Is it true that if $G$ is a coanalytic family of separable Banach spaces, which is not Borel, and $H\subset G$ is not Borel, then H is coanalytic?
This is something I have come across. I am reading a proof that the class of Banach spaces with separable dual are coanalytic and not Borel, and the proof also says the class of separable reflexive spaces are also coanalytic but not Borel. However it only proves the non borelness of the separable reflexive spaces.
Edit: I'm not sure if the question was written too well. What I mean to say is, why was it sufficient to only show that the class of spaces with separable dual are conalytic? Does this imply that the class of reflexive spaces is coanalytic?
There are at most $2^{\aleph_0}$ subsets of a standard Borel space, or of a separable metric space. The same bound applies to coanalytic subsets. But usually, you have a lot more of non Borel subsets (for instance, in the case your family ${G}$ is uncountable).