It is well known that the completely regular spaces $X$ are characterized as those topological spaces whose topology is recovered from $C(X)$, the set of continuous functions $X\to \mathbb R$. In other words, $C(X)$ determines the topology on $X$. Is there a similar result where $C(X)$ is replaced by $C_U(X)$, the set of uniformly continuous functions? Thus, for which uniform spaces is it true that the set of uniformly continuous functions to the reals determines the uniformity? I'm also interested in the same question for quasi-uniform spaces. References are welcome if this is well-known material.

  • $\begingroup$ You might want to ask this on mathOverflow. mathoverflow.net $\endgroup$ – 9301293 Nov 5 '15 at 4:36

It seems the answer is "Yes.", a corollary of a result in Bourbaki (Theorem IX.I.4.I):

Given a uniformity $\mathcal{U}$ on a set $X$, there is a family of pseudometrics on $X$ such that the uniformity defined by this family is identical with $\mathcal{U}$.

Now pseudometrics aren't quite functions from $X$ to $\mathbb{R}$ (they're functions from $X\times X$ to $\mathbb{R}$), but this can easily be fixed: Let $\mathcal{D}$ be a collection of pseudometrics on $X$ that induces $\mathcal{U}$. Then, $$ \left\{ d_x:d\in \mathcal{D},\ x\in X\right\} , $$ where $d_x\colon X\rightarrow \mathbb{R}$ is defined by \begin{equation} d_x(y):=d(x,y), \end{equation} is a collection of real-valued uniformly-continuous functions on $X$ from which you can recover the uniformity $\mathcal{U}$.


They don't determine the uniformity.

Let $X$ be a uniform space and let $X'$ be $X$ with the weak uniformity coming from the uniformly continuous maps $X\to R.$ I understand the question to be whether the map $X\to X'$ is an isomorphism of uniformly continuous spaces.

One can describe the entourages $V$ of $X'$ as supersets of finite intersections $\bigcap_{i=1}^n\{(x,y)\mid |f_i(x)-f_i(y)|<2\}$ for some uniformly continuous $f_1,\dots,f_n:X\to\mathbb R.$ Equivalently, $V\supseteq \{(x,y)\mid \|f(x)-f(y)\|_\infty<2\}$ for some uniformly continuous $f:X\to\mathbb R^n.$

So $X$ is covered by the countable union $V[f^{-1}(\mathbb Z^n)].$ If $X$ is $\ell^\infty(\mathbb N),$ the entourage $V=\{(x,y)\mid \|x-y\|<1\}$ cannot be of this form - you can't cover the space by countably many unit balls. Another counterexample is to take $X$ to be an uncountable set with the discrete uniformity, and take $V$ to be the diagonal - you can't cover $X$ with a countable union of singletons.

(If $X$ is a topological group with its usual uniform structure, and $X\to X'$ is not an isomorphism, we have the same topological group with two different uniformities. This is not a contradiction: the group multiplication in $X'$ must fail to be uniformly continuous.)

If $X$ is furthermore a length space, and $X\to X'$ is an isomorphism, then we get the nice pair of conditions $$d(f(x),f(y))\leq C(d(x,y)-1)$$ $$d(x,y)\leq \tfrac12(d(f(x),f(y))-1)$$ where $C$ comes from the uniform continuity of $f$ applied to a chain $x=x_1,x_2,\dots,x_k=y$ with each $d(x_i,x_{i+1})<1$ and $k\leq d(x,y)+1,$ and the $\tfrac12$ comes from considering a chain $f(x)=x'_1,x'_2,\dots,x'_m=f(y)$ with each $d(x'_i,x'_{i+1})<2$ and $m\leq \tfrac12 d(f(x),f(y))+1.$ These make $f$ a quasi-isometric embedding. So two-dimensional hyperbolic space is a counterexample. If you are interested in this aspect, a good reference is "Metric Geometry" by Burago, Ivanov, Burago.


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