Existence of an unique uniform space, generated by a family of coverings. It is known, that given a family $M$ of subsets of a set $X$, there exists a unique, minimal topology $\tau$, such that $M \subset \tau$, we get this topology as the intersection of all topologies that have $M$ as a subset.
Is the analogical claim in uniform spaces true, when using the uniform cover definition?
https://en.wikipedia.org/wiki/Uniform_space#Uniform_cover_definition
That is, given a family $N$ of coverings of $X$, does there exist a unique, minimal uniformity (set of coverings) $\mathcal U$, such that $N \subset \mathcal U$?
I've tried to show that such a set of coverings exists, using similar approach as with topological spaces, but failed, because of the need for a star refinement for any covering, and I can't think of a counter-example.
 A: The answer depends whether it is required that $\mathcal U$ is a family of open covers of a topological space $X$ or it is a family of covers of the set $X$. In the second case a cover of the set $X$ by singletons (that is, one-point sets) is a star refinement of an arbitrary cover of the set $X$. In the first case, according to Engelking’s “General topology” 
Thus each non-paracompact $T_1$ space raises a counterexample.
A: Not for any family of covers, no. But for "normal families of covers" yes: 
A set of covers $\mu$ is a normal family, if for every cover in $\mu$ there is some other cover of $\mu$ exists that star-refines it.
For two covers $\mathcal{U}, \mathcal{V}$ of $X$ define 
$\mathcal{U} \wedge \mathcal{V} = \{U \cap V: U \in \mathcal{U} ,V \in \mathcal{V}\}$ which is the largest cover that refines them both. Similarly for finite meets.
Note that the definition of covering uniformity can also (a bit different from the Wikipedia one, a bit more like a filter) be stated as: a non-empty family of coverings of $X$, $\mu$, is called a covering uniformity iff


*

*$\forall \mathcal{U}, \mathcal{V} \in \mu: \mathcal{U} \wedge \mathcal{V} \in \mu$. (closed under "meets")

*$\forall \mathcal{U} \in \mu$ : for every cover of $X$ with $\mathcal{U} \prec \mathcal{V}$, then $\mathcal{V} \in \mu$ (closed under enlargements, i.e. upper refinements) 

*$\forall \mathcal{U} \in \mu: \exists \mathcal{V} \in \mu: \mathcal{V} \prec^\ast \mathcal{U}$. (normal family).


If $\mathcal{M}$ is any normal family of covers then the set of covers $\mu$ defined by
$$\mathcal{U} \in \mu \Leftrightarrow \exists \mu_1,\ldots, \mu_n \in \mu \exists \mathcal{U}_1 \in \mu_1,\ldots, \mathcal{U}_n \in \mu_n: \bigwedge_{i=1}^n \mathcal{U}_i \prec \mathcal{U}$$
is the smallest covering uniformity that contains $\mathcal{M}$.
But there can be no such containing uniformity if $\mathcal{M}$ has a cover without a star-refinement. In a paracompact space all family of all open covers is a normal family (this is a restatement of classical theorems on paracompactness, as Alex mentions). So if $X$ is Tychonoff (so uniformisable) and not paracompact (e.g. $\omega_1$ in the order topology), let $\mu$ be the uniformity that induces the topology of $X$, and $\mathcal{U}$ be an open cover without an open star-refinement.
Then the set of covers $\mu \cup \{\mathcal{U}\}$ does not generate a covering uniformity.
