Let $X$ be a set and $\mathcal{U}$ a cover of this set.
A preuniformity $\mu$ on $X$ is a family of covers with the following conditions satisfied:
- $\{X\}\in \mu$
- Any cover in $\mu$ has a star-refinement
- Any two covers in $\mu$ have a common refinement
- A covering, of which refinement is in $\mu$, is also in $\mu$
A normal sequence is a sequence of coverings $\mathcal{U}_n$ such that $\mathcal{U}_{n+1}$ is a star-refinement of $\mathcal{U}_n$.
Under which conditions there is a smallest preuniformity $\mu$ on $X$ with $\mathcal{U}\in\mu$?
Let $\mathcal{U}_n$ be a normal sequence with $\mathcal{U}_1 = \mathcal{U}$. Then $\mu[\mathcal{U}_n] = \{\mathcal{V}:\mathcal{U}_n\text{ refines }\mathcal{V}\}$ is a preuniformity containing $\mathcal{U}$. Conversely, if a smallest preuniformity exists then it must be of this form. Our task is then equivalent to finding a normal sequence $\mathcal{U}_n$, $\mathcal{U}_1= \mathcal{U}$ with the property that for any normal sequence $\mathcal{U}_n'$, $\mathcal{U}_1' = \mathcal{U}$ and any $n$ there is $m$ such that $\mathcal{U}_m'$ refines $\mathcal{U}_n$.
Consider the cover $\mathcal{U} = \{\{n, n+1\}: n\in\mathbb{Z}\}$ of $\mathbb{Z}$. There are two star-refinements of $\mathcal{U}$ which are also partitions of $\mathbb{Z}$, so together they form a normal sequence. Those are the partitions $\mathcal{U}' = \{\{2n, 2n+1\}:n\in\mathbb{Z}\}$ and $\mathcal{U}'' = \{\{2n+1, 2n+2\}: n\in\mathbb{Z}\}$. Since those two normal sequences $\mathcal{U}, \mathcal{U}', \mathcal{U}', ...$ and $\mathcal{U}, \mathcal{U}'', \mathcal{U}'', ...$ exist, this shows that the sequence $\mathcal{U}_n$ would have to be constant and $\mathcal{U}$ would have to be its own star-refinement, but it clearly isn't. Hence there is no smallest preuniformity on $\mathbb{Z}$ containing the refinement $\mathcal{U}$.
One instance when such preuniformity exists is when $\mathcal{U}$ is its own star-refinement. Are there any simpler conditions for existence of such preuniformity that I'm not seeing?