A set is countable if it has a bijection with the natural numbers, and is computably enumerable (c.e.) if there exists an algorithm that enumerates its members.
Any non-finite computably enumerable set must be countable since we can construct a bijection from the enumeration.
Are there any examples of countable sets that are not computably enumerable? That is, a bijection between this set and the natural numbers exists, but there is no algorithm that can compute this bijection.
Are there any examples of countable sets that are not enumerable?
Yes. All subsets of the natural numbers are countable but not all of them are enumerable. (Proof: there are uncountably many different subsets of $\mathbb{N}$ but only countably many Turing machines that could act as enumerators.) So any subset of $\mathbb{N}$ that you already know is not recursively enumerable is an example – such as the set of all numbers coding Turing machines that halt for every input.
Yes, every undecidable (not semi-decidable) language has this property.
For example, consider the set $L = \{(x,M) \mid M \text{ does not halt on input } x \}$.
Suppose we have an algorithm that can enumerate the members of this set. If such an algorithm existed, we could use this to solve the halting problem with inputs $x,M$, with the following algorithm:
$M$ either halts, or does not halt on $x$. If it halts, eventually we will find an $n$ where we reach a halting state. If it doesn't halt, then eventually we will reach $(M,x)$ in our enumeration.
Thus we have a reduction, and we can conclude that no such enumeration exists.
Note that such enumerations can exist for semi-decidable problems. For example, you can enumerate the set of all halting machine-input pairs by enumerating all possible traces of all Turing Machine executions after $n$ steps, and filter out ones that do not end in a halting state.
In computability theory we deal with subsets of $\Sigma^*$, where $\Sigma = \{0,1\}$. This language is countably infinite, and so any subset $L \subseteq \Sigma^*$ is countable. Furthermore, there are many undecidable but recursively enumerable languages whose complements are not recursively enumerable. These languages are subset of $\Sigma^*$ and hence are countable.
I want to give another concrete example. The set of encodings of always halting Turing machines $R$ is countable (since there are only countably many Turing machines, halting or non-halting). But $R$ itself is not computably enumerable.
Assume $R$ were computably enumerable. Then given index $i$ we can find the encoding of the always halting Turing machine $T_i$ in $R = \{T_1, T_2, \dots\}$.
Let's define a Turing machine $D$ that accepts the i-th word if and only if $T_i$ rejects it and rejects the i-th word if and only if $T_i$ accepts it. This Turing machine $D$ is also an always halting Turing machine, since we can compute the i-th encoding of the always halting Turing machine $T_i$ and simulate it on the i-th word. After $T_i$ halts, we just reverse the decision of $T_i$.
However, $D$ does not appear in the enumeration of always halting Turing machines $R$, since for each machine $T_i$ it decides differently at least on the i-th input word. This is a contradiction and therefore the assumption that $R$ is computably enumerable is false.
So even when we focus on always halting Turing machines can we define a set that is countable but not computably enumerable.
