Uniform Space induced by pseudo metric

Question

Let $X$ and $Y$ be uniform spaces where the uniform structures are induced by pseudo metrics $d$ and $d^{'}$ respectively. Then a function $f:X \to Y$ is uniformly continuous if $\forall \epsilon >0$ $\exists \delta >0 $ such that $d(x,y)< \delta $ implies $ d^{'} (f(x),f(y))< \epsilon $ for all $x,y \in X$?

Answer 1

This is quite immediate from the definitions:

The uniformity $\mathcal{U}_d$ on $X$ has as a base all sets $U(r), r>0$ where $U(r)=\{(x,x') \in X^2: d(x,y) < r\}$. Similarly $\mathcal{U}_{d'}$ on $Y$ is generated by all $V(s), s>0$, similarly defined on $Y^2$ using $d'$.

Let $f$ be uniformly continuous from $(X, \mathcal{U}_d)$ to $(Y, \mathcal{U}_{d'})$, and let $\varepsilon >0$. Then $V(\varepsilon)=\{(y,y') \in Y^2: d'(y,y') < \varepsilon\} \in \mathcal{U}_{d'}$ and so by uniform continuity of $f$ we have that $(f \times f)^{-1}[V(\varepsilon)] \in \mathcal{U}_d$ so that there is some $\delta>0$ such that $$U(\delta) \subseteq (f \times f)^{-1}[V(\varepsilon)]$$

This inclusion can be restated as (because $(f \times f)(x,x')=(f(x), f(x'))$ etc.):

$$\forall x,x' \in X: (d(x,x') < \delta) \to d'(f(x),f(x')) < \varepsilon\tag{1}$$

This shows one inclusion. If $(1)$ holds we can similarly show that when we consider a basic $V(\epsilon)$ of $\mathcal{U}_{d'}$ then $U(\delta)$ (for the promised $\delta$ for that $\varepsilon$) obeys $U(\delta) \subseteq (f \times f)^{-1}[V(\varepsilon)]$ and so $(f \times f)^{-1}[V(\varepsilon)] \in \mathcal{U}_{d}$ and as this holds for basic uniformities of $\mathcal{U}_{d'}$, it holds for all of them, so $f$ is uniformly continuous.