Construct a specific base for Fine uniformities in the diagonal(Entourages) case

Question

For every uniformizable space $X$ there is a finest uniformity on $X$ compatible with the topology of $X$ called the fine uniformity or universal uniformity.

To construct Fine uniformities, Let the family $\{\mathcal{V}_i:i\in I\}$ of all compatible uniformities on $X$ and generates the initial uniformity $\mathcal U$ w.r.t. $id_i:X\to(X,\mathcal{V}_i)$ on $X$. therefore, $\mathcal U$ is finer than each $\mathcal{V}_i$ for $i\in I$. Let us note that, completely regularity is used to show that $\{\mathcal{V}_i:i\in I\}\not=\emptyset$, since $X$ is a completely regular(uniformizable) space and have a compatible uniformity.

Exercise 36H. from Willard, General Topology says that,

The fine uniformity $\mathcal D_F$ on a uniformizable space is the uniformity having for a base $\beta$, the open sets $D\supset\Delta$ such that for some sequence $D_1, D_2,...$ of open sets containing $\Delta$, $D_n\circ D_n\subset D_{n-1}$ for all $n$ and $D_1= D$.

Is $\mathcal U$ and $\mathcal D_F$ exactly the same uniformity on a uniformizable space $X$? If the answer is yes, then how we can construct a base $\beta$ for $\mathcal U$ ?

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Edit: I am trying to solve this question. $\beta$ is a base for a uniformity on $X$ and the uniformity $\mathcal D_F$ generated by $\beta$ is finer than $\mathcal U$. now it is enough to show that, $\mathcal D_F\subset \mathcal U$. If the $\mathcal D_F$ is compatible with the topology of $X$, then $\mathcal D_F\subset \mathcal U$, since $\mathcal U$ is the finest uniformity on $X$ compatible with the topology of $X$.

Now my problem is to show that $\mathcal D_F$ is compatible with the topology of $X$. thanks in advice

Answer 1

Given a uniformity $\cal U$ on a set $X$, let $\tau_{\cal U}$ be the topology generated by (or compatible with) $\cal U$.

Let $\text{Uni}(\tau_{\cal U})$ denote the collection of uniformities compatible with $\tau_{\cal U}$. So trivially, ${\cal U}\in\text{Uni}(\tau_{\cal U})$.

Set $${\cal B} = \big\{\bigcap_{i=1}^k U_i: U_i \in \bigcup \text{Uni}(\tau_{\cal U}) \text{ and } k\in\mathbb{N}\big\}.$$ It is easy to verify that this forms a fundamental system for a uniformity, that is:

1. $B_1,\ldots, B_n\in {\cal B} \implies \bigcap_{i=0}^n B_i \in {\cal B}$,
2. $\forall B\in{\cal B}:\Delta_X = \{(x,x): x\in X\} \subseteq B$,
3. $\forall B\in {\cal B} \exists B'\in {\cal B} : B'^{-1}\subseteq B$,
4. $\forall B\in {\cal B} \exists A\in {\cal B}: A^2\subseteq B$.

Then the fine uniformity is the uniformity ${\cal U}_{\text{f}} := \{U\subseteq X\times X: \exists B\in{\cal B}(B\subseteq U)\}$.

It suffices to verify that for $x\in X$ and $B\in{\cal B}$ we have that $B(x)$ is a neighborhood of $x$ in the topological space $(X,\tau_{\cal U})$. This is easily proved by observating that a finite intersection of neighborhoods of $x$ is again a neighborhood of $x$ in $(X,\tau_{\cal U})$.

So we get $\tau_{{\cal U}_{\text{f}}} \subseteq \tau_{\cal U}$. By construction we trivially have $\tau_{\cal U}\subseteq \tau_{{\cal U}_{\text{f}}}$, so we get $\tau_{{\cal U}_{\text{f}}} = \tau_{\cal U}$, which answers the question.