Uniformity inducing Topology If $\mu$ is any covering uniformity on a set $X$ inducing the topology $\mathcal{T}$ on $X$ and if $(\mathcal{B_n})_{n \in \mathbb{N}}$ is a normal sequence of open covers of $X$ then $\mu \cup \{\mathcal{B_1},\mathcal{B_2}, \mathcal{B_3}\ldots\}$, being a normal family of open covers, is  a subbase for a uniformity , say $\nu$. How to show that $\nu$ generates the same topology $\mathcal{T}$ on $X$ ?
 A: The uniformity $\nu$ can be described as
$$\mathcal{U} \in \nu \text { iff } \exists \mathcal{V} \in \mu \exists n: \mathcal{B_n} \wedge \mathcal{V} \prec \mathcal{U}$$.
as always with subbases, as I also used in my other answer, e.g.
Now if $O$ is open w.r.t. $\nu$, then let $x \in O$. Then there is some cover $\mathcal{U} \in \nu$ such that $\operatorname{st}(x,\mathcal{U}) \subseteq O$. Find $\mathcal{V}$ and $\mathcal{B_n}$ as in the above characterisation.
Then (by standard facts about stars of covers):
$$ \operatorname{st}(x,\mathcal{V}) \cap \operatorname{st}(x,\mathcal{B_n}) = \operatorname{st}(x, \mathcal{B_n} \wedge \mathcal{V}) \subseteq \operatorname{st}(x,\mathcal{U})$$ 
The left hand side is open in $\mathcal{T}$ (as $\mu$ induces it, and as the covers $\mathcal{B}_n$ are open covers w.r.t. $\mathcal{T}$). This shows that $x \in \operatorname{Int}_{\mathcal{T}}(O)$, and as this holds for all $x \in O$, $O \in \mathcal{T}$. So $\mathcal{T}_\nu \subseteq \mathcal{T}$.
$\mathcal{T} = \mathcal{T}_\mu \subseteq \mathcal{T}_\nu$ is obvious as $\mu \subseteq \nu$, and so $\mathcal{T} = \mathcal{T}_\mu = \mathcal{T}_\nu$.
