[FOM] Borel sets in ZF

[Harvey Friedman]

Solovay wrote in response to Shipman:

>
>  > > Choice is essential here, you need AC both to get a non-Borel set and to
>>  > get a set of reals of cardinality aleph-one.
>>  >
>	This is wrong. One can prove the existence of a non-Borel set in
>ZF [by essentially the Cantor diagonal arguement.]


Maybe Shipman is using the definition of Borel set of reals as the 
least sigma algebra containing the open sets of reals. Then every 
countable set of reals is a Borel set, and therefore every countable 
union of countable sets of reals is a Borel set, according to this 
definition. But it is consistent with ZF that some countable union of 
countable sets of reals is all reals. Hence it is consistent with ZF 
that every set of reals is the countable union of countable sets, by 
intersecting.

Come to think of it, what is a Borel set in ZF, officially? Of 
course, under the well founded tree definition, you can easily prove 
the existence of a non-Borel set in ZF, as Solovay pointed out.