$M$ is some Turing machine, $\left<M\right>$ is the code of the Turing machine.$L =\{\left<T\right> | L(T) \ne \emptyset\}$How to see intuitively that $L$ is partially decidable?We can try running a given $M$ on all strings and accept if $M$ accepts.However, what if the simulated $M$ gets into an infinite loop?
Language is non-empty is recursively enumerable (intuitive way)
computability;turing machines
A language $L$ is partially computable if there is a machine $M$ that on input $x$:If $x \in L$ then $M$ halts.If $x \notin L$ then $M$ doesn't halt.This is equivalent to the other definitions.To answer your question, what we do is to run the input machine $M$ on all inputs in parallel. If it never accepts on any input, then we never halt, but that's ok.In order to run $M$ on all inputs in parallel, we use the technique of dovetailing:We run $M$ for one step on the first input.We run $M$ for one step on the first two inputs.We run $M$ for one step on the first three inputs.And so on.