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Let $(X,\mathcal D)$ be a uniform space and $A,B\subseteq X$. Let's say $A$ is uniformly inside $B$ and write $A\le B$ iff there's some entourage $D$ for which $$(\forall a\in A)(D[a]\subseteq B)$$

A uniform space $(X,\mathcal D)$ is uniformly normal iff for each open set $U$ and any closed set $C\subseteq U$, there's some open set $W$ such that $$C\le W\subseteq \overline W \le U$$

Do you have an example of a normal uniform space that is not uniformly normal?!

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  • $\begingroup$ This is among the most routine straightforward elementary exercises possible. $\endgroup$ – Włodzimierz Holsztyński Sep 26 '13 at 0:04
  • $\begingroup$ @WlodzimierzHolsztynski: It is said somewhere deleted user posts are deleted if they are downvoted. $\endgroup$ – user35674 Sep 26 '13 at 3:32
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The definition of uniform normality given is not a property of the underlying topology of your uniform space, but the definition of uniform normality is a property of the underlying proximity on your uniform space. Furthermore, the definition of uniform normality is correct since the uniformly normal spaces are precisely the uniform spaces where Urysohn's lemma holds for uniformly continuous functions.

The reader is referred primarily to [2] for more information on proximity spaces. One should consult [1] for more basic information on proximity spaces.

If $(X,\mathcal{U})$ is a uniform space, then $(X,\mathcal{U})$ becomes a proximity space where $R\delta S$ if and only if $D[R]\cap D[S]\neq\emptyset$ for each $D\in\mathcal{U}$. Said differently, $R\overline{\delta}S$ if and only if $D[R]\cap D[S]=\emptyset$ for some $D\in\mathcal{U}$. The definition of $\preceq$ is the same as defined before. In particular, $R\preceq S$ if and only if $D[R]\subseteq S$.

$\textbf{Proposition}$ Let $(X,\mathcal{U})$ be a uniform space. Then the following are equivalent.

  1. $(X,\mathcal{U})$ is uniformly normal (as defined in the question)

  2. Whenever $C$ is a closed set, $U$ is an open set and $C\subseteq U$, then $C\preceq U$.

  3. If $C_{1},C_{2}$ are two disjoint closed sets, then $C_{1}\overline{\delta}C_{2}$.

  4. $R\overline{\delta}S$ if and only if $\overline{R}\cap\overline{S}=\emptyset$.

  5. $R\preceq S$ if and only if $\overline{R}\preceq S^{\circ}$.

Furthermore, if $(X,\mathcal{U})$ satisfies any of the above conditions, then $X$ is normal as a topological space.

Recall that any compact space has a unique uniformity and a unique proximity. Therefore when talking about uniformities and proximities on compact spaces, there is no need to mention the uniformity on proximity in question. The following proposition can be found in [2][Cor. 12.12].

$\textbf{Proposition}$ If $(X,\mathcal{U})$ is a uniform space and $C$ is a compact Hausdorff space, then a function $f:(X,\mathcal{U})\rightarrow C$ is a proximity map if and only if it is uniformly continuous.

Using the above characterization of the bounded real-valued uniformly continuous functions, we obtain a few more characterizations of uniform normality by replacing "uniformly continuous" by proximity map.

$\textbf{Proposition}$ Let $(X,\mathcal{U})$ be a uniform space. Then the following are equivalent.

  1. $(X,\mathcal{U})$ is uniformly normal.

  2. If $C_{1},C_{2}\subseteq X$ are disjoint closed sets, then there is a uniformly continuous map $f:X\rightarrow[0,1]$ with $f|_{C_{1}}=0$ and $f|_{C_{2}}=1$.

  3. $X$ is normal and every continuous map $f:X\rightarrow[0,1]$ is uniformly continuous.

  4. $X$ is normal, and for every compact Hausdorff space $C$, every continuous map $f:X\rightarrow C$ is uniformly continuous.

With all these facts mentioned, it must be said that except for rare occasions, normal spaces typically can be given compatible uniformities which are not uniformly normal. The only spaces that I can think of that are normal and all of whose uniformities are uniformly normal are similar to ordinal spaces.

$\textbf{Proposition}$ Let $X$ be a normal space. Then the following are equivalent.

  1. Every compatible uniformity on $X$ is uniformly normal.

  2. The space $X$ has a unique compatible proximity.

  3. The space $X$ has a unique compactification.

  4. $X$ is compact or $X$ is locally compact and the one-point-compactification of $X$ coincides with the Stone-Cech compactification of $X$.

  5. $X$ is locally compact and $A\cap B\neq\emptyset$ whenever $A,B$ are non-compact closed subsets of $X$.

  6. $X$ has a unique compatible uniformity.
  1. Willard, Stephen. General Topology. Reading, MA: Addison-Wesley Pub., 1970.

  2. Naimpally, S. A., and B. D. Warrack. Proximity Spaces. Cambridge Eng.: University, 1970.

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  • $\begingroup$ You answered what I could ask (but I did not because uniform normality was just a raw idea to reach an example that could help construct a counterexample for something that hardly is related to normality) . The last proposition is very good for characterizing uniform normality (although for proving it I'll never resort to proximities, because they proximity spaces have less generality). btw, I have some ideas that may complete this characterzation: $\endgroup$ – user31967 Sep 24 '13 at 21:20
  • $\begingroup$ I think intuitively uniform normality has close connection to divisibility $\endgroup$ – user31967 Sep 24 '13 at 21:22
  • $\begingroup$ And because you used proximities it may be true that a space is uniformly normal iff its precompact reflection is. $\endgroup$ – user31967 Sep 24 '13 at 21:23
  • $\begingroup$ And I think intuitively that every paracompact Hausdorff uniform space is uniformly normal $\endgroup$ – user31967 Sep 24 '13 at 21:25
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    $\begingroup$ Paracompact uniform spaces are generally not uniformly normal. $\endgroup$ – Joseph Van Name Sep 24 '13 at 21:42
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Yes, since you're using a bad definition. $\;\;\;$ Let $\:\langle X\hspace{.02 in},\hspace{-0.03 in}\mathcal{D}\rangle\:$ be the Euclidean plane with its
usual uniform structure. $\;\;\;$ Let $C$ be the closed lower half-plane, and let $U$ be the subset of
the plane that is below the graph of $\:x\mapsto \frac1{1+x^{\hspace{.02 in}2}}\;$. $\;\;\;$ $U$ is open and $C$ is closed and $\:C\subseteq U\;$.
For any open set $W$, if $\:C\leq W\:$ then there exists a positive $\epsilon$ such that $W$ contains the line $\:y=\epsilon\;$.
Since no such line is a subset of $U$, there is no open set $W$ such that $\: C\leq W\subseteq \overline{W} \leq U \;$.
Therefore $\:\langle X\hspace{.02 in},\hspace{-0.03 in}\mathcal{D}\rangle\:$ does not satisfy your definition of being uniformly normal.

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  • $\begingroup$ Do you know a good definition?! $\endgroup$ – user31967 Sep 24 '13 at 13:57
  • $\begingroup$ I don't see anything wrong with that definition of uniformly normal. It just does not coincide with normality. $\endgroup$ – Joseph Van Name Sep 24 '13 at 15:02
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    $\begingroup$ I would think it should specify $\:\: C\leq U \;\;$. $\;\;\;\;\;$ $\endgroup$ – user5810 Sep 24 '13 at 17:53
  • $\begingroup$ yeah it seems more rational but I wonder if it implies normal normality. with my definition every uniformly normal space is normal. $\endgroup$ – user31967 Sep 24 '13 at 18:24

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