Example of a Borel set that is neither FσF_\sigma nor GδG_\delta

I’m looking for subset A of R such that A is a Borel set but A is neither Fσ nor Gδ.

Answer

There are many examples. Here’s one:

Observe first that the rational numbers Q are an Fσ. This is because they are a countable union of points. The irrational numbers RQ=qQR{q} are thus a Gδ. Since both Q and RQ are dense and disjoint it follows from the Baire category theorem that Q cannot be a Gδ. [Edit: See also this thread here containing several proofs that Q can’t be a Gδ in R. These proofs explicitly avoid Baire].

The same reasoning shows that F=Q0 is an Fσ in [0,), but isn’t a Gδ, and that G=R0Q0 is a Gδ in (,0], but isn’t an Fσ. Their union FG is then an example of a Borel subset of R which is neither an Fσ nor a Gδ because if it were an Fσ then the same would hold for G=(FG)(,0), for example. I leave it as an exercise to show that FG is both an Fσδ and a Gδσ.

That’s probably the easiest example. A few more (both more interesting but also more involved ones) can be found in this MO thread.


For a much more in-depth discussion of such ideas, I recommend looking into one of the following books:

Specifically, look up the sections on the Borel hierarchy.

Attribution
Source : Link , Question Author : Babak Miraftab , Answer Author : Community

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