When is the internal covering number of a metric space monotonic?

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).

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

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!