Suppose f is a computable function from a recursively enumerable set U to the natural numbers and that L,K are r.e. subsets of U. Is f(L-K) a difference of r.e. subsets? The motivation comes from

Primes occurring as orders of elements of a finitely presented group

A positive answer would mean that the theorem proposed in HW's nice answer is 100% correct. Otherwise the $\epsilon$-clarification in my answer is actually needed.


Is f injective? If so, yes. If not, no. In the second case, you could achieve any $\Sigma^0_2$ set.

  • $\begingroup$ It is not injective. Are the sets I am talking about always $\Sigma_2^0$? $\endgroup$ – Benjamin Steinberg Feb 9 '13 at 5:03
  • $\begingroup$ Don't you need $f$ to be infinite-to-one everywhere to get every $\Sigma^0_2$-set in this way? $\endgroup$ – François G. Dorais Feb 9 '13 at 5:05
  • $\begingroup$ 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. $\endgroup$ – Benjamin Steinberg Feb 9 '13 at 5:08
  • $\begingroup$ That is sets of primes $\endgroup$ – Benjamin Steinberg Feb 9 '13 at 5:09
  • 1
    $\begingroup$ @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 $\Delta^0_2$. $\endgroup$ – Noah Schweber Feb 9 '13 at 7:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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