Let $X$ be a compact metric space, and let $K\_X$ be the set of non-empty closed subsets of $X$, equipped with the $\sigma$-algebra
$$ \mathcal{B}(K\_X) \ := \ \sigma(\{C \in K\_X : C \cap U = \emptyset\} \, : \, \text{open } U \subset X ) . $$
(This is precisely the Borel $\sigma$-algebra of the Hausdorff metric.) Define the equivalence relation $\,\sim\,$ on $K\_X$ to be topological equivalence (i.e. $C\_1 \sim C\_2$ if and only if there exists a homeomorphism $\,h \colon C\_1 \to C\_2$), and let
$$ \pi \colon K\_X \to \tfrac{K\_X}{\sim} $$
be the natural projection.



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> Does there exist a standard $\sigma$-algebra on $\frac{K\_X}{\sim}$ such that $\pi$ is measurable?
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Can the set of compact metrisable topologies naturally be equipped with the structure of a standard Borel space?
In most cases the answer is no:
A much stronger result was proved in
J. Zielinski: The complexity of the homeomorphism relation between compact metric spaces, Adv. Math. 291 (2016), 635–645.
It is shown that the homeomorphism equivalence relation on compact subspaces of the Hilbert cube X=Q is Borel bireducible to the "universal orbit" equivalence relation and hence it can not be Borel reduced to the equality of points in a standard Borel space.


Also it is known that if $X$ is an uncountable compact metric space then the homeomorphism equivalence relation between its compact subspaces is complicated as well (as complicated as isomorphism relation of countable graphs). Hence the answer is no in this case too.


In the case $X$ is countably infinite compact metric space it is homeomorphic to a countable ordinal number and thus, there are only countably types of closed subspaces of $X$ up to homeomorphism. In this case the answer to your question is clearly positive.


