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



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> **Question**: Is there an $L$-random real $r$ which is a minimal cover over another real $x$?
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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 $x\in L[z]\wedge z\in L[r]\wedge r\not\in L[z]\wedge z\not\in L[x]$.



Minimal cover v.s random reals
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 $L[x] \subsetneq L[z] \subsetneq L[r]$.


