The following are three definitions of a totally bounded uniform spaces on a set $U$:

  1. For every entourage $E$ there exists a finite cover $S$ of $U$ such that $\forall A\in S:A\times A\subseteq E$.
  2. For every entourage $E$ there exists a finite set $B$ such that $E[B]=U$ (here $E[B]$ is the image of the set $B$ by the relation $E$).
  3. Every proper filter on $U$ contains a Cauchy filter.

I know the proof that $(1)\Leftrightarrow(2)$ (see my book).

But how to prove equivalence with the third (for any uniform spaces)?

  • 1
    $\begingroup$ Side remark: In section 1.4 in your book your motivation to study funcoids instead of topological spaces is very poor. That way you won't gain any readers. Instead, you should answer the following questions: What are, specifically, the drawbacks of topological spaces? Why are they ugly, as you claim? And in how far are funcoids better? $\endgroup$ – Martin Brandenburg Feb 25 '14 at 14:36
  • $\begingroup$ @MartinBrandenburg: It is an advise which calls for a good thing, but is difficult to accomplish. If you want help me to rewrite my book, become a co-author and write yourself. I have no better idea on how to accomplish your advise $\endgroup$ – porton Feb 25 '14 at 15:02
  • 2
    $\begingroup$ No offense, but when you cannot answer these questions, the whole theory has no motivation and others will regard it as useless. You have observed this quite a lot on mathoverflow. $\endgroup$ – Martin Brandenburg Feb 25 '14 at 20:05

Point 3. should be formulated as: "every filter on $U$ has a Cauchy refinement", where a refinement of a filter is a larger filter, a superset.

Suppose $U$ is a totally bounded uniform space. Let $\mathcal{F}$ be a proper filter on $U$ and let $\mathcal{F}'$ be a refining ultrafilter. I claim $\mathcal{F}'$ is Cauchy:

let $D$ be an entourage of $U$ and let $E$ be a symmetric entourage of $U$ such that $E \circ E \subset D$. Then there exists a finite subset $F$ of $U$ such that $E[F] = U$, by total boundedness. Then each of the sets $E[x], x \in F$ is $D$-small and they form a finite cover of $U$, so one of them is in the ultrafilter $\mathcal{F}'$, as required. So $\mathcal{F}'$ is Cauchy.

On the other hand, suppose that 3. is true, and $U$ is not totally bounded. The latter means that there exists an entourage $D$ such that $X \setminus D[F]$ is non-empty, for every finite subset $F$ of $U$. All such sets $D[F]$ are closed under finite unions, so all sets $X \setminus D[F]$ form a filter base. Let $\mathcal{F}$ be the filter generated by this base, and by assumption $\mathcal{F}'$ is a Cauchy filter that contains $\mathcal{F}$.

Let $M$ be a $D$-small (for this $D$) member of $\mathcal{F}'$. Let $F$ be a finite subset of $U$. Then $X \setminus D[F] \in \mathcal{F} \subset \mathcal{F}'$ and so $M$ intersects this set, say in $x$. Then $M \subset D[x]$ (as $M \times M \subset D$, as $M$ is $D$-small) and so $M \subset D[F \cup \{x\}]$. But this means that $M \in \mathcal{F}'$ does not intersect $X \setminus D[F \cup \{x\}] \in \mathcal{F} \subset \mathcal{F}'$, which is a contradiction. This shows that $U$ is totally bounded.

  • $\begingroup$ I don't understand why $M \subset D[x]$ $\endgroup$ – porton Feb 27 '14 at 20:53
  • $\begingroup$ If $p \in M$ is arbitary, $(x,p) \in M \times M \subset D$, and $(x,p) \in D$ means $p \in D[x]$. $\endgroup$ – Henno Brandsma Feb 27 '14 at 21:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.