I am currently reading the Gödel's Incompleteness Theorem explained by Vladimir A. Uspensky. And I came across the following theorem which I think has contradiction with itself. Since I am a physics student I have no grounding in Logic therefore I am asking for someone to explain to me if I am right or wrong.
Definition of the Problem: Let $\left<L,T\right>$ be a fundamental pair, let $\left<P, \bar P, \delta\right>$ be a deductive system over $L$, and let $Q$ be the set of all provable words. Suppose that $V\subseteq L^\infty$. We say that the deductive system $\left<P, \bar P, \delta\right>$ is: (A) Consistent relative to $V$ if $V\cap Q \subseteq V\cap T$ (B) Complete relative to $V$ is $V\cap T\subseteq V\cap Q$.
The Theorem is then:
Theorem: Suppose that $V$ is an enumerable subset of $L^\infty$, and the set of true statements in $V$ is not enumerable. Then there is no deductive system which is both consistent and complete relative to $V$.
The problem I am facing is that if a set $A$ is enumerable, which means that there exists a 1 to 1 resemblence for the set elements and the natural numbers, then how can $V$ be both enumerable and still have a subset (the set of true statements in $V$) that is not enumerable?
This is a terminology issue; see the bottom of page $247$. Uspensky is using "enumerable" to mean "computably enumerable" (also called "recursively enumerable," "semidecidable," or "recognizable" - yeah, logic has a terminology problem), not "countable" (which is how you're reading it). Indeed, since the alphabet $L$ is assumed to be finite the whole set $L^\infty$ is countable, so every subset of $L^\infty$ is also countable.