1
$\begingroup$

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).

$\endgroup$
7
  • 1
    $\begingroup$ 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 $\infty$. $\endgroup$ Apr 1 '20 at 18:19
  • 2
    $\begingroup$ 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
    Apr 1 '20 at 18:53
  • 1
    $\begingroup$ @user44191, a very clean example, nice -- punctured contained if the full (unpunctured) disk. (Your first word "punctured" was a typo). $\endgroup$
    – Wlod AA
    Apr 1 '20 at 19:09
  • 3
    $\begingroup$ Ultrametric spaces satisfy 2. (en.wikipedia.org/wiki/Ultrametric_space) $\endgroup$ Apr 1 '20 at 19:21
  • 1
    $\begingroup$ Yes it seems that only ultrametric spaces satisfy condition 2 --- it sufficient to check 2-point sets in 3-point sets. $\endgroup$ Apr 4 '20 at 0:54
2
$\begingroup$

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!

$\endgroup$
1
  • 1
    $\begingroup$ 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
    Apr 2 '20 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.