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(base theory = ZFC)

Are any Hamel bases for the vector space $\mathbb{R}^{\omega}$ in the

1. analytical hierarchy?
2. projective hierarchy?

In any of the above cases where the answer is not simply "no", is anything known about what levels they are or can be in?


My knowledge of descriptive set theory is basically just what's on wikipedia, so I probably won't know other theorems even if they are proved in every textbook on the subject. However, I suspect the answers will be
"1. no; 2. none are below $\Delta^1_n$, if V=L then they are in $\Delta^1_n$, if projective determinacy then no"
with n a small natural number explicitly known but not to me.

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1 Answer 1

up vote 4 down vote accepted

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.

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Stefan, your first few $\Delta^2_1$s should be $\Delta^1_2$. –  Joel David Hamkins Sep 20 '10 at 10:42
    
@Joel: You are right. I am editing this. –  Stefan Geschke Sep 20 '10 at 16:14
    
@Torsten Ekedahl: Yes, a basis for $\mathbb R$ as a $\mathbb Q$-vector space gives you a homomorphism from the reals onto $\mathbb Q$ and hence a nonmeasurable set. But here we are looking for a Hamel basis of the vector space $\mathbb R^\omega$ over $\mathbb R$, if I understand the question correctly. You don't get a homomorphism from $\mathbb R$ onto $\mathbb Q$ that is somehow easily definable from the Hamel basis in this situation, right? –  Stefan Geschke Sep 20 '10 at 16:24

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