$T$ be a theory that claims that both $P$ and $\neg P$ are infinite. Prove that $T$ is a complete theory Let $\mathscr{L}=\{P\}$ a language with one unary predicate. Let $T$ be a theory that claims that both $P$ and $\neg P$   are infinite. Prove that $T$   is a complete theory.
I tried to prove by structural induction on some formula $\varphi$:
Let $\mathcal M=\langle M,\sigma \rangle$ be a model for $T$.
If $\varphi = (c_{1} = c_{2})$, then $\sigma(c_{1})=\sigma(c_{2})$ iff $\mathcal M\vDash \varphi$,
Therefore only one of $T\cup{\{\varphi\} }$ or $T\cup{\{\neg \varphi\} }$ is consistent..
If $\varphi = P(c)$ then $\sigma(c)\in \sigma(P)$ iff $\mathcal M\vDash \varphi $,
Therefore only one of $T\cup{\{\varphi\} }$ or $T\cup{\{\neg \varphi\} }$ is consistent..
But I don't know how to procceed, and don't know if it's even a good way to start.
 A: Let us prove that $T$ has quantifier elimination: every formula $\phi(x_1,\ldots,x_n)$ is equivalent to some quantifier-free formula $\psi(x_1,\ldots,x_n)$ modulo $T$. To prove this, it suffices to prove that every formula $\phi$ of the form $\exists y_1,\ldots,y_m. \bigwedge_{i\in I} R_i$ where the $R_i$ are literals $P$,$\neg P$, or $\neq$, is equivalent to a quantifier-free formula. Using the fact that $P$ and $\neg P$ are both infinite in models of $T$, one sees that such a formula is always equivalent to $\bigwedge_{i\in J} R_i$, where one only keeps the literals that involve only variables from $\{x_1,\ldots,x_n\}$. 
Let us now prove that given models $\mathbf A,\mathbf B$ of $T$, they both contain a common substructure (which is not necessarily a model of $T$). Here, any model of $T$ contains the substructure that consists of only one point, where this point is in $P$. This implies that $T$ is complete: if $\mathbf C$ is this one-point structure, we have by quantifier elimination that $\mathbf A_C\equiv \mathbf B_C$, where $\mathbf A_C$ is the structure obtained by adding one constant symbols which is interpreted by the point in $\mathbf C$. In particular, $\mathbf A\equiv\mathbf B$.
Another way to prove that $T$ is complete is Vaught's criterion: if $T$ is $\kappa$-categorical for some cardinal $\kappa$ at least as large as the cardinality of the language, then $T$ is complete. Indeed, if $\mathbf A$ is the unique up-to-isomorphism model of $T$ of size $\kappa$ and $\mathbf B$ is an arbitrary model of $T$, we have by Löwenheim-Skolem that either $|\mathbf B|<\kappa$ and there exists an elementary extension $\mathbf C$ of $\mathbf B$ of size $\kappa$, or $|\mathbf B|\geq\kappa$ and $\mathbf B$ has an elementary substructure $\mathbf C$ of size $\kappa$. In both cases, $\mathbf C$ is isomorphic to $\mathbf A$, so that $\mathbf B\equiv\mathbf A$ for all $\mathbf B$.
Then, if you can prove that $T$ is $\omega$-categorical, using Vaught's criterion we have that $T$ is complete. Now, $\omega$-categoricity has many equivalent formulations when the language is at most countable (see the Ryll-Nardzewski, Engelev, Svenonius theorem). For example, if there exists only finitely many inequivalent formulas with $n$ free variables for all $n\geq 1$, then $T$ is $\omega$-categorical.
So given a formula with $n$ free variables, one obtains an equivalent quantifier-free formula with $n$ free variables by quantifier elimination. Since the language is finite, there are only finitely many inequivalent such formulas, and $T$ is $\omega$-categorical.
