The following set theoretical question is inspired by a question from recursion theory:

Question: Is there an $L$-random real $r$ which is a minimal cover over another real $x$?

Where a minimal cover $r$ over $x$ means that $x\in L[r] \wedge r\not\in L[x]$ but there is not real $z$ so that $x\in L[z]\wedge z\in L[r]\wedge r\not\in L[z]\wedge z\not\in L[x]$.

  • $\begingroup$ Adding a single random real is like iterating $\omega$ of them, if my memory serves me right, so the answer should probably be no. $\endgroup$
    – Asaf Karagila
    Mar 3 '17 at 14:25
  • $\begingroup$ I don't see why it should probably be no. $\endgroup$
    – 喻 良
    Mar 4 '17 at 6:25
  • $\begingroup$ If B is a measure algebra and A is a complete subalgebra of B, then each one of B and the quotient B/A is either trivial or isomorphic to a measure algebra. It follows that the answer is no. What is the recursion theoretic analogue for Martin-Lof random reals? $\endgroup$
    – Ashutosh
    Mar 5 '17 at 1:21
  • $\begingroup$ @ Ashutosh Thanks. The recursion theoretic question asks what's the measure of the Turing degrees which can be a minimal cover. It seems also unknown for hyperdegrees. $\endgroup$
    – 喻 良
    Mar 5 '17 at 2:12
  • $\begingroup$ Is the category analogue known: Is every sufficiently generic real a minimal cover of some real? $\endgroup$
    – Ashutosh
    Mar 5 '17 at 2:27

Suppose $r$ is random over $L$, $x \in 2^{\omega} \cap L[r]$ and $r \notin L[x]$. For $y \in 2^{\omega}$, let $y_0, y_1 \in 2^{\omega}$ denote the even and odd parts of $y$ - So $y = y_0 \oplus y_1$. Let $B$ denote the random algebra and $A$, the complete subalgebra of $B$ generated by $x$. Let $G_A$ be an $A$-generic filter over $L$ such that $L[x] = L[G_A]$. Then $L[r]$ is obtained by adding a $B/G_A$-generic filter over $L[x]$. Since this is same as adding a random real over $L[x]$, we can choose a real $y \in L[r]$ such that $y$ is random over $L[x]$ and $L[x][y] = L[r]$. Let $z = x \oplus y_0$. Then $L[x] \subsetneq L[z] \subsetneq L[r]$.

  • 1
    $\begingroup$ Thanks for the proof. So you actually proved that for every random real $r$ and real $x<_L$r, $r$ is $L$-equivalent to the join of $x$ with an $x$-random real. This obviously fails for Martin-L\" of randomness. $\endgroup$
    – 喻 良
    Mar 5 '17 at 7:35

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