Consider a metric space $(X,d)$ and let $\kappa$ be a cardinal. We say that $(X,d)$ is $\kappa$-**homogenous**, if every (surjective) isometry $h:X\_1 \to X\_2$ between subspaces of $(X,d)$ of size $< \kappa$ extends into an automorphism of $(X,d)$, i.e., into an isometry $f:(X,d) \to (X,d)$ such that $f\upharpoonright X\_1 = h$. For instance, the plane $\mathbb{R}^2$ is (at least) $\omega$-homogenous when equipped with the standard Euclidean metric.


**Question:** Does there exists a *compact* metric space $(X,d)$ which is $\omega$-homogenous but not $\omega\_1$-homogenous? In other words, does there exists a $\omega$-homogenous compact metric space $(X,d)$ and countably infinite sets $X\_1,X\_2 \subseteq X$ which are isometric via $h$ but $h$ does not extend into an automorphism of $(X,d)$?


My guess is that the answer should be yes, but I'm having a hard time coming up with a concrete example. Part of the problem is that I can't think of any homogenous compact metric spaces besides $S^{n-1}$.



$\omega$-homogenous space which is not $\omega_1$-homogenous
No: if $X$ compact metric is $\omega$-homogeneous then it is $\omega\_1$-homogeneous.


Indeed let $f:X\_1\to X\_2$ be a bijective isometry, with $X\_1$ countable. Write $X\_1$ as increasing union of finite subsets $F\_n$. So $f\_{|F\_n}$ extends to a bijective isometry $g\_n$ of $X$. Since the isometry group of $X$ is compact metrizable under uniform convergence, up to extract, we can suppose that $g\_n$ uniformly converges to some bijective isometry $g$. Then on each $F\_n$, $g\_n$ eventually equals $f\_{|F\_n}$, so $g$ extends $f\_{|F\_n}$ for each $n$. Thus $g$ extends $f$.


