End point compactification for metric spaces

Question

Freundenthal introduced ends of topological spaces and the end point compactification of locally compact topological spaces adding one point for each end of the topological space (see here). For example the end point compactification of $\mathbb{R}$ is homeomorphic to the unit interval $[0,1]$ because "$\mathbb{R}$ has two ends".

If you take the set $A=\{ (x, 1/x)\ |\ x>0 \}\cup\{(x, 0)\ |\ x\in\mathbb{R}\}$, it is homeomorphic to $\mathbb{R}\sqcup\mathbb{R}$ so its end point compactification will be homeomorphic to $[0,1]\sqcup[0,1]$.

But seeing $A$ as a metric space one wants to say that "the two ends at the right are the same" and to compactify $A$ with only three ends, giving a connected topological space homeomorphic to $[0,1]$.

Is there in the litterature such a notion of "metric end point compactification" which would compactify $A$ with only three ends?

(I’m not asking how to define such a compactification, I already have a definition which seems to work for my purpose, I just want to know if something like that is already known)

Thank you

Edit: Here is a more precise definition of the compactification I need (I’m not sure this is exactly the correct definition, but it should be something like that): let’s say that a metric space $(E,d)$ is 0-connected if for all $\varepsilon>0$ and $x,y\in E$, there exists a finite sequence $(u_0, \dots, u_n)$ such that $u_0 = x$, $u_n = y$ and $d(u_i,u_{i+1})<\varepsilon$ for all $0\leqslant i\leqslant n-1$ (every connected space is 0-connected, but the space $A$ introduced before is an example of space which is non connected but 0-connected) (by the way, if this notion of 0-connectedness has already a name in the litterature, I will be happy to know it).

Then, take the definiton of the Freudenthal end point compactification (see link above) but replace "$U_i$ is a connected component of …" by "$U_i$ is a 0-connected component of …".

In particular $A$ has three ends, and the complex plane or the hyperbolic space have only one end.

Answer 1

I think what you are looking for is boundary at infinity. For example, the boundary at infinity of the hyperbolic plane $\mathbb H^2$ is its "ideal boundary" circle, and adding it to $\mathbb H^2$ yields a topological disc.

There are many (non-equivalent) definitions of the term to choose from, depending on how weird your spaces are and what asymptotic properties you want to capture. Probably the most "classic" one is Gromov's, which works perfectly for your example.

Fix a point $o\in A$ (the "origin", on which the result does not depend). To each point $p\in A$, associate a function $f_p:A\to\mathbb R$, defined as follows: $$ f_p(x) = dist(x,p)-dist(o,p) $$ The map $p\mapsto f_p$ is a quite reasonable representation of $A$ in the space of (1-Lipschitz, fixed at the origin) functions on $A$. The desired compactification is the closure of the range $\{f_p:p\in A\}$ in the compact open topology. It is indeed compact if $A$ is a proper metric space (i.e., all closed balls are compact).

The functions that are in the closure but not in the original set form the "boundary at infinity" of $A$. They are similar to Busemann functions of geodesic rays (and actually are Busemann functions if $A$ is a complete Riemannian manifold).

Answer 2

Another possibility is to use proximities - or equivalently (totally bounded) uniformities: in the metric case one defines $A$ and $B$ to be 'close' (usually denoted $A\mathrel\delta B$) if $d(A,B)=0$. The relation $\delta$ satisfies the proximity axioms and it determines a compactification $\tilde X$ of the space $X$ with the property that for subsets $A$ and $B$ of $X$ one has $A\mathrel\delta B$ iff the closures of $A$ and $B$ in $\tilde X$ intersect. The Freudenthal compactification comes from the proximity defined by declaring $A$ and $B$ to be 'far apart' iff they have disjoint open neighbourhoods $U$ and $V$ respectively such that $X\setminus (U\cup V)$ is compact, that is, they are separated by a compact set.