# Semidecidable properties of computable reals

By computable real I mean $$x\in\mathbb{R}$$ such that there is some computable total function $$p_x$$ that takes a natural number $$n$$ and returns a dyadic rational $$r$$ such that $$|x-r|<2^{-n}$$. I think this is probably equivalent to any other reasonable definition but I want to fix this to avoid ambiguity.

By a semi-decidable set $$S$$ of computable reals, I mean that there should be a program that, when given the Godel code for some $$p_x$$, halts exactly when $$x\in S$$.

One easy class of examples of semi-decidable sets of reals is given by open intervals $$(a,b)$$ where $$a$$ and $$b$$ are both computable.

Is it true that every semi-decidable set is open?

I think this should be true, essentially because if a computable process $$P$$ takes $$p_x$$ and then decides that $$x\in S$$, it must have done so using at most finitely many calls to $$p_x$$. But how do I know that $$P$$ did not do something sneaky by looking at the code for $$p_x$$? This should be impossible, but e.g. Rice's theorem does no apply directly because I am not assuming that $$P$$ decides some property of arbitrary programs, just those that happen to represent computable reals (and note that these are not even a r.e. set).

• Following this answer, it is enough to find a computable sequence $x_i$ such that $x_i\notin S$ but $\lim x_i$ is computable and in $s$. If $S$ is not open, by definition (almost) there exists a Cauchy sequence $x_i$ which lies outside of $S$ and converges to a boundary point of $S$. The problem is making the sequence constructive somehow, in order to apply a reduction from the complement of the halting problem. – Ariel Nov 12 '18 at 13:09