Is the convergence of a sequence independent of the chosen metric? Given a metric $\rho$ on $X$ and a sequence $x_n$ in $X$. Does the convergence of $x_n\to x$ under $\rho$ also imply the convergence to the same limit under any other metric $\sigma$?
I don't know th answer, intuitively i think it's true.
I didn't find working counterexamples.
I tried to derive that $[\rho(x_n,x)\to 0]\Rightarrow[\sigma(x_n,x)\to0]$ but I have no idea how to get there using the properties of a metric.
EDIT Thanks for your answers. Just to understand my wrong intuition: I feel that discontinuity is important here. Could one also construct a counterexample with a continous metric $\rho$ and a simply connected $X\subset\mathbb{R}^n$?
 A: Without the metric being specified, $X$ is just a set of elements, no order, no topology, nothing but the set. Therefore limits have to depend on the metric. To stress this point, consider the following metric on $\mathbb R$:
Let $\tau\colon\mathbb R\to\mathbb R$ be the bijection given by
$$
t \mapsto
\begin{cases}
0 & \text{if $t=10$,} \\
10 & \text{if $t=0$,} \\
t & \text{otherwise.}
\end{cases}
$$
Define a metric $d\colon\mathbb R\times\mathbb R\to\mathbb R$ by $d(x,y)=|\tau(x)-\tau(y)|$. This metric is just the standard metric on $\mathbb R$ with $0$ and $10$ interchanged, so for example the sequence $(1/n)_{n\in\mathbb N}$ will converge to $10$ under this metric.
However, if two metrics give rise to the same topology on $X$ (i.e. they are topologically equivalent) limits will indeed be preserved, since limits can be defined using only the topology.
Regarding your recent edit on continuity of the metric: Continuity on a metric space is usually defined using the metric, so saying a metric is continous implies you already have some other metric (or topology) on the set and want continuity with respect to this topology. This still doesn't imply topological equivalence, for example take $d(x,y)=0$ for all $x,y$. Under this pseudometric every point will be a limit of every sequence. (This will not happen for non-pseudo metrics) Still $d$ is continous with respect to the standard topology on $\mathbb R$!
A: Your intuition here is dead wrong. Don't worry, there's nothing wrong with that, though:P.
Define the metric on $\mathbb R$ like this: $d(x,y)=\cases{1\text{ if }x\neq y\\0\text{ if }x=y}$. In this metric, $\frac1n$ does not converge to $0$ as $n\rightarrow \infty$. In fact, only constant (from some point on) sequences converge in this metric.
A: A metric $\rho$ on $X$, gives the notion of distance between elements of $X$. If you change it, then the notion of distance changes. Consequently, everything that depends on the notion of distance (like sequence convergence, open and closed sets) changes.
So, the definition tells you that a sequence converges, when there is a $x\in X$ such that the distance between $x$ and the members of the sequence becomes more and more small. If you change the notion of distance, then such a $x$ might change or might not exist at all.
For example, say there is a metric $\rho(x,y)=|x-y|$ for which a sequence $(a_n)_{n\in\mathbb{N}}\in\mathbb{R}$ converges. You define a new metric
$$\rho'(x,y)=\cases{\frac{1}{\rho}\text{ if }x\neq y\\0\text{ if }x=y}$$ on the real numbers. You can check that $\rho'$ indeed is a metric. but you'll see that with $\rho'$ the sequence no longer converges. Another example is the discrete metric.
