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Suppose two metrics $d_1$ ( Total variation distance) and $d_2$ (Wasserstein metric) defined on the same metric space. I don't know the definition of $d_1$ being stronger than $d_2$. I searched on google and only found things related to uniform metric.

More specifically, I'm interested in characterization via sequences. So

  1. What does it mean for $d_1$ to be stronger than $d_2$?
  2. What is true?: \begin{equation}\label{abc1}\tag{I} X_n \to_{d_2} X \implies X_n \to_{d_1} X \end{equation} or \begin{equation}\label{abc2}\tag{II} X_n \to_{d_1} X \implies X_n \to_{d_2} X \end{equation}
uniquesolution
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André Goulart
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    Do you know anything about topology? I think topology provides the most succinct explanation. – Ian Nov 25 '22 at 21:15
  • I know that metrics define topologies, but the reciprocal is not true. Having said that, which is true (I) or (II)? – André Goulart Nov 25 '22 at 21:27
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    It's (II), same concept as the induced topology of $d_1$ being stronger than the induced topology of $d_2$. – Ian Nov 25 '22 at 21:29

1 Answers1

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As you mentioned in a comment, a metric defines a topology. A topology $\tau$ on a set $X$ is a collection of sets (called open sets) that satisfy the axioms of topology. If you have two topologies $\tau_1,\tau_2$ on the same set, then $\tau_1$ is said to be stronger than $\tau_2$, if every member of the collection $\tau_2$ is also a member of the collection $\tau_1$. In other words - every set open in $\tau_2$ is also open in $\tau_1$. Another terminology often used is $\tau_1$ is finer than $\tau_2$.

In terms of the metric - if someone tells you that a metric $d_1$ is stronger than a metric $d_2$, then every open set in the metric $d_2$ is also open in the set $d_1$.

Now assume you have a sequence $\{x_n\}$ converging in the metric $d_1$, to some limit $x_0$. To prove that also $d_2(x_n,x_0)\to 0$, we must show that for every $\varepsilon>0$ there exists some $N$ such that for all $n>N$, $d_2(x_n,x_0)<\varepsilon$. Since any open $d_2$-ball around $x_0$ is also open in the $d_1$ metric, the set of point $\{x| d_2(x_n,x_0)<\varepsilon\}$ is open in the $d_1$ metric, hence contains some $d_1$-ball of some positive radius $r>0$ around $x_0$, and since $d_1(x_n,x_0)\to 0$ there exists some $N$ such that for all $n>N$ $d_1(x_n,x_0)<r$, and all these $x_n$'s are also contained in the $d_2$ ball around $x_0$ of radius $\varepsilon$.

uniquesolution
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