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Let $(X,\mathcal D)$ be a normal (diagonal) uniform space and $G$ be the set of all homeomorphisms $f:X\to X$. Let $\Delta$ be the uniformity on $X^X$ (inherited by $G$) by subbase

$$\Lambda =\{ \{(f,g)\in X^X\times X^X\mid (\forall a\in A) \big((f(a),g(a))\in E\big)\} \mid A\in \mathcal A, D\in \mathcal D\}$$ where $\mathcal A$ is the set of all closed sets.

Let $m,n:(D,\le)\to G$ be two nets converging to $f,g\in G$ respectively. How can I prove $$n_x\circ m_x \to f\circ g$$ ?

Something that may solve my problem is how to describe a normal uniform space directly. Every uniform space is completely regular but how can a normal uniform space be described directly by adding some property to entourages?

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I am unsure if the following result is exactly what you wanted, but I did it anyways since it seemed like a fun exercise to do.

$\mathbf{Theorem}$ Let $(X,\mathcal{U}),(Y,\mathcal{V}),(Z,\mathcal{W})$ be uniform spaces. Let $f_{d}:X\rightarrow Y,g_{d}:Y\rightarrow Z$ be uniformly continuous maps with $f_{d}\rightarrow f,g_{d}\rightarrow g$ uniformly for some uniformly continuous functions $f,g$. Then $f_{d}g_{d}\rightarrow g$ uniformly.

$\mathbf{Proof}$ Suppose that $R\in\mathcal{W}$. Then there is a symmetric entourage $S\in\mathcal{W}$ with $S\circ S\subseteq R$. Then by uniform convergence, there is some $d\in D$ where if $e\geq d$, then $(g_{e}(y),g(y))\in S$. Furthermore, there is some symmetric entourage $T\in\mathcal{V}$ where if $y_{1},y_{2}\in T$, then $(g(y_{1}),g(y_{2}))\in S$. Again, by uniform convergence, there is some $d'\geq d$ where if $e\geq d$, then $(f_{e}(x),f(x))\in T$. In particular, if $e\geq d'$, and $x\in X$, then $(gf_{e}(x),gf(x))\in S$ and $e\geq d$, so $(g_{e}f_{e}(x),gf_{e}(x))\in S$. Therefore, we have $(g_{e}f_{e}(x),gf(x))\in S\circ S\subseteq T$. We conclude that $g_{d}f_{d}\rightarrow gf$ in the uniformity of uniform convergence. $\mathbf{QED}$

It does not seem like normality has a nice characterization in terms of uniform spaces since uniform spaces have much more structure than simply a topology. However, there are close connections between uniform properties and topological properties that are similar to complete regularity. For instance, it is well known that below the first measurable cardinal, the realcompact spaces coincide with the spaces with a compatible complete uniformity.

Furthermore, there is a connection between paracompactness and uniform properties. If $(X,\mathcal{U})$ is a uniform space, then let $H(X)$ be the set of closed subspaces of $X$. Then for each entourage $E\in\mathcal{U}$, let $\hat{E}$ be the binary relation on $H(X)$ where $(C,D)\in\hat{E}$ if and only if $C\subseteq E[D]=\{y|(x,y)\in E\,\textrm{for some}\,x\in D\}$ and $D\subseteq E[C]$. Then $H(X)$ becomes a uniform space with a uniformity generated by the entourages $\{\hat{E}|E\in\mathcal{U}\}$. However, the uniform space $H(X)$ is generally not complete even if $X$ is complete. On the other hand, if $H(X)$ is a complete uniform space, then $X$ is complete. We say that a uniform space $X$ is supercomplete if and only if the hyperspace $H(X)$ is complete. It is well known that a uniform space $(X,\mathcal{U})$ is supercomplete if and only if $X$ is paracompact and the locally fine coreflection of $X$ is the fine uniformity on $X$. In particular, since each paracompact space is normal, every supercomplete uniform space is normal as well.

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  • $\begingroup$ if $\mathcal A=\{X\}$ the first theorem is the same proposition as mine. But $\mathcal A$ is the set of all closed sets and this needs some aid from normality. $\endgroup$ – user31967 Sep 21 '13 at 12:00
  • $\begingroup$ It does not matter whether you take all closed sets of just the set $X$. You still obtain the same uniformity. $\endgroup$ – Joseph Van Name Sep 21 '13 at 15:14
  • $\begingroup$ in fact I thought that topology is the closed-based topology. But it is not! $\endgroup$ – user31967 Sep 25 '13 at 3:34

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