Does every compact metric space continuously embed into a Hilbert space (possibly with large distortion)?

4$\begingroup$ All compact metric spaces embeds into the Hilbert cube, which itself embeds into the Hilbert space. $\endgroup$ – Aleksei Kulikov Jun 15 at 22:57

$\begingroup$ Thanks! Reference? $\endgroup$ – Aryeh Kontorovich Jun 15 at 23:06

3$\begingroup$ Honestly, both are Wikipedia:) but the second one is an explicit construction while the first one is: wlog the diameter of $K$ isless than $1$; choose countable dense set $x_1, x_2, \ldots$ and map $x$ to $(d(x, x_1), d(x, x_2), \ldots)$. This is a continuous injection from compact set into Hilbert cube, hence homeomorphism. $\endgroup$ – Aleksei Kulikov Jun 15 at 23:10

$\begingroup$ @AlekseiKulikov This seems to only require separability and boundedness  is compactness really necessary? $\endgroup$ – Aryeh Kontorovich Jun 16 at 6:47

3$\begingroup$ @AryehKontorovich Compactness is not needed, actually every Polish space embeds into the Hilbert cube (even more: a space is Polish iff it is homeomorphic to a $G_\delta$ subset of the Hilbert cube). Boundedness is irrelevant because any metric space is homeomorphic to a bounded metric space $\endgroup$ – Alessandro Codenotti Jun 16 at 8:23
You may also send $X$ into a space $L_2(X,\mu)$ via the FréchetKuratowski isometry $x\mapsto d(\cdot,x)\in C^0(X),\\cdot\_\infty$, followed by the bounded linear inclusion $C^0(X)\to L_2(X,\mu)$, where $\mu$ is a probability measure on $X$. If $\text{supp}(\mu)=X$ (e.g. $\mu$ is a series of deltas $\sum_{k=1}^\infty2^{k}\delta_{q_k}$ for a dense set $\{q_k\}_{k\ge1}$), then $C^0(X)\to L_2(X,\mu)$ is injective, and the composition $j:X\to L_2(X,\mu)$ is a continuous embedding (a homeo of $X$ with $j(X)$).

$\begingroup$ this appear to be the "0 variant" of the answer already given in comments. There must be trivial objects too, anyway $\endgroup$ – Pietro Majer Jun 15 at 23:40