[FOM] Borel sets in ZF

Robert M. Solovay solovay at math.berkeley.edu
Wed Sep 11 17:11:28 EDT 2002

```Thanks. I stand corrected.

--Bob

On Wed, 11 Sep 2002, Harvey Friedman wrote:

> 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.
>
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```

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