Uniformly continuous independent of metrics?

Question

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 $e'$?

Answer 1

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 $(\mathbb R,d')\to(\mathbb R,d)$.

Answer 2

Let $X=\Bbb Z^+$, and let $d$ be the usual metric on $X$. For $m,n\in X$ let

$$d'(m,n)=\left|\frac1m-\frac1n\right|\;;$$

$d$ and $d'$ both induce the discrete topology on $X$. Let $Y=X$ and $e=e'=d$, and let $f:X\to Y$ be the identity map.

• $f$ is uniformly continuous with respect to $d$ and $e$, since $\delta=1$, for instance, ‘works’ for each $\epsilon>0$.

• However, no choice of $\delta$ ensures that $e'\big(f(m),f(n)\big)=|m-n|<1$ whenever $d'(m,n)<\delta$, so $f$ is not uniformly continuous with respect to $d'$ and $e'$.