Descriptive complexity of Hamel bases of R^ω - MathOverflow most recent 30 from http://mathoverflow.net2013-05-19T15:12:47Zhttp://mathoverflow.net/feeds/question/39359http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/39359/descriptive-complexity-of-hamel-bases-of-rDescriptive complexity of Hamel bases of R^ωRicky Demer2010-09-20T05:07:59Z2010-09-20T17:36:58Z
<p>(base theory = ZFC)</p>
<p>Are any Hamel bases for the vector space $\mathbb{R}^{\omega}$ in the
<br><br>1. analytical hierarchy?<br>2. projective hierarchy?</p>
<p>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?</p>
<p><br></p>
<p>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<br>"1. no; 2. none are below $\Delta^1_n$, if V=L then they are in $\Delta^1_n$, if projective determinacy then no"<br>with n a small natural number explicitly known but not to me.</p>
http://mathoverflow.net/questions/39359/descriptive-complexity-of-hamel-bases-of-r/39367#39367Answer by Stefan Geschke for Descriptive complexity of Hamel bases of R^ωStefan Geschke2010-09-20T07:57:35Z2010-09-20T16:15:07Z<p>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. </p>
<p>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.</p>
<p>By the way, the question what a basis for $\mathbb R$ over $\mathbb Q$ looks like has been studied quite a bit.</p>