Is the union of computable enumerable sets computably enumerable

Let $$\{A_n : n \in \mathbb{N}\}$$ be a collection of c.e. (computably enumerable) sets. Is $$\bigcup_n A_n$$ c.e.?

That is, is the union of c.e. sets c.e.? Otherwise, under what conditions will this be true?

Every language can be written as a countable union of singletons $$\{x\}$$.

On the other hand, the union of finitely many c.e. languages is c.e.

• We can be more general, the union of a c.e. set of c.e. sets is c.e itself, not just finite. – orlp Apr 3 at 13:25
• Thanks but how does one go about proving this. Is this a standard theorem etc. Any links online where proof is available – Amit wadhwa Apr 3 at 14:24
• Take it as an exercise. – Yuval Filmus Apr 3 at 15:16

For any c.e. collection of c.e. sets, the union is c.e., however if the collection itself is not c.e, then not (a counterexample is $$S_x = \{x + n : n \in \mathbb{N}\}$$, then $$\bigcup_{x\in\mathbb{R}} S_x$$ is not c.e. even though it's a union of only c.e. sets).

To see the above by explicit construction, first if you allow duplicates in your enumeration (this isn't strictly the union of sets) then we have an explicit map from $$\mathbb{N} \to \mathbb{N}^2$$. See for example the inverse of the Cantor pairing function. Then we can first select which c.e. set to pick, and then which element from that set. You can explicitly filter out the duplicates afterwards to get a proper enumeration.

• I am bit confused. Did you mean the answer to the question I posted as true or false ? ie. The union of ce sets here it's over natural numbers till inifinity. So is the ce ? – Amit wadhwa Apr 3 at 14:34
• How do you even define c.e for a set of sets? – nir shahar Apr 3 at 14:48
• A collection $A_n$ of c.e. sets is uniformly c.e. if there is a Turing machine that on input $n$, enumerates all words in $A_n$. We can relax that even further in various ways. For example, we could have a Turing machine which enumerates pairs $(n_i,w_i)$, which correspond to the collection $A_n = \{ w_i : n_i = n \}$. – Yuval Filmus Apr 3 at 15:15
• @Amitwadhwa It depends on whether you have a Turing machines that can generate the elements of $A_n$. IMO the standard definition of a c.e. set is unwieldy and confusing and an (equivalent) definition that simply asks for a Turing machine $M$ that maps integer $n$ to an encoding of the $n$th element of the set (in some arbitrary order without duplicates) is much simpler to think about and compose to answer questions like these. If you can construct a Turing machine computing a bijective mapping between $\mathbb{N}$ and some set $S$, then $S$ is r.e. – orlp Apr 3 at 15:48