A question about the proof of Godel's Incompleteness Theorem The article A Computability Proof of Gödel’s First Incompleteness Theorem, by Jørgen Veisdal on Cantor's Paradise contains the following passage:

The second property regards the complement of a set $E$, that is, all the strings which are not in set $E$. First, notice that if $E$ is decidable, so is the complement of $E$ (we can construct a set $F$ of all the strings that are shown to not belong to $E$). As such, if the set $E$ can be constructed by a mechanical process (is computably enumerable), so too must its complement.

If a set $E$ is enumerable, why does its complement also have to be enumerable? What if its complement is infinite? Moreover, if a set $E$ is decidable, why does its complement have to be decidable? 

Decidable means there's a decision procedure for determining whether an element is in the set or not. By symmetry, this holds for the complement too: we can determine whether the element is not in the set or not.
However, it is false that the complement of a computably enumerable set is computably enumerable. In fact, this fact is very important to Gödel's theorem: one example of such a set is the set of all Gödel numbers of theorems. (Another famous one is the halting set.)
If a set $S$ is computably enumerable and so is its complement, then it is decidable, since we can decide if $x$ is in $X$ by enumerating both $S$ and it complement and seeing which list $x$ appears in. The converse is clearly true as well, so one characterization of decidable sets is those that are c.e. and whose complements are also c.e.
