Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\Omega$ be the halting probability (see (http://en.wikipedia.org/wiki/Chaitin's_constant) and R. Downey, and D. Hirschfeldt (2010), Algorithmic Randomness and Complexity for reference). If A is Martin-Löf random with respect to $\Omega$ (so A is 2-random), does A compute an infinite set $G\subseteq \Omega$ or $G\subseteq 2^\omega-{\Omega}$ ? This was motivated by the idea of constructing models using random sets to avoid $SRT_2^2$ in the reverse mathematics context.

share|improve this question
    
Your link got cut off I believe. –  Jason Rute Jun 12 '13 at 18:04
    
Thanks! Just fixed it. –  Zhang Jing Jun 13 '13 at 1:54

1 Answer 1

up vote 6 down vote accepted

No. Since $A$ is ML-random relative to $\Omega$, by van Lambalgen's theorem it follows that $\Omega$ is ML-random relative to $A$. If $A$ could compute such a set $G = \{g_0 < g_1 < g_2 < \dots \}$, then $A$ could form the ML-test $(V_n)_{n \in \omega}$, where $V_n = \{ X : (\forall i < n )[X(g_i) = 1]\}$. Note that $\mu(V_n) = 2^{-n}$. This would capture $\Omega$, a contradiction.

Replace $X(g_i) = 1$ with $X(g_i) = 0$ for the case $G \subseteq 2^{\omega} - \Omega$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.