Base of Fine Uniformity

Question

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?

Answer 1

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).

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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 $\mu_F$.