Given a productive set, there is a collection of c.e. sets union of which is the productive set, as we know that every c.e. set is with a c.e. function with a index.
My question: is the set of the indices of c.e.sets that cover a productive set also productive one?



Is set of the indices of c.e.sets that cover a productive set also productive one?
No, it isn't. This is a consequence of the fact that we have a choice of infinitely many equivalent indices for each c.e. set. Let $E$ be a set of indices such that


$$(e, j \in E \land e \neq j) \implies W\_e \neq W\_j $$
$$ P = \bigcup\_{e \in E} W\_e$$


For any $e$ let $e' \neq e$ with $W\_e = W\_{e'}$ (note that if $e \in E$ then $e'$ isn't by assumption). Now we build $\hat{E}$ by meeting requirements $R\_i$ ensuring that $\phi\_i$ isn't a production function for $\hat{E}$.


Let $e\_i$ be an enumeration of elements of $E$ *in order*. Suppose at the start of stage $s$ we've have $\hat{E}\_s = \lbrace \hat{e}\_0, \ldots, \hat{e}\_{n} \rbrace$ Let $j\_s$ be such that $W\_{j\_s} = \hat{E}\_s$ (must exist as it's finite). Now compute $\phi\_s(j\_s)$. If it doesn't converge or $\phi\_s(j\_s) \in \hat{E}\_s$, or $\phi\_s(j\_s) = e\_k, k \leq n$ then we've met $R\_i$ as it's either not in $\hat{E}$ or is in $W\_{j\_s}$. So set $\hat{e}\_{n+1} = e\_{n + 1}$.


If $\phi\_s(j\_s)$ isn't of the form $e\_k$ or $e'\_k$ for some $k$ it similarly won't be in $\hat{E}$ so do the same.


If $\phi\_s(j\_s) = e\_k$ or $\phi\_s(j\_s) = e'\_k$ for $k > n$ then simply set $\hat{e}\_k$ to be equal to whichever one disagrees with $\phi\_s(j\_s)$ so $\phi\_s(j\_s) \neq \hat{e}\_k$ and set $\hat{e}\_l, n < l < k$ to be $e\_l$ giving us $E\_{s+1}$.


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However, note that if $E$ is the set of all $e$ with $W\_e$ contained in $P$ then I'm pretty sure it will be productive since we can just consider the set of all indexes for the empty c.e. set and notice that we can m-reduce $\bar{0'}$ to that set.


