# Isometry into the Hilbert cube

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.

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