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Given a radius $r > 0$, the internal covering number of a subset $T$ of a metric space $(X, d)$ is denoted $N_r(T)$ and is defined to be the smallest number of balls of radius $r$ (under $d$) with centres in $T$ such that $T$ is contained in the union of the balls.

Given another subset of $X$, $U$, which is a superset of $T$, it is not necessarily true that $N_r(T) \leq N_r(U)$.

My question:

  1. Are there well known examples of sets for which $T \subseteq U$ but $N_r(T) > N_r(U)$?
  2. Are there necessary/sufficient conditions on $X$ or $d$ such that the internal covering number is monotonic, i.e. $T \subseteq U \implies N_r(T) \leq N_r(U)$?

In case it is relevant, my application is to cases in which $X$ is generated by $T$ under some (infinite) set of transformations (e.g. a Lie group).

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    $\begingroup$ Do you mean balls of a particular radius? Otherwise the internal covering number of a nonempty bounded set is $1$ and the internal covering number of an unbounded set is $\infty$. $\endgroup$
    – Robert Israel
    Commented Apr 1, 2020 at 18:19
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    $\begingroup$ An easy example: if $U$ is the Euclidean $n$-dimensional punctured disc of radius $r$, and $T$ the punctured disk, then $N_r(T) = n + 1$ while $N_r(U) = 1$. For general $X$, this idea implies that for any $x \in X, r \in \mathbb{R}_{>0}$ there is some $x_r$ such that $d(x, y) < r \implies d(x_r, y) < r$ (or $\leq r$, depending on whether you mean the open or closed ball). $\endgroup$
    – user44191
    Commented Apr 1, 2020 at 18:53
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    $\begingroup$ @user44191, a very clean example, nice -- punctured contained if the full (unpunctured) disk. (Your first word "punctured" was a typo). $\endgroup$
    – Wlod AA
    Commented Apr 1, 2020 at 19:09
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    $\begingroup$ Ultrametric spaces satisfy 2. (en.wikipedia.org/wiki/Ultrametric_space) $\endgroup$
    – Anton Petrunin
    Commented Apr 1, 2020 at 19:21
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    $\begingroup$ Yes it seems that only ultrametric spaces satisfy condition 2 --- it sufficient to check 2-point sets in 3-point sets. $\endgroup$
    – Anton Petrunin
    Commented Apr 4, 2020 at 0:54

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The most obvious example: Suppose $U$ is a closed ball of radius $1$ in $\mathbb R^d$, and $T$ is the corresponding sphere. Then if $r = 1$, $N_r(U) = 1$ but $N_r(T) > 1$.

EDIT: Let $X$ be any metric space such that there are three points $a,b, c$ with $d(a,b) \le d(a,c) < d(b,c)$. Then take $d(a,c) \le r < d(b,c)$, $T = \{b,c\}$ and $U = \{a,b,c\}$. We have $N_r(U) = 1$ but $N_r(T) = 2$. Asking that this example does not exist is a rather severe restriction on the metric space!

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    $\begingroup$ I think that that restriction is equivalent to the space being ultrametric, as said in the comments to the question. Examples include p-adic spaces. $\endgroup$
    – user44191
    Commented Apr 2, 2020 at 17:28

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