I am taking an undergraduate course in topology and am stuck with trying to prove the following Theorem from Willard:
36.11 Theorem. Every normal family of covers of X is a subbase for some uniformity on X; the converse fails
A normal family of uniform covers is defined as a family of covers in which each each cover in the family has a star refinement that is also in the family.
Thank you in advance for your time!!
Let $\nu$ be a normal family of covers of $X$; define $\mu$ by the set of all covers $\mathcal{U}$ of $X$ such that there are finitely many covers $\mathcal{V}_1, \mathcal{V}_2,\ldots,\mathcal{V}_n \in \nu$ such that $$\bigwedge_{i=1}^n \mathcal{V}_i \prec \mathcal{U}$$
Claim: this defines a uniformity $\mu$ and $\nu$ is clearly a subbase for it by definition.
Checking the axioms for a covering uniformity is straightforward: Let $\mathcal{U}_1,\mathcal{U}_2 \in \mu$, let $\mathcal{V}_1,\ldots, \mathcal{V}_n$ be the witnessing covers from $\nu$ for $\mathcal{U}_1$ and $\mathcal{V}_{n+1},\ldots, \mathcal{V}_{n+m}$ those for $\mathcal{U}_2$. For each $\mathcal{V}_i$ we can find a $\mathcal{W}_i \in \nu$ such that $\mathcal{W}_i \prec^\ast \mathcal{V}_i$ as $\nu$ is a normal family. Then $\nu \ni \bigwedge_{i=1}^{n+m} \mathcal{W}_i \prec^\ast \mathcal{U}_j$ for $j=1,2$ so $\mathcal{U}_1$ and $\mathcal{U}_2$ have a common star-refinement in $\mu$. The enlargement axiom (b) is trivially fulfilled.
For the converse, which I interpret as "every subbase for a uniformity is a normal family", this is indeed false:
Let $X$ be any uniformisable space which is not paracompact in its induced topology (like $\omega_1$ in the order topology), and give it its fine uniformity $\mu$.
By Willard 36.16 the set $\nu$ of all open covers of $X$ is a base (so certainly a subbase too) of $\mu$.
But $\nu$ is not a normal family: as $X$ is not paracompact, by 20.15 it has an open cover without an open star-refinement.