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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.

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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 $s\mathbb R$.

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    $\begingroup$ @James-Hanson Now I am trying to write down the proof. It is rather long (more than 2 pages of a4 format). $\endgroup$ – Taras Banakh May 14 '19 at 5:35
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    $\begingroup$ @James-Hanson I have just sent you the proof of this theorem by the e-mail (written in your web-page). $\endgroup$ – Taras Banakh May 14 '19 at 9:10
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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 $\bigcap_{i<m} \{x \in S : d(x,D_i) \geq \frac{\varepsilon_i}{5}\}$ 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 $F = \bigcap_{i<m} \{x \in sS : f_{D_i}(x) \geq \frac{\varepsilon_i}{5}\}$. 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 $x \notin \overline{D}$. $\square$

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