# Convergence in probability only depends on topology?

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?

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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.
Indeed, that could be a good place to point out that a.s convergence is not topological (i.e. there is no topology on RV yielding it). This is because for any topological convergence we have that $x_n \to x$ if and only if any subsequence has a subsubsequence converging to $x$ (which is not the case for a.s. convergence). – Ori Gurel-Gurevich Nov 29 '11 at 19:14