Could a non-separable metric space (uncountable) be Borel-isomorphic to a compact Polish space (uncountable too)?

Or, put differently, what would be sufficient conditions on this non-separable metric to be Borel isomorphic to a compact Polish space?


No, it cannot. Indeed, assume that $X$ is a non-separable metric space and $f:X\to K$ is a Borel isomorphism between $X$ and a Polish compact space $K$. Since the space $X$ is non-separable and metric, it contains an uncountable closed discrete (even $\varepsilon$-separated for some $\varepsilon>0$) subset $A$. Then $f(A)$ is an uncountable Borel subset of the space $K$. The Perfect Set Theorem for Borel Sets (Alexandrov, Hausdorff [13.6, Kech]) claims each Borel subset of a Polish space either is countable or else it contains a Cantor set. Thus $f(A)$ contains a Cantor set $C$. Let $B=f^{-1}(C)$. Since each subset of $B$ is Borel (in $X$) then each subset of $C$ is Borel (in $C$), which is a contradiction.


[Kech] Alexander S. Kechris. Classical Descriptive Set Theory , Springer, 1995.

  • $\begingroup$ Thank you! May I only ask about the last sentence: " Since each subset of $B$ is Borel (in $X$) then each subset of $C$ is Borel (in $C$), which is a contradiction." Why each subset of $B$ is then Borel? Also, do you mean that "...then each subset of $C$ is Borel in $K$"? (not "in $C$") $\endgroup$ – Kass Aug 10 '16 at 13:06
  • $\begingroup$ @kaktus $B$ is a closed discrete subset of the space $X$, so each subset of $B$ is closed discrete (and hence Borel) in $X$. To obtain the contradiction, it suffices to consider subsets of $C$ and it is more natural than to consider subsets of $K$, because it is clear what is $C$. $\endgroup$ – Alex Ravsky Aug 10 '16 at 15:58

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