0
$\begingroup$

How to show that all normally open covers form a base for the fine uniformity $\mu_F$? If $\mathcal{B}$ is the collection of all normally open covers, we first need to show that $\mathcal{B}$ is a subcollection of $\mu_F$. Then we have to show that every $\mathcal{U}$ in $\mu_F$ is refined by some cover from $\mathcal{B}$. How to go about the proof?

$\endgroup$
  • $\begingroup$ I fleshed out the Willard proof some more. Time to accept / upvote some answers maybe? $\endgroup$ – Henno Brandsma Jul 14 '17 at 17:56
0
$\begingroup$

Following Willard, General Topology, 36.15 (which is what you seem to be doing as well):

some definitions recap: A sequence $(\mathcal{U}_n)_{n \ge 1}$ of covers of $X$ is said to be a normal sequence, when for all $n\ge 1$: $\mathcal{U}_{n+1}\prec^\ast \mathcal{U}_n$.

A cover $\mathcal{U}$ is said to be a normal cover, if there is a normal sequence $(\mathcal{U}_n)_{n \ge 1}$ as above with $\mathcal{U}_1 = \mathcal{U}$, so it can be star-refined as often we like.

Note that by the definition of covering uniformities, all covers in a covering uniformity are normal covers. Such covers need not be open and they often aren't.

A family of covers is called a normal family, if every member of the family is star-refined by some member of the family. The set of members of a normal sequence is a normal family. Any normal family of covers generates a unique smallest uniformity containing that family, and then this family is called a "subbase" for the generated uniformity.

An open cover $\mathcal{U}$ of $X$ is said to be "normally open" when there is a normal sequence $(\mathcal{U}_n)_{n \ge 1}$ of open covers with $\mathcal{U}_1 =\mathcal{U}$. Such a cover is clearly normal but in a special way, as it can be star-refined by open covers (instead of just covers).


We start in a uniformisable space $(X,\mu)$ with induced topology $\mathcal{T}_\mu$. Let $\mu_F$ be the corresponding "fine" uniformity, which is the largest (by inclusion) covering uniformity, that induces $\mathcal{T}_\mu$. We constructed it here

So let $\mathscr{B}$ be the collection of all normally open (in said topology) covers of $X$.

Now take some (fixed for now) normally open $\mathcal{U}$ from $\mathscr{B}$, and construct the promised normal sequence of open covers $(\mathcal{U}_n)_{n \ge 1}$ with $\mathcal{U}_1 = \mathcal{U}$. Then in this answer I showed that $\mu \cup \{\mathcal{U}_n \mid n \in \mathbb{N}\}$ is a normal family that induces a uniformity $\mu'$ such that $\mathcal{T}_{\mu'} = \mathcal{T}_\mu$. As $\mathcal{U} \in \mu'$ and $\mu'$ is a uniformity inducing $\mathcal{T}_\mu$, we also know $\mu' \subseteq \mu_F$ by maximality and so $\mathcal{U} \in \mu_F$.

We have shown (as $\mathcal{U} \in \mathscr{B}$ was arbitrary), that indeed $\mathscr{B} \subseteq \mu_F$.

Now we use the following

Fact: (e.g. Willard, General Topology; 36.7) For any covering uniformity $\mu$, the open uniform covers (i.e. open covers that happen to be members of $\mu$) form a base for $\mu$.

Proof sketch of Fact: let $\mathcal{U} \in \mu$ and let $\mathcal{V} \in \mu$ be such that $\mathcal{V} \prec^\ast \mathcal{U}$. Then note that $\mathcal{O}=\{\operatorname{st}(x,\mathcal{V}): x \in X\}$ is an open cover of $X$ (in the induced topology) (note that $\mathcal{V}\prec \mathcal{O}$, so that $\mathcal{O} \in \mu$, as required) that refines $\mathcal{U}$.

This Fact implies

Fact 2: every open cover in a uniformity $\mu$ is normally open.

Proof of Fact 2: Let $\mathcal{U} \in \mu$ be an open cover Proof: Define $\mathcal{U}_1 = \mathcal{U}$. Having defined $\mathcal{U}_n$ for some $n$, such that $\mathcal{U}_n$ is an open cover from $\mu$, let $\mathcal{V}$ be any cover in $\mu$ such that $\mathcal{V}\prec^\ast \mathcal{U}_n$, which can be done as $\mathcal{U}_n \in \mu$ and then by the above fact there is an open cover $\mathcal{O}$ in $\mu$ such that $\mathcal{O} \prec \mathcal{V}$. Standard facts about refinements learn us that:

$$\mathcal{O} \prec \mathcal{V} \prec^\ast \mathcal{U}_n \implies \mathcal{O} \prec^\ast \mathcal{U}_n$$

allowing us to go in the recursion by defining $\mathcal{U}_{n+1} = \mathcal{O}$ keeping everything open and in $\mu$, so we can continue. This recursively defined sequence shows that $\mathcal{U}$ is indeed normally open.

Now take any $\mathcal{U} \in \mu_F$. Then by the above fact, there is an open cover $\mathcal{O} \in \mu$ refining it. By Fact 2, it is normally open so a member of $\mathscr{B}$. This shows $\mathscr{B}$ is a base for $\mu_F$.

$\endgroup$

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.