Explicit Hamel basis of real numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:54:23Z http://mathoverflow.net/feeds/question/46063 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46063/explicit-hamel-basis-of-real-numbers Explicit Hamel basis of real numbers Buschi Sergio 2010-11-14T19:24:14Z 2010-11-17T03:36:22Z <p>Is there an explicit construction of a Hamel basis of the vector space of real numbers \$\mathbb R \$ over the field of rational numbers \$\mathbb Q \$?</p> http://mathoverflow.net/questions/46063/explicit-hamel-basis-of-real-numbers/46065#46065 Answer by Joel David Hamkins for Explicit Hamel basis of real numbers Joel David Hamkins 2010-11-14T19:47:07Z 2010-11-17T03:36:22Z <p>It is consistent with the axioms of ZFC that there is a Hamel basis of \$\mathbb{R}\$ over \$\mathbb{Q}\$ with complexity \$\Delta^1_2\$ in the descriptive set theoretic hierarchy. This is true, for example, in the constructible universe \$L\$, where there is a \$\Delta^1_2\$ well-ordering of the reals, as I explain in <a href="http://mathoverflow.net/questions/5303/basis-of-linfinity/8647#8647" rel="nofollow">this MO answer</a>, which is closely related to this question. Complexity \$\Delta^1_2\$ is a surprisingly low complexity, since such a set (and its complement) can both be obtained by starting with a certain closed set in \$\mathbb{R}^3\$, projecting it to \$\mathbb{R}^2\$, taking the complement, and projecting down to \$\mathbb{R}\$, and so it would seem to count as fairly explicit. </p> <p>Meanwhile, there can never be a Hamel basis of \$\mathbb{R}\$ over \$\mathbb{Q}\$ that is Borel, that is, with complexity \$\Delta^1_1\$, since from any Borel Hamel basis one can produce a non-Lebesgue measurable set of the same complexity by the Vitali argument (remove an element, take the span of the other elements, and consider its cosets). But of course every Borel set is Lebesgue measurable.</p> <p>At the same time, it is a consequence of the existence of large cardinals that every projective set of reals is Lebesgue measurable, and in this case, there can be no projective Hamel basis for \$\mathbb{R}\$ over \$\mathbb{Q}\$, again by the Vitali argument. The projective hierarchy of sets arises by closing the Borel sets under continuous images, as well as complements, countable unions and intersections. Thus, in such a situation, there can be no easily-described Hamel basis for \$\mathbb{R}\$ over \$\mathbb{Q}\$.</p> http://mathoverflow.net/questions/46063/explicit-hamel-basis-of-real-numbers/46066#46066 Answer by Pietro Majer for Explicit Hamel basis of real numbers Pietro Majer 2010-11-14T19:50:03Z 2010-11-14T21:07:40Z <p>If you have such a basis, you also have a subspace of co-dimension 1, and this turns out to be a <a href="http://en.wikipedia.org/wiki/Vitali_set." rel="nofollow">Vitali set</a>, that is quite a non-constructible object. For details, e.g. check <a href="http://mathoverflow.net/questions/32720/non-borel-sets-without-axiom-of-choice" rel="nofollow">this answer</a> and <a href="http://mathoverflow.net/questions/23202/explicit-big-linearly-independent-sets/23206#23206" rel="nofollow">this</a>. </p> http://mathoverflow.net/questions/46063/explicit-hamel-basis-of-real-numbers/46084#46084 Answer by Andreas Blass for Explicit Hamel basis of real numbers Andreas Blass 2010-11-14T23:05:09Z 2010-11-14T23:05:09Z <p>Arnie Miller has shown that if V=L then one can do a bit better than what Joel said; there will be a <code>\$\Pi^1_1\$</code> Hamel basis for the reals over the rationals. The reference is "Infinite combinatorics and definability," Annals of Pure and Applied Logic 41 (1989) 179-203. This paper also improves the complexity bound from <code>\$\Delta^1_2\$</code> to <code>\$\Pi^1_1\$</code> for several other constructions under the hypothesis V=L.</p>