# Which points in the Samuel compactification of a metric space $X$ are limits of uniformly discrete subsets of $X$?

Given a metric space $$(X.d)$$ the Samuel compactification of $$X$$, written $$sX$$, is the unique compactification with the property that if $$Y$$ is an arbitrary compact Hausdorff space and $$f:X\rightarrow Y$$ is a uniformly continuous map (with the unique compatible uniformity on $$Y$$) then $$f$$ factors through a unique map $$\overline{f}:sX\rightarrow Y$$. One way to construct this is to let $$I=U(X,[0,1])$$ be the collection of all uniformly continuous functions from $$X$$ to $$[0,1]$$ and then let $$sX$$ be the topological closure of the natural image of $$X$$ in $$[0,1]^I$$. Another way is as the Gelfand spectrum of the $$C^*$$-algebra of bounded uniformly continuous functions on $$X$$.

A subset $$Q\subseteq X$$ is uniformly discrete if there is an $$\varepsilon>0$$ such that for any $$x,y\in Q$$ with $$x\neq y$$, $$d(x,y) > \varepsilon$$.

So broadly the question is this:

Let $$(X,d)$$ be a complete metric space. Under what conditions is it true that for every $$x \in sX\setminus X$$ there is a uniformly discrete set $$Q\subseteq X$$ such that $$x\in \overline{Q}$$?

I believe this is true for $$\mathbb{R}^n$$ and any space that is 'uniformly locally compact' in the sense that there is some $$\varepsilon>0$$ such that every closed $$\varepsilon$$-ball is compact and for every $$\delta>0$$ there is an $$n$$ such that every closed $$\varepsilon$$-ball can be covered by at most $$n$$ open $$\delta$$-balls. On the other hand I anticipate difficulty with spaces that are not locally compact or just not uniformly locally compact.

A metric space $$X$$ is called isometrically homogeneous if for any points $$x,y\in X$$ there exists a bijective isometry $$f:X\to X$$ such that $$f(x)=y$$.

For isomemtrically homogeneous spaces this problem has the following answer (predicted by James Hanson).

Theorem. An isometrically homogeneous complete metric space $$X$$ is locally compact if and only if every $$x\in sX$$ is contained in the closure of some uniformly discrete set $$D\subset X$$ in $$sX$$.

The proof of this theorem is a bit long (2 pages). So, I will write it down as a paper, post it to the arXiv and will add a link.

Corollary. Every point $$x\in s \mathbb R^n$$ is contained in the closure of some uniformly discrete set $$D\subset \mathbb R^n$$ in $$s\mathbb R$$.

• @James-Hanson Now I am trying to write down the proof. It is rather long (more than 2 pages of a4 format). – Taras Banakh May 14 '19 at 5:35
• @James-Hanson I have just sent you the proof of this theorem by the e-mail (written in your web-page). – Taras Banakh May 14 '19 at 9:10

The statement is false in any unit sphere of an infinite dimensional Banach space. A compactness argument together with the form of Dvoretsky's theorem stated here (theorem 1.2), gives:

Proposition 1: For any $$\varepsilon > 0$$ and $$n<\omega$$ there is a $$K(\varepsilon, n)$$ such that, for any Banach space $$X$$ of dimension $$\geq K(\varepsilon ,n)$$ with unit sphere $$S$$ and any $$1$$-Lipschitz $$f: S \rightarrow [0,2]$$, there is an $$n$$-dimensional subspace $$Y \subseteq X$$ such that for all $$x,y \in Y\cap S$$, $$|f(x)-f(y)| < \varepsilon$$.

Proof: Assume that the statement is false for some $$\varepsilon >0$$ and $$n<\omega$$. Then for every sufficiently large $$k$$ we can find a Banach space $$X_k$$, with unit sphere $$S_k$$, and a $$1$$-Lipschitz function $$f_k : S_k \rightarrow [0,2]$$ such that for every $$n$$ dimensional subspace $$Y$$ of $$X_k$$, there are $$x,y \in Y \cap S_k$$ such that $$|f(x)-f(y)|\geq \varepsilon$$. Take an ultraproduct of the sequence $$(X_k, f_k)$$ to get $$(X,f)$$, with $$X$$ the Banach space ultraproduct and $$f$$ defined in the obivous way. You can check that $$f$$ is well-defined, $$1$$-Lipschitz, and takes values in $$[0,2]$$. It will also be infinite dimensional and therefore give a counterexample to theorem 1.2 in the reference (a certain form of Dvoretzky's theorem). $$\square$$

($$[0,2]$$ is just the range of the metric when restricted to the unit sphere of an infinite dimensional Banach space.)

Proposition 2: Let $$S$$ be the unit sphere of an infinite dimensional Banach space $$X$$ and let $$sS$$ be its Samuel compactification. There is an $$x \in sS$$ that is not in the closure of any uniformly discrete subset of $$S$$.

Proof: Let $$D\subset S$$ be a uniformly discrete subset that is $$(>\varepsilon)$$-separated. Assume without loss that $$\varepsilon < 1$$. By proposition 1, for any $$n\geq 2$$, for any subspace $$Y\subseteq X$$ of dimension at least $$K(\frac{\varepsilon}{5}, n)$$, there is a subspace $$Z \subseteq Y$$ of dimension $$n$$ such that for any $$x,y \in Z \cap S$$, $$|d(x,D)-d(y,D)|< \frac{\varepsilon}{5}$$.

Now suppose for the sake of contradiction that there is $$x \in Z \cap S$$ such that $$d(x,D) < \frac{\varepsilon}{5}$$. Let $$u \in D$$ be such that $$d(x,u) < \frac{\varepsilon}{5}$$. By the reverse triangle inequality, $$d(-x,u) > 2-\frac{\varepsilon}{5} > \frac{\varepsilon}{2}$$, so since $$n\geq 2$$, $$Z \cap S$$ is connected and thus there is a $$y \in Z \cap S$$ such that $$d(y,u) = \frac{\varepsilon}{2}$$. By construction $$d(y,D) < \frac{\varepsilon}{5} + \frac{\varepsilon}{5}$$, so there must be some $$v\in D \setminus \{u\}$$ such that $$d(y,v) < \frac{2\varepsilon}{5}$$, but this implies that $$d(u,v) < \frac{\varepsilon}{2} + \frac{2\varepsilon}{5} < \varepsilon$$, which contradicts that $$D$$ is $$(>\varepsilon)$$-separated. Therefore $$d(x,D) \geq \frac{\varepsilon}{5}$$ for every $$x p\in Z \cap S$$.

Therefore in particular, for any $$(>\varepsilon)$$-separated set $$D \subset S$$, there are arbitrarily large subspaces $$Y \subseteq X$$ such that $$Y\cap S \subseteq \{x \in S : d(x,D) \geq \frac{\varepsilon}{5}\}$$.

This implies, again by proposition $$1$$, that for any finite sequence $$D_0,D_1,\dots,D_{m-1}$$ uniformly discrete sets which are $$(>\varepsilon_0)$$-, $$(>\varepsilon_1)$$-, ..., and $$(>\varepsilon_{m-1})$$-separated, respectively, the set $$\bigcap_{i is non-empty.

In the Samuel compactification $$sS$$, for any uniformly discrete set $$D$$, the function $$d(x,D)$$ has a unique continuous extension $$f_D$$ to $$sS$$ and in particular the closure $$\overline{D}$$ is precisely the zeroset of the function $$f_D$$.

So now for the same sequence of uniformly discrete sets as above, consider the closed set $$F = \bigcap_{i. We already showed that this set is non-empty and by the above statement about $$\overline{D}$$, $$F$$ is disjoint from each of the sets $$\overline{D}$$.

By compactness of $$sS$$, this implies that the intersection $$\bigcap\{\{x \in sS : f_{D}(x) \geq \frac{\varepsilon}{5}\}:D\subset S \text{ }(>\varepsilon)\text{-separated} \}$$ is non-empty and so contains some $$x$$ such that for any $$(>\varepsilon)$$-separated set $$D \subset S$$, $$f_{D}(x) \geq \frac{\varepsilon}{5}$$, so in particular $$x \notin \overline{D}$$. $$\square$$