Following


[About uniform continuity](https://mathoverflow.net/questions/402661/about-uniform-continuity)


Let $E$ be a topological space, for all $a \in E$, we associate an open set of $E$, $U(a)$ containing $a$.


We will say that $\{U (a), a \in E\}$ is a uniform covering of $E$ if


For any dense $B$ in $E$ the covering $\{U (b), b \in B\}$ covers $E$.


Let $(E, d) $ and $(E', d') $ 2 metrics spaces, and $f$ continuous function of $E$ to $E'$.


Is it true that :


$f$ uniformly continuous


iff


for any uniform cover of $E'$ $U$, for all $b$ in $E$, $V (b) = f^{-1}(U (f (b))$ $V$ is a uniform cover of $E$?



Uniform covering and uniform continuity
Let $(E',d')$ be the Euclidean plane $\mathbb{R}^2$ with the usual metric.
Let $(E,d)$ be the interval $[-1,1]$ with the usual metric.


Let $f$ be the inclusion map of $[-1,1]$ to the interval $[-1,1]\times \{0\}$; it is an isometry to its image and is uniformly continuous.


Consider the following cover of $E'$:
$$ U(x) = \begin{cases}
B\_1(0) & x\in (-1,1)\times \{0\} \\
E' & \text{otherwise} \end{cases}. $$
Notice that any dense subset $B'\subseteq E'$ must contain at least one point that is not on $(-1,1)\times \{0\}$ and hence $U$ is a uniform cover by your definition.


But let $B\subsetneq E$ be the set of all irrational numbers in that interval. If $x\in B$ then $U(f(x)) = B\_1(0)$ and $f^{-1}(U(f(x)) = (-1,1)$, and hence we see that $\{V(b): b\in B\} = \{(-1,1)\}$ and does not cover $E$.


