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Feb 9, 2013 at 15:49 vote accept Benjamin Steinberg
Feb 9, 2013 at 13:09 comment added Benjamin Steinberg @Francois, but I am allowing any computable image of a difference of re sets.
Feb 9, 2013 at 8:00 comment added Noah Schweber (In the above, technically in defining $f(X)_s$ I need to restrict attention to $y<s$ as possible witnesses to keep the approximation effective, but that doesn't wind up affecting anything.)
Feb 9, 2013 at 7:57 comment added Noah Schweber @Francois: I think you're right. If we insist $f$ be finite-to-one, then it seems that $X\in\Delta^0_2\impliesf(X)\in\Delta^0_2$, by the limit lemma: let $X_s$ be the approximation to $X$ at stage $s$, and approximate $f(X)$ by saying that $n\in f(X)_s$ if there is at stage $s$ some $y$ which appears to be in $X$ such that $f(y)=n$. Then for each $n$ and each $y\in f^{-1}(n)$, cofinitely often we have $y\in X_s\iff y\in X$; since $f^{-1}(n)$ is finite, we then get that cofinitely often, $n\in f(X)_s\iff n\in f(X)$, so $f(X)$ is $\Delta^0_2$.
Feb 9, 2013 at 5:19 comment added François G. Dorais Benjamin, $\Sigma^0_2$-sets are the sets which are r.e. relative to the halting set, this is much larger than the differences of r.e. sets.
Feb 9, 2013 at 5:09 comment added Benjamin Steinberg That is sets of primes
Feb 9, 2013 at 5:08 comment added Benjamin Steinberg I am thinking of f,U,L,K as things I can vary and I would like to know the full range of primes that can be picked up.
Feb 9, 2013 at 5:05 comment added François G. Dorais Don't you need $f$ to be infinite-to-one everywhere to get every $\Sigma^0_2$-set in this way?
Feb 9, 2013 at 5:03 comment added Benjamin Steinberg It is not injective. Are the sets I am talking about always $\Sigma_2^0$?
Feb 9, 2013 at 4:58 history answered Dan Turetsky CC BY-SA 3.0