What do you call this kind of metric space?

Question

Let $M$ be a metric space and let $S\subset M$. We define the $\varepsilon$-neighbouhood of $S$ as the set $U_\varepsilon = \{x\in M\ :\ d(x, S) < \varepsilon\}$, where the distance between a point and a set is defined as $d(x, S) = \inf\{d(x, s)\ :\ s\in S\}$.

If $B_r(x)\subset M$ is the open ball of radius $r$ centered around $x$, we have $U_\varepsilon(B_r(x))\subset B_{r+\varepsilon}(x)$. Is there a name or anything to be read on metric spaces that have the property $B_{r+\varepsilon}(x) = U_\varepsilon(B_r(x))\ \forall r > 0, \varepsilon > 0, x\in M$?

Edit: Not all spaces satisfy this condition. Consider the two point subset of the line $\{0,1\}$. We have

$$U_{\frac{1}{2}}(B_{1}(0)) = U_{\frac{1}{2}}(\{0\}) = \{0\} \neq B_{\frac{3}{2}}(0) = \{0,1\}$$

Motivation: I'm reading a Mikhael Gromov's paper on Groups of Polynomial Growth (http://www.numdam.org/item/?id=PMIHES_1981__53__53_0). On section 3 he defined a space with this property, which he names connectivity. I'm interested in learning more about this kind of space.

Answer 1

This is closely related to a class of metric spaces which I created, and termed "interpolation spaces" (unfortunately the name is taken by something in functional analysis already) but you won't find anything about this online yet, so I'll give the definition briefly:

A metric space $(M,d)$ is called an interpolation space if given any $x,y\in M$ and any $r\in\mathbb{R}$ such that $r> -d(x,y)$, we can find a $z\in M$ such that \begin{align} d(y,z) &= |r| \\ d(x,z) &= d(x,y)+ r \end{align}

Those last two conditions require that the triangle inequality is an equality for the 3 points, so these are spaces where you can find a point between and beyond two given points at a given distance, so that all $3$ points are "colinear".

These are related to your spaces (connectivity is a bad name in my opinion, seeing as these spaces do not have to be connected, but I will stick with it for now)

Every interpolation space has the connectivity property

The proof is rather straightforward, we need to show the inclusion (we can assume $\varepsilon < r$)

$$B_{r+\varepsilon}\subseteq U_{\varepsilon}(B_r(x)) $$

Thus let $$y\in B_{r+\varepsilon}(x)$$ but $$y\notin B_{r}(x)$$ then we can define $$ r_0:= - \frac{d(y,B_r(x))+\varepsilon}{2}> -\varepsilon$$ (This last inequality is not trivial, you should prove it from the interpolation property)

And then by the interpolation property there is some $z\in M$ such that

\begin{align} d(y,z) &= |r_0| < \varepsilon \\ d(x,z) &= d(x,y) - \frac{d(y,B_r(x))+\varepsilon}{2} < r \end{align}

(This last inequality follows since $d(x,y) = r + d(y,B_r(x))$, this too is a non-trivial consequence of the interpolation property)

Then $z\in B_r(x) $ and $y\in B_{\varepsilon}(z) $ and hence by the triangle inequality,

$$ y\in B_{r+\varepsilon}(x) $$

Which concludes the proof.

However, the two notions are not equivalent, the rationals, $\mathbb{Q}$ have connectivity, but they do not have the interpolation property. There are other nice properties possessed by interpolation spaces that these spaces do not possess as well, for example a closed ball is compact if and only if the boundary of every closed ball contained in the original closed ball, is compact.