>>12477806This is a basic result of computable analysis anon. I don't know enough about the topic to explain why your counterexample doesn't work, but my guess is that functions with jump discontinuities are not computable because of something involving convergence and compactness.
See the subsection on topology and computability
https://en.wikipedia.org/wiki/Computable_analysis>>12477818It would depend on what you mean. First of all, the result only hold on the reals, but I'm sure there some analog for the integers. But if you are talking about the sign function on the integers with the usual (i.e. discrete) topology, then the sign function f actually is continuous, because, e.g. if we take an open ball B(1) around the integer 1, then for any integer n such that f(n)=1, it trivially follows that there exists an open ball B(n) containing n such that for all $x \in B(n)$ we have $f(x) \in B(1)$, since the singleton set {x} is an open ball in the integers.
