A pseudometric on a set $X$ is a metric except it need not be Hausdorff. A gauge $\mathcal D_X$ on $X$ is a an ideal of pseudometrics, i.e.:
- $\mathcal D_X \neq \emptyset$
- $d_1,d_2 \in \mathcal D_X \Rightarrow d_1 \vee d_2 \in \mathcal D_X$, where $d_1\vee d_2$ is the maximum of $d_1$ and $d_2$
- $d_1$ is a pseudometric and $d_1 \leq d_2 \in \mathcal D_X \Rightarrow d_1 \in \mathcal D_X$
The pair $(X, \mathcal D_X)$ is called a gauge space (this is virtually the same as a uniform space, a notion that I am not familiar with). We denote the open ball around $x\in X$ of distance $\varepsilon$ w.r.t. to $d$ by $B_d(x,\varepsilon)$.
The gauge space is totally bounded, if for all $d\in \mathcal D_X$ and $\varepsilon > 0$ there are $x_1,\dots, x_n \in X$, s.t. $X = \bigcup_{i=1}^n B_d(x_i, \varepsilon)$.
A function $f : X\to Y$ between gauge spaces is uniformly continuous, if for all $d_Y\in \mathcal D_Y$ and $\varepsilon > 0$ there exist $d_X\in \mathcal D_X$ and $\delta > 0$, such that: $$\forall x,x'\in X : d_X(x,x') < \delta \Rightarrow d_Y(f(x),f(x')) < \varepsilon$$
A filter $F$ of $X$ is Cauchy, if it is proper and for all $d\in \mathcal D_X$ and $\varepsilon > 0$ there exists an $A\in F$ with $\operatorname{diam}_d(A) := \sup_{x,y\in A} d(x,y) \leq \varepsilon$.
A function $f : X\to Y$ between gauge spaces is Cauchy-continuous if for all Cauchy filters $F$, $f(F)$ is Cauchy, where $f(F)$ is the filter on $Y$ generated by $\{f(A) : A\in F\}$.
I'm struggling to prove the following:
Let $f : X\to Y$ be Cauchy-continuous between gauge spaces, such that $X$ is totally bounded. Then $f$ is uniformly continuous.
How do I go about proving this?