If a sequence converges according to one metric, must it converge according to all metrics?

Question

Given a sequence of points in a metric space, with a given metric $d$, will the same sequence of elements converge if you change the metric of that metric space?

Answer 1

No. Let $X=\mathbb Q$. If $d$ is the usual metric, the sequence $(1/n)_{n\in\mathbb N}$ is convergent. But if you change the metric to the discrete metric $d'$: $$ d'(a,b)=\begin{cases}1,&\ a\ne b\\0,&\ a=b\end{cases} $$ the same sequence is not Cauchy.

Answer 2

No. The sequence $\left(\frac1n\right)_{n\in\mathbb N}$ converges in $[0,1]$ (to $0$) with respect to the usual metric, but not with respect to the discrete metric.

However, the answer will be affirmative if both metrics induce the same topology.

Answer 3

Here's an example that doesn't involve the discrete metric:

Consider the half-open interval $[0,2\pi)$, once with the metric $d_1(x,y)=|x-y|$, and once with the metric $d_2(x,y)=2\sin\frac{|x-y|}{2}$. And consider the sequence $a_n=2\pi-\frac{1}{n}$.

According to $d_1$, this sequence does not converge in the interval (it can't converge to $2\pi$ because that's not in the interval), but according to $d_2$ is converges to $0$.