Let $(M,d)$ be a metric space, $\varepsilon>0$ be a real number. Fix $p \in M$ and denote
$$
U_\varepsilon(p) = \left\{x \in M : d(p,x) < \varepsilon \right\}.
$$
Now define the order $\leq$ on the set $K = \left\{ U_{\varepsilon}(p) : \varepsilon\in \mathbb{R} \wedge \varepsilon>0 \right\}$ as the following:
$$
U_{\varepsilon_1}(p) \leq U_{\varepsilon_2}(p) \quad\Longleftrightarrow\quad U_{\varepsilon_1}(p) \subseteq U_{\varepsilon_2}(p).
$$
I'm wondering if this order on $K$ will have the same order type as the natural order on the positive reals.
If yes, then each point on an arbitrary metric space would induce an order on the set of all open-balls around it that is order-isomorphic to the order of the positive reals.
Thanks
For a general metric space this is not true.
For example, let $M=[0,1]$, with the usual metric, and let $K$ be the collection of open balls you defined, around the point 0.
Then, since $U_{\epsilon_1}(0)=U_{\epsilon_2} (0)$ for all $\epsilon_1,\epsilon_2>1$, and since containment is strict only when the radius is smaller than $1$, it follows that the order type of $K$ is the same as the half-open interval $(0,1]$ (which is not the same as $(0,\infty)$ since it has a largest element).
However, you will always get a subset of $(0,\infty)$ for the order type.
An interesting question might be: which order types are realizable in this way? One can definitely get a disconnected thing (for example, in the case of $M=\{0,1\}$, where $K$~$\{0,1\}$, or $X=[0,1] \cup [2,3]$, where $K$~$[0,1]\cup[2,3]$). However, I don't think that $\mathbb{Q}\cap (0,\infty)$ is realizable like this (just a hunch).