Suppose $(X,d)$ is a separable metric space with $d\leq 1$, and let $(x_n)$ and $(x'_n)$ be two dense sequences. Define embeddings $$\varphi,\varphi':X\to[0,1]^\mathbb{N}$$ by putting $\varphi(x)=(d(x,x_n))_{n}$ and $\varphi'(x)=(d(x,x'_n))_{n}$ respectively. Then $\mathcal{X}=\overline{\varphi(X)}$ and $\mathcal{X}'=\overline{\varphi'(X)}$ are two metrizable compactifications of $X$.

It seems to me that $\mathcal{X}$ and $\mathcal{X}'$ always homeomorphic. In order to define a maps $\mathcal{X}\xrightarrow\theta\mathcal{X}'$ and $\mathcal{X}'\xrightarrow{\theta'}\mathcal{X}$ we define $\theta$ on $\varphi(X)$ as $\varphi'\circ\varphi^{-1}$ (I'm cutting corners here), and extend it by uniform continuity (and similarly for $\theta'$). I haven't checked the details, but this will produce (uniformly) continuous maps $\mathcal{X}\xrightarrow\theta\mathcal{X}'$ and $\mathcal{X}'\xrightarrow{\theta'}\mathcal{X}$ satisfying $\theta\circ\varphi=\varphi'$ and $\theta'\circ\varphi'=\varphi$ so that $\theta\circ\theta'$ coïncides with the identity function on the dense subset $\varphi(X)\subset\mathcal{X}$, and $\theta'\circ\theta$ coïncides with the identity function on the dense subset $\varphi'(X)\subset\mathcal{X}'$. And thus we should get that $\theta$ and $\theta'$ are inverse homeomorphisms of one another.

Question : Is the above reasoning sound, and if so, what is the name of this compactification ? Does it have a universal property ? Maybe something similar to the Stone-Cech compactification but in the category of compact metric spaces ?


Neat question! I haven't seen this construction before, but here are some thoughts.

First, I think separability of $X$ (and subsequently metrizability of $\mathcal{X}$) is really a red herring here. For any bounded metric space $(X,d)$, you could construct a compactification in the same way, taking the closure $\mathcal{X}$ of the image of the map $X\to [0,N]^X$ that sends $x$ to the function $d(x,-)$ (where $N$ is some upper bound on the metric). Equivalently, you could just take closure of the image of $X\to [0,N]^Q$ where $Q$ is any dense subset of $X$. In the case that $X$ is separable, you can choose $Q$ to be countable, and so $\mathcal{X}$ will be metrizable, but this is not essential to what is going on here.

Here are some other ways to describe this compactification $\mathcal{X}$. It is the smallest compactification of $X$ to which the function $d(x,-)$ extends continuously for each $x\in X$. Alternatively, it is the spectrum of the closed subalgebra of $C_b(X)$ generated by the functions $d(x,-)$ for each $x\in X$ (these functions form a subspace of $C_b(X)$ which is canonically isometric to $X$).

Note that this compactification is highly sensitive to the choice of metric $d$, and not just how $d$ behaves at small scales. For instance, consider the discrete space $X=\mathbb{N}$ with a metric $d$ such that $d(x,y)$ is always either $1$ or $2$ if $x\neq y$. The points of $\mathcal{X}\setminus X$ correspond to all functions $\mathbb{N}\to\{1,2\}$ that are pointwise accumulation points of the functions $d(x,-)$. So, by choosing $d$ appropriately, you can arrange for $\mathcal{X}\setminus X$ to be any nonempty closed subspaces of $\{1,2\}^\mathbb{N}$. (This takes a bit of work to prove, since you have to be careful about the restriction that $d$ is symmetric. Basically, you can choose $d$ freely "below the diagonal" of $\mathbb{N}\times\mathbb{N}$, and that is enough to control the accumulation points of the functions $d(x,-)$.)

  • 1
    $\begingroup$ I think it also may be sensitive to the choice of metric on the Hilbert cube $[0,1]^\mathbb{N}$. $\endgroup$ – Nate Eldredge Jan 12 '18 at 21:51
  • $\begingroup$ No, you can just use the product topology on $[0,1]^\mathbb{N}$ to define the closure. Of course, if you want to endow the compactification with a metric in the case that $X$ is separable, that metric will depend on what metric you use on $[0,1]^\mathbb{N}$ (and also probably on your choice of countable dense subset). $\endgroup$ – Eric Wofsey Jan 12 '18 at 21:54
  • $\begingroup$ Thank you for your answer ! Please excuse my taking nearly a week to acknowledge it. Do you think there will be some universal property characterizing $\mathcal{X}$ ? Can you think of criteria that will ensure that $X$ is an open subset of $\mathcal{X}$ ? Could you elaborate your last claim that any closed subset of $\{1,2\}^\mathbb{N}$ can be realized this way ? You are right that the set up is superfluous, and one can work directly with arbitrary dense subsets of $X$, or with $X$ directly. $\endgroup$ – Olivier Bégassat Jan 17 '18 at 13:21
  • $\begingroup$ I doubt there is any nice universal property. Here's a not-so-nice one: a map $f:X\to Y$ for compact Hausdorff $Y$ extends (uniquely) to $\mathcal{X}$ iff for every $g:Y\to\mathbb{C}$, $g\circ f$ can be uniformly approximated by sums and products of the functions $d(x,-)$ and constant functions. $X$ is open in $\mathcal{X}$ iff $X$ is locally compact; this is true more generally for any compactification (assuming all spaces are Hausdorff). $\endgroup$ – Eric Wofsey Jan 17 '18 at 18:01
  • $\begingroup$ Given any nonempty closed subset $A\subset\{1,2\}^\mathbb{N}$, there exists a sequence $(s_n)$ of points of $\{1,2\}^\mathbb{N}$ such that the set of accumulation points of $(s_n)$ is exactly $A$. Now define $d(n,m)=s_n(m)$ whenever $n>m$ (which by symmetry determines $d$). For each $n$, then, the function $d(n,-)$ agrees with $s_n$ for its first $n$ inputs, which for $n$ large means that $d(n,-)$ is getting very close to $s_n$. You can then verify that this implies the set of accumulation points the functions $d(n,-)$ is the same as the set of accumulation points of $(s_n)$. $\endgroup$ – Eric Wofsey Jan 17 '18 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.