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Given a uniformizable (w.r.t. entourage uniformity) space $X$ there is a finest uniformity on $X$ compatible with the topology of $X$ called the fine uniformity or universal uniformity. A uniform space is said to be fine if it has the fine uniformity generated by its uniform topology.

why the fine uniformity is exists and How we can construct it? the completely regularity is needed to construct the fine uniformity?

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  • $\begingroup$ What definition of uniformity are you using? Entourages or covers? Or pseudometrics? $\endgroup$
    – Henno Brandsma
    Commented Nov 23, 2015 at 19:16
  • $\begingroup$ @HennoBrandsma ,Entourages. $\endgroup$
    – M.A.
    Commented Nov 24, 2015 at 3:37

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why the fine uniformity exists

It is supremum of all compatible uniformities on $X$.

and How we can construct it?

I guess its base should be a family of all open entourages of the diagonal $U$ such that there exists a sequence $\{U_n\}$ of open entourages such that $U_0=U$ and $U_{n+1}^2\subset U_n$ for each $n$.

the completely regularity is needed to construct the fine uniformity?

Yes, a topological space is uniformizable space iff it is completely regular.

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    $\begingroup$ Dear Professor Ravsky, I am trying to understand the fine uniformity by your notes. At first i peruse the initial uniformity $\mathfrak{U}$ on a set $X$; Then choose the family $\{\mathfrak{V}_i:i\in I\}$ of all compatible uniformities on $X$ and generates the initial uniformity $\mathfrak{U}$ w.r.t. $id_i:X\rightarrow (X, \mathfrak{V}_i )$ on X. therefore $\mathfrak{U}$ is finer than each $\mathfrak{V}_i$ for $i\in I$. But i cant construct a base for $\mathfrak{U}$ like the base in your notes. Is my argument correct? $\endgroup$
    – M.A.
    Commented Nov 26, 2015 at 6:42
  • $\begingroup$ And the completely regularity is needed to construct the fine uniformity; $\{\mathfrak{V}_i:i\in I\}\neq \emptyset$ since $X$ is a completely regular space and have a compatible uniformity. $\endgroup$
    – M.A.
    Commented Nov 26, 2015 at 6:45
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    $\begingroup$ @M.A. See Wikipedia article related to uniformizable spaces. The initial uniformity has a base of the described form, because $D_{f,\varepsilon}\circ D_{f,\varepsilon}\subset D_{f,2\varepsilon}$ for each continuous function $f$ on the set $X$ and $\varepsilon>0$. But the initial uniformity is not always the fine uniformity. The fine uniformity is the supremum of the family $\{{\frak V_i}:i\in I\},$ which is non-empty, because it contains the initial uniformity. $\endgroup$
    – Alex Ravsky
    Commented Nov 28, 2015 at 5:10

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