Let $f:R \to (R,U^{'})$ be a mapping defined by: $f(x)$ =0 if $x\in Q$ and $f(x)$ = 1 if $x\in Q^{c}$. Find the uniform structure induced by $f$ , if $ (R,U^{'}) $ is uniform space induced by pseudo metric (discrete uniform space , indiscrete uniform space)?
Where $U_d=\{v\subseteq X\times X : v_\epsilon \subseteq v\} $ and $ v_\epsilon =\{(x,y)\in X \times X: d(x,y)< \epsilon \} $ is uniformity induced by pseudo metric, $U=\{v\subseteq X\times X : id_{x} \subseteq v\}$ is discrete uniform structure and $U=\{X\times X\}$ is indiscrete uniform structure.
It suffices to consider a base for the codomain uniformity $\mathcal{U}'$ and find all sets $f \times f)^{-1}[U]$ for $U$ in that base, and we get a base for the induced uniformity $\mathcal{U}$ on $\Bbb R$.
For the indiscrete uniformity we only have to consider $U=\Bbb R^2$, with inverse image $\Bbb R^2$ as well, so we also get the indiscrete uniformity on the domain. $f$ doesn't have much influence.
For the discrete uniformity we can pick a one set base $\Delta_{\Bbb R}$ and then $$(f \times f)^{-1}[\Delta_{\Bbb R}]=\{(x,x') \in \Bbb R^2: f(x)=f(x')\}= \Bbb Q^2 \cup \Bbb P^2$$ where $\Bbb P$ stands for the set of irrationals.
So this induces the uniformity $\{A \subseteq \Bbb R^2: \Bbb Q^2 \cup \Bbb P^2 \subseteq A\}$.
Something similar holds for the metric uniformity generated by all $v_\epsilon$, with $\epsilon < 1$ WLOG. There $(f(x),f(x')) < \epsilon$ iff $f(x)=f(x')$ too, so all their inverse images are also $\Bbb Q^2 \cup \Bbb P^2$ and we get the same uniformity as the previous one.