# Set that is recursively enumerable but is NOT decidable

I was trying to find a set that is recursively enumerable i.e $$\exists f\; \text{computable and a program P that computes}\; f$$ such that $$A = \{ x\; :\; P(x)\downarrow \}.$$ But it is not decidable so it's characteristic function is not computable. I tried with using the Halting problem and taking $$A$$ as $$A = \{ x\; :\; \text{ the x-th computable function stops }\}$$ but I don't think that A is recursively enumerable. Thank you.

The set $$S_h=\{ (x,y) | \text{program x halts when run on input y} \}$$ is recursively enumerable since a universal program that takes input $$(x,y)$$ and emulates program $$x$$ running on input $$y$$ will halt for exactly the inputs in $$S_h$$. However, as the halting problem shows, $$S_h$$ is not decideable because its complement is not recursively enumerable.