Basically the title. Let $f$ be the function from $\Bbb R$ under any metric $\rho$ to $\Bbb R$ under the euclidean metric defined by $f(x)=x$. Is $f$ guaranteed to be continuous? It feels sensible that if I take $\rho(x, y)$ to be arbitrarily small, I should somehow be able to make $|x-y|$ arbitrarily small, but I'm having trouble coming up with an explicit epsilon-delta argument. Is there some counterexample I'm not seeing?
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Let $g(x)=x$ for $x \neq 0,1$. Let $g(0)=1$ and $g(1)=0$. Let $D(x,y)=|g(x)-g(y)|$. Then $D$ is metric on $\mathbb R$. If $f$ is continuous w.r.t this metric then there exists $\delta >0$ such that $|x-y| <\frac 1 2$ whenever $|g(x)-g(y)| <\delta$. Take $x=0$ and $y=1+\frac {\delta} 2$ to get a contradiction.
Kavi Rama Murthy
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@WimC $(\frac 1 2 , 2)$ is open in the Euclidean topology but $1$ is a point in this set which is not an interior point w.r.t. the metric $D$. (Points close to $0$ belong to any ball around $1$). – Kavi Rama Murthy Sep 23 '20 at 05:44