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Given a set of points $S$ in some metric space, a pair of points $x, x'$ will be termed $\epsilon$-connected if they are connected by a series of points $x_1, \ldots, x_m \in S$ such that $d(x, x_1)$, $d(x_1, x_2), \ldots, d(x_{m-1}, x_m)$, $d(x_m, x')$ are all at most $\epsilon$.

Suppose $n$ points are drawn from a fixed probability distribution $\mathcal{D}$ on some metric space. I would like to say that $D$ is [SOMETHING] if for any fixed pair of points $p, p'$ and any $\epsilon>0$ as $n \to \infty$ the probability that $p$ and $p'$ will be $\epsilon$-connected goes to 1.

Is there a common term for this type of thing?

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  • $\begingroup$ I will modify your definition of $\epsilon$-connected here to allow the bound to be saturated. Given $d$, let $u$ be the maximal subdominant ultrametric induced by $d$ (see VI.C of dx.doi.org/10.1103%2FRevModPhys.58.765). Then $x$ and $x'$ are $\epsilon$-connected w/r/t $d$ iff $u(x,x') \le \epsilon$. $\endgroup$
    – Steve Huntsman
    May 21, 2014 at 18:00
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    $\begingroup$ This doesn't really have anything to do with the distribution. Assuming every open set has positive measure, then as $n\to\infty$ you will with probability going to 1 choose points arbitrarily close to any particular finite set of points, so if $p$ and $p'$ are $\epsilon$-connected with respect to the entire space then they will be (say) $2\epsilon$-connected with respect to your $n$ points. $\endgroup$
    – Eric Wofsey
    May 21, 2014 at 23:41
  • $\begingroup$ I don't know of a name for the underlying property of metric spaces, though, and it seems like an interesting property. It is implied by connectedness and equivalent to it if the space is compact, but it also holds for spaces like $\mathbb{Q}$ or a union of two asymptotic curves in $\mathbb{R}^2$. $\endgroup$
    – Eric Wofsey
    May 22, 2014 at 0:04
  • $\begingroup$ Of course this happens if every set has positive measure. But it isn't true in many important cases. e.g. If the support of the distribution is a union of two disjoint balls. $\endgroup$
    – Amit Moscovich
    May 22, 2014 at 11:16
  • $\begingroup$ If a nonempty open set has measure zero, then obviously your condition will fail if you choose $p$ to be in that open set. If you restrict $p$ and $p'$ to lie in the support of the distribution, then you may as well replace your metric space with the support and then my argument still applies. Either way, it only depends on the support, not the distribution itself. $\endgroup$
    – Eric Wofsey
    May 22, 2014 at 22:08

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