erné [Ern01]). To keep our exposition self-contained we now provide the reader with a series of lemmas that lead to a proof of this fact.
Lemma 3.14. Let 𝒳 be a Boolean space. The unique compatible uniform structure on 𝒳 admits
(i) as a base the entourages of the form 𝑈(𝐶𝑖) = ⋃𝑖𝐶𝑖× 𝐶𝑖 where (𝐶𝑖) is a finite partition of 𝒳into clopen sets;
(ii) as a subbase the entourages of the form𝑈𝐶 = (𝐶 × 𝐶) ∪ (𝒳 ∖ 𝐶 × 𝒳 ∖ 𝐶) where 𝐶 is a clopen set of 𝒳.
Proof. Since𝒳is compact and Hausdorff, the unique uniform structure which is compatible with the topology of 𝒳 consists of the neighbourhoods of the diagonal of 𝒳 ([Bou06, Theorem 1, II.27]). Clearly each 𝑈(𝐶𝑖) with (𝐶𝑖) a clopen partition of 𝒳 is a clopen neighbourhood of the diagonal. Conversely let 𝑈 be a neighbourhood of the diagonal of 𝒳. Since 𝒳 is zero-dimensional, there exists for each𝑥 ∈ 𝒳a clopen𝐶𝑥 such that𝑥 ∈ 𝐶𝑥and𝐶𝑥×𝐶𝑥 ⊆ 𝑈. By compactness of 𝒳, there exist 𝑥1, … , 𝑥𝑛 such that 𝐶𝑥
1, … , 𝐶𝑥
𝑛 covers 𝒳. We can then construct a partition 𝐶1, … , 𝐶𝑛 of 𝒳in clopen such that 𝐶𝑖 ⊆ 𝐶𝑥
𝑖. It follows that 𝑈(𝐶𝑖) ⊆ 𝑈 which concludes the proof.
Lemma 3.15. Let 𝒳be a Boolean space and let 𝐹 be a closed subspace of 𝒳.
Then the clopen sets of 𝐹 coincide with the blocks of 𝐹.
1The space𝑆is also the Stone dual of the Boolean algebraBlocks(𝑆), cf. Subsection4.1.3.
Proof. By definition, if 𝐵 is a block of 𝐹 then 𝐵 is clopen in 𝐹. Conversely, let 𝐶 be clopen in 𝐹. The blocks of 𝐹 form a clopen base of 𝐹, and since 𝐶 is open, it is a union of a family ℬ of blocks of𝐹. Since 𝐹 is compact and 𝐶 closed in 𝐹, there is a finite subset ℬ′ of ℬ for which 𝐶 = ⋃ ℬ′. Since the blocks of 𝐹 form a Boolean algebra, it follows that 𝐶 is a block.
Lemma 3.16. Let 𝒳be a Boolean space, and let 𝑆 ⊆ 𝑇 ⊆ 𝒳. Then Blocks(𝑆) = {𝐵 ∩ 𝑆 ∣ 𝐵 ∈ Blocks(𝑇 )}.
Proof. Let 𝐵 ∈ Blocks(𝑇 ) and let 𝐶 be clopen in 𝒳 such that 𝐵 = 𝐶 ∩ 𝑇. Then 𝐵 ∩ 𝑆 = 𝐶 ∩ 𝑆 is a block of 𝑆. Conversely, if 𝐵 is a block of 𝑆 then there is𝐶 clopen in 𝒳with 𝐵 = 𝐶 ∩ 𝑆. It follows that𝐵 = (𝐶 ∩ 𝑇 ) ∩ 𝑆 where 𝐶 ∩ 𝑇 ∈ Blocks(𝑇 ).
Lemma 3.17. Let 𝒳 be a Boolean space and let𝑆 be a subset of𝒳. Then the uniform structure induced on 𝑆 by𝒳 admits
(i) as a base the entourages of the form 𝑈(𝐶
𝑖) = ⋃
𝑖𝐶𝑖× 𝐶𝑖 where (𝐶𝑖) is a finite partition of 𝑆 into blocks;
(ii) as a subbase the entourages of the form𝑈𝐶 = (𝐶 × 𝐶) ∪ (𝑆 ∖ 𝐶 × 𝑆 ∖ 𝐶) where 𝐶 is a block of 𝑆.
Proof. Applying Lemmas 3.14 and 3.15 we obtain that the uniform subspace 𝑆 admits as a subbase the entourages of the form 𝑈𝐶 where 𝐶 is a block of 𝑆.
Moreover, the uniform subspace 𝑆 of 𝒳 is identical to the uniform subspace 𝑆 of 𝑆. It follows that entourages of the form (𝑆 × 𝑆) ∩ 𝑈𝐶 where 𝐶 is a block of 𝑆 constitute a subbase for the uniform space 𝑆. On the one hand (𝑆 × 𝑆) ∩ 𝑈𝐶 = {(𝑠, 𝑡) ∈ 𝑆 × 𝑆 ∣ 𝑠 ∈ 𝐶 ↔ 𝑡 ∈ 𝐶} = 𝑈𝐶∩𝑆. On the other hand, {𝐶 ∩ 𝑆 ∣ 𝐶 ∈ Blocks(𝑆)} = Blocks(𝑆) by Lemma 3.16. We have thus obtained that the entourages of the form𝑈𝐵 where 𝐵 ∈ Blocks(𝑆) constitute a subbase for the uniform space 𝑆.
Proof of Proposition 3.13. ⇒: Suppose 𝑓 is uniformly continuous and let 𝑓 ∶̂ 𝑆 → 𝑇 be its continuous extension. Then for all clopen𝐶 of𝒴the set𝑓−1(𝐶 ∩ 𝑇 ) =𝑓−1̂ (𝐶) ∩ 𝑆 is a block of𝑆.
⇐: Suppose 𝑓 ∶ 𝑆 → 𝑇 preserves blocks by preimage. By Lemma 3.17, it is enough to show that for each block 𝐵of 𝑇 the preimage of 𝑈𝐵 by𝑓 × 𝑓 is an entourage of𝑆. In fact, (𝑓 × 𝑓)−1(𝑈𝐵) = 𝑈𝑓−1(𝐵).