Isometry into the Hilbert cube

Question

Let $(X,d)$ be a compact metric space. Does there exist a metric $d'$ on the Hilbert cube $H = [0, 1]^\mathbb{N}$, compatible with its topology, such that $(X,d)\hookrightarrow (H,d')$ is an isometric embedding?

(1). It is know that there exists topological embedding $(X,d)\rightarrow H$.

(2). If $X$ is compact then $X$ isometrically embeds into $C(X,\mathbb{R})$ (real valued continuous functions on $X$), where $C(X,\mathbb{R})$ is given supremum metric. The function is given by associating to each point $x_0\in X$, the map $\varphi_{x_0}(y)=d(x_0,y)$. See Kuratowski-Wojdysławski embedding.

But $C(X,\mathbb{R})$ is not even locally compact, but $H$ is.

I think answer is true.

Answer 1

By theorem of Hausdorff, every continuous metric on a closed subspace extends to a continuous metric on the whole space.

Thus, treating $(X, d)$ as a subspace of $H$, there exists a continuous metric $d'$ on $H$ such that $d'\restriction_{X\times X} = d$.

A continuous metric on a compact metrizable space generates its topology, see e.g. van Mill, appendix, exercise A.5.15.

Thus $d'$ is a metric on $H$ generating the same topology.

So for any compact metric space $(X, d)$ we can find a metric $d'$ on $H$ such that the topological embedding $(X, d)\to H$ can be made into an isometry $(X, d)\to (H, d')$.

• F . Hausdorff, Erweiterung einer Homöomorphie Fund. Math., 16 (1930), 353-360.

• J. van Mill, Infinite-dimensional topology of functions spaces