Uniformly continuous independent of metrics? Let $(X,d)$ and $(Y,e)$ be metric spaces. A map $f:X\to Y$ is uniformly continuous if for each $\epsilon>0$ there exists $\delta >0$ such that whenever $d(x,y)<\delta$ we have $e(f(x),f(y))<\epsilon$.
Suppose $f:X\to Y$ is uniformly continuous w.r.t. $d$ and $e$. My question is, if $d'$ and $e'$ are metrics on $X$ and $Y$, resp., which induce the same topologies as $d$ and $e$, then is $f$ still uniformly continuous w.r.t. $d'$ and $e'$?

No. Let $d$ be the usual metric on $\mathbb R$, and let $d'$ be the metric $d'(x,y)=|\arctan(x)-\arctan(y)|$. It's not hard to check that $d'$ induces the same topology on $\mathbb R$ (use the continuity of $\arctan$). Then the identity function on $\mathbb R$ is uniormly continuous when considered as a function $(\mathbb R,d)\to(\mathbb R,d)$, but not when considered as a function $(\mathbb R,d')\to(\mathbb R,d)$.
