Suppose $(S,d)$ is a Polish space, and $X$, $(X\_n)$ are random variables such that $X\_n \to X$ in probability in $(S,d)$. Now suppose $d'$ is another metric on $S$, giving the same topology. Does $X\_n \to X$ in probability in $(S,d')$?


I believe the answer is yes, and the following is a sketch proof. Since $S$ is Polish, each of the laws of $X$, $X\_n$ is tight (or "Radon") so for each $n$ there is a compact $K \subset S$ such that both $X$ and $X\_n$ are in $K$ with high probability. The metrics $d$ and $d'$ must be uniformly equivalent on $K$ (since they are continuous with respect to each other), and so making $X$ and $X\_n$ close under $d$ is the same as making them close under $d'$. By the convergence in probability hypothesis, we can do this.


If my reasoning is right, this must be something very standard. Can someone confirm the result for me?



Convergence in probability only depends on topology?
A sequence converges in probability to, say, $f$ iff every subsequence has a further subsequence that converges almost surely to $f$. This second condition is independent of the metric that gives the topology, hence so is the first. 


