Unary predicates theory with no principal types

Question

Definition

Take a satisfiable set of $L$-sentences $\Sigma$ and variables $x = x_1, \ldots, x_n$. Denote by $S_x(\Sigma)$ the set of all $\Sigma$-realizable $x$-types in $L$.

A type $p(x) \in S_x(\Sigma)$ is principal if it contains a $\Sigma$-complete formula. Equivalently, the singleton $\{p(x)\}$ is an open set in $S_x(\Sigma)$, or principal $x$-types are exactly the isolated points of $S_x(\Sigma)$.

Problem

I want to find a theory with no principal types realizable in $T$. I was suggested theory of infinitely many independent unary predicates is such a theory.

Attempt

I realize I'd have to prove that no formula in $\mathcal{L}$ is $T$-complete. To that end, I think I would need to show that if and only if $\phi(x)$ is a single-variable formula not containing a unary predicate symbol $U_n$ then both $\phi(x) \wedge U_n(x)$ and $\phi(x) ∧ \neg U_n(x)$ are realized in $T$. It seems intuitive, yet I do not know how to prove it.

How could I go about proving it?

Answer 1

You have the right idea, so I will just work out your argument. Basically $T$ just says nothing. So any $\mathcal{L}$-structure is a model. So suppose for a contradiction that a type $p(x)$ is isolated by $\phi(x)$ and let $a \in M$ realise $p(x)$. Let $U_n$ be a predicate that does not occur in $\phi(x)$. Now let $\mathcal{L}'$ be the language $\mathcal{L}$ without the predicate $U_n$. Let $M'$ be the $\mathcal{L}'$-reduct of $M$. Note that still $M' \models \phi(a)$. We define an $\mathcal{L}$-structure $M_1$ by taking $M'$ and interpreting $U_n$ as just $\{a\}$. We define another $\mathcal{L}$-structure $M_2$ by taking $M'$ and interpreting $U_n$ as the empty set. Now $M_1 \models \phi(a) \wedge U_n(a)$ and $M_2 \models \phi(a) \wedge \neg U_n(a)$, contradicting that $\phi(a)$ isolates $p(x)$.