## Return to Answer

2 Fixed typos

A projective Hamel basis under V=L should be easy: Take a $\Delta_1^2$ \Delta^1_2$wellordering of$\mathbb R^\omega$and prove the existence of a Hamel basis using this wellordering. That will give you a projective Hamel basis low ($\Delta_1^2$?)$\Delta^1_2$?) in the projective hierarchy. Negative results are often proved by constructing sets without the Baire property or nonmeasurable sets (which then tells that the constructed set is at least$\Delta_2^1$) from your assumption. My guess would be that your suggested answer is true with n=2$, $n=2$, but I don't quite see how to construct a "weird" set from the Hamel basis yet.

By the way, the question what a basis for $\mathbb R$ over $\mathbb Q$ looks like has been studied quite a bit.

1

A projective Hamel basis under V=L should be easy: Take a $\Delta_1^2$ wellordering of $\mathbb R^\omega$ and prove the existence of a Hamel basis using this wellordering. That will give you a projective Hamel basis low ($\Delta_1^2$?) in the projective hierarchy.

Negative results are often proved by constructing sets without the Baire property or nonmeasurable sets (which then tells that the constructed set is at least $\Delta_2^1$) from your assumption. My guess would be that your suggested answer is true with n=2$, but I don't quite see how to construct a "weird" set from the Hamel basis yet. By the way, the question what a basis for$\mathbb R$over$\mathbb Q\$ looks like has been studied quite a bit.