Proof of the "second half of the Heine-Cantor theorem"

Question

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?

Answer 1

You can consider completions, as follows. Let $\mathcal U, \mathcal V$ be the uniformities induced by the given gauges for $X,Y$ respectively. Let $(\hat X, i), (\hat Y, j)$ be completions of $X,Y$ respectively (you can also directly construct the completion for gauge spaces in an “isometric” fashion, as I describe in the comment below). Then, because $f$ is Cauchy continuous, the map $ j \circ f \circ i^{-1}: i(X) \to \hat Y$ is Cauchy continuous, hence there is a Cauchy continuous extension $\hat f$ to all of $\hat X$. Moreover, $\hat X$ is compact because $X$ is totally bounded. Hence $\hat f$ is uniformly continuous, which implies its restriction $\hat f|_{i(X)} = j \circ f \circ i^{-1}$ is also uniformly continuous. By composition with the uniformly continuous maps $j^{-1}$ and $i$, uniform continuity of $f$ follows.

Comment. Since the OP said they were not familiar with uniform spaces, I sketch a few ideas on how one could do this with gauge spaces. Let $(X, \mathcal U)$ be a uniform space, and fix a gauge $P= \{p_{\alpha}\}_{\alpha \in A}$ for $X$ (you may also consider $X$ directly as a gauge space in what follows, without explicit mention of the uniformity by instead considering the analogous concepts for gauge spaces as you define in your question). Then, there is a complete (and possibly non-hausdorff) uniform space (gauge space) $\hat X$ and a uniform isomorphism (onto its image) $i:X \to \hat X$ such that $i(X)$ is dense in $\hat X$; moreover, $\hat X$ may be equipped with a gauge $\hat P=\{\hat p_{\alpha}\}_{\alpha \in A}$ such that $\hat p_{\alpha}(i(x), i(y))= p_{\alpha}(x,y)$ for each $x,y \in X$. Moreover, $\hat X$ is unique up to isomorphism, in the sense that if $(\tilde X, i’)$ is another completion, there is a uniform isomorphism $I: \hat X \to \tilde X$, though $I$ need not be unique; however, if $X$ is Hausdorff, $I$ is uniquely determined.

Note that there is also the notion of a Hausdorff completion (that is, $\hat X$ can be taken to be Hausdorff), but $i$ will not generally be injective in this case; when $X$ is Hausdorff, both notions of the completion coincide. Bourbaki gives a proof for the existence of Hausdorff completions.

A sketch for the “isometric” completion is as follows (a more detailed account may be found in Schecter). For every $\alpha$, consider the pseudometric space $(X, p_{\alpha})$, and let $(X’, p’_{\alpha})$ be a complete pseudometric space which is an isometric completion of $(X, p_{\alpha})$, in the sense that there is a map $j_{\alpha}: (X, p_{\alpha}) \to (X’, p’_{\alpha})$ which is an isometric isomorphism onto its image, which is dense in $X’$. (The existence of isometric pseudometric completions is proven in Schecter, chapter 19 for instance). Then consider the embedding into the product $$i: X \to \prod_{\alpha \in A} (X’, p’_{\alpha}): x \mapsto (j_{\alpha}(x))_{\alpha}.$$ Equip the product with the product uniformity; then it is complete being the product of complete spaces. Put $\hat X = \overline{i(X)}$. Define then $\hat p_{\alpha}(x,y) = p’_{\alpha}(\pi_{\alpha}(x), \pi_{\alpha}(y))$ whenever $x,y \in \hat X$, and $\pi_{\alpha}$ is the projection. Details that this gives the desired result are left to the reader.