[Solved]: Is the infinite union of computable sets computable?

The claim as stated is easy to refute, so let’s ask a (slightly) more difficult question: is the countable union of computable languages computable? The answer is again negative: the language $L = { 1^e : text{ program $e$ halts on the empty input}}$ is not computable, but is the union of countably many singletons. What if we require the countably many languages to be uniformly computable? This means that not only is there a one-parameter program $P_i$ for the $i$th language $L_i$ such that $P_i(x) = T$ iff $x in L_i$, but also there is a two-parameter program $P$ such that $P(i,x) = T$ iff $x in L_i$ (both $P_i$ and $P$ are required to always halt). In this case, the union still need not be computable. For example, let $L_n = { e : text{ program $e$ halts on the empty input within $n$ steps}}$. These programs are uniformly computable, but their union $L$ is not computable. A union of a sequence of uniformly computable languages is known as recursively enumerable (r.e.) or computationally enumerable (c.e.), and is a fundamental concept in computability theory. The relation between computable languages and c.e. ones is the same as that between P and NP.